Proof. First let $w>0$. From [Reference Gradshteyn, Ryzhik, Zwillinger and Moll26, page 428, Formula 3.723.5],

(3.3)$$\begin{eqnarray}\mathfrak{R}(y,w)=-{\textstyle \frac{1}{2}}(e^{-wy}\overline{\text{Ei}}(wy)+e^{wy}\text{Ei}(-wy)),\end{eqnarray}$$

where $\text{Ei}(x)$ is the exponential integral defined for $x>0$ by [Reference Jahnke and Emde34, page 1] $\text{Ei}(-x)=-\int _{x}^{\infty }e^{-t}/t\,\text{d}t$. Thus the exponential integral function is related to logarithmic integral $\text{li}(x)=\int _{0}^{x}\frac{\text{d}t}{\log (t)}$ by

(3.4)$$\begin{eqnarray}\text{Ei}(-x)=\text{li}(e^{-x}).\end{eqnarray}$$

Also [Reference Jahnke and Emde34, page 3],

(3.5)$$\begin{eqnarray}\overline{\text{Ei}}(x)=\text{li}(e^{x}).\end{eqnarray}$$

Thus from (3.3)–(3.5), we see that

(3.6)$$\begin{eqnarray}\mathfrak{R}(y,w)=-\frac{1}{2}(e^{-wy}\text{li}(e^{wy})+e^{wy}\text{li}(e^{-wy})).\end{eqnarray}$$

Now Dixon and Ferrar [Reference Dixon and Ferrar23, page 165, Equation (5.4)] have proved that

(3.7)$$\begin{eqnarray}e^{x}\text{li}(e^{-x})+e^{-x}\text{li}(e^{x})=\unicode[STIX]{x1D70B}^{3/2}K_{/\frac{1}{2}}(x),\end{eqnarray}$$

where [Reference Dixon and Ferrar22, Equation (3.12)]

(3.8)$$\begin{eqnarray}K_{/\unicode[STIX]{x1D708}}(z)=\frac{1}{\unicode[STIX]{x1D70B}}\mathop{\sum }_{k=0}^{\infty }\frac{(z/2)^{2k}\{2\log (z/2)-\unicode[STIX]{x1D713}(k+1)-\unicode[STIX]{x1D713}(k+\unicode[STIX]{x1D708}+1)\}}{\unicode[STIX]{x1D6E4}(k+1)\unicode[STIX]{x1D6E4}(k+\unicode[STIX]{x1D708}+1)}.\end{eqnarray}$$

Let $\unicode[STIX]{x1D708}=1/2$ in (3.8), use the duplication formula for the gamma function and then combine the resulting identity with (3.6) and (3.7) to obtain (3.1) for $w>0$. Since both sides of (3.1) are analytic for $\text{Re}(w)>0$, we obtain (3.1) in this region by analytic continuation. To prove (3.2), divide both sides of (3.1) by the first term of the right-hand side and note that $\unicode[STIX]{x1D713}(1)=-\unicode[STIX]{x1D6FE}$ as well as $\lim _{y\rightarrow 0^{+}}(wy)^{2k}(\unicode[STIX]{x1D713}(2k+1)-\log (wy))/((2k)!(-\unicode[STIX]{x1D6FE}-\log (wy)))=0$ for $k\geqslant 1$.◻

Proof. Here, we use the analogue of Watson’s lemma for Laplace transform in the setting of Fourier transforms [Reference Olver50], [Reference Dai and Naylor17, Equations (1.3), (1.4)]. It says that if the form of $h(t)$ near $t=0$ is given as a series of algebraic powers, that is,

(3.16)$$\begin{eqnarray}h(t)\sim \mathop{\sum }_{n=0}^{\infty }b_{n}t^{n+\unicode[STIX]{x1D706}-1}\end{eqnarray}$$

as $t\rightarrow 0^{+}$, then under certain restrictions on $h$ (see [Reference Olver50], [Reference Dai and Naylor17, Section 2] for the same),

(3.17)$$\begin{eqnarray}\int _{0}^{\infty }e^{ist}h(t)\,\text{d}t\sim \mathop{\sum }_{n=0}^{\infty }b_{n}e^{i(n+\unicode[STIX]{x1D706})\unicode[STIX]{x1D70B}/2}\unicode[STIX]{x1D6E4}(n+\unicode[STIX]{x1D706})s^{-n-\unicode[STIX]{x1D706}}\end{eqnarray}$$

as $s\rightarrow \infty$. Let $h(t)=t/(t^{2}+w^{2})$. Then it is easy to see that $h(t)$ satisfies (3.16) with $\unicode[STIX]{x1D706}=1$ and

$$\begin{eqnarray}b_{n}=\left\{\begin{array}{@{}ll@{}}\displaystyle 0\quad & \text{for }n\text{ even},\\ \displaystyle (-1)^{\frac{n-1}{2}}w^{-n-1}\quad & \text{for }n\text{ odd}.\end{array}\right.\end{eqnarray}$$

Now invoking (3.17) twice, once with $s=y$ and then with $s=-y$, and then adding the resulting two identities, we arrive at

$$\begin{eqnarray}\displaystyle \mathfrak{R}(y,w) & {\sim} & \displaystyle \frac{1}{2}\mathop{\sum }_{n=0}^{\infty }b_{n}e^{i(n+1)\unicode[STIX]{x1D70B}/2}\unicode[STIX]{x1D6E4}(n+1)y^{-n-1}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{2}\mathop{\sum }_{n=0}^{\infty }b_{n}e^{i(n+1)\unicode[STIX]{x1D70B}/2}\unicode[STIX]{x1D6E4}(n+1)(-y)^{-n-1}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{b_{2n-1}(2n-1)!\cos (n\unicode[STIX]{x1D70B})}{y^{2n}}=-\mathop{\sum }_{n=1}^{\infty }\frac{(2n-1)!}{w^{2n}y^{2n}}.\nonumber\end{eqnarray}$$

This completes the proof. ◻

Proof. It is important to note that the above double integral does not converge absolutely and hence Fubini’s theorem is inapplicable, that is, we cannot change the order of integration. First assume $u>0$. We prove (3.18) in the form $\int _{0}^{\infty }\mathfrak{R}(y,w)\,\text{d}y=0$, where $w=2\unicode[STIX]{x1D70B}u$ and $\mathfrak{R}(y,w)$ is defined in (2.5). First of all, (3.2) and (3.15) imply that this integral exists. Now let $N$ be a positive integer and consider the integral

(3.19)$$\begin{eqnarray}I(w,N):=\int _{0}^{\infty }e^{-\frac{y}{N}}\mathfrak{R}(y,w)\,\text{d}y.\end{eqnarray}$$

With the help of Fubini’s theorem, one can write

(3.20)$$\begin{eqnarray}\displaystyle I(w,N) & = & \displaystyle \int _{0}^{\infty }\frac{t}{w^{2}+t^{2}}\,\text{d}t\int _{0}^{\infty }e^{-\frac{y}{N}}\cos (yt)\,\text{d}y\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{N}\int _{0}^{\infty }\frac{t}{(w^{2}+t^{2})(\frac{1}{N^{2}}+t^{2})}\,\text{d}t\nonumber\\ \displaystyle & = & \displaystyle \frac{N}{1-N^{2}w^{2}}\int _{0}^{\infty }\biggl(\frac{t}{w^{2}+t^{2}}-\frac{t}{\frac{1}{N^{2}}+t^{2}}\biggr)\,\text{d}t\nonumber\\ \displaystyle & = & \displaystyle \frac{N}{2(1-N^{2}w^{2})}\lim _{A\rightarrow \infty }\bigg\{[\log (w^{2}+t^{2})]_{0}^{A}-\biggl[\log \biggl(\frac{1}{N^{2}}+t^{2}\biggr)\biggr]_{0}^{A}\bigg\}\nonumber\\ \displaystyle & = & \displaystyle \frac{N}{2(1-N^{2}w^{2})}\lim _{A\rightarrow \infty }\bigg\{\log \biggl(\frac{w^{2}+A^{2}}{\frac{1}{N^{2}}+A^{2}}\biggr)+2\log \biggl(\frac{1}{Nw}\biggr)\bigg\}\nonumber\\ \displaystyle & = & \displaystyle \frac{N\log (Nw)}{N^{2}w^{2}-1},\end{eqnarray}$$

since $\lim _{A\rightarrow \infty }\log (\frac{w^{2}+A^{2}}{\frac{1}{N^{2}}+A^{2}})=0$. Now note that $e^{-\frac{y}{N}}\mathfrak{R}(y,w)\rightarrow \mathfrak{R}(y,w)$ pointwise as $N\rightarrow \infty$ and $|e^{-\frac{y}{N}}\mathfrak{R}(y,w)|\leqslant \mathfrak{R}(y,w)$. Also, as mentioned before, the fact that $\mathfrak{R}(y,w)$ is integrable, as a function of $y$ from $0$ to $\infty$, is clear from (3.2) and (3.15). Hence letting $N\rightarrow \infty$ on both sides of (3.19) and employing Lebesgue’s dominated convergence theorem, we see that

$$\begin{eqnarray}\displaystyle \lim _{N\rightarrow \infty }I(w,N) & = & \displaystyle \int _{0}^{\infty }\lim _{N\rightarrow \infty }e^{-\frac{y}{N}}\mathfrak{R}(y,w)\,\text{d}y\nonumber\\ \displaystyle & = & \displaystyle \int _{0}^{\infty }\mathfrak{R}(y,w)\,\text{d}y,\nonumber\end{eqnarray}$$

whereas (3.20) implies that $\lim _{N\rightarrow \infty }I(w,N)=0$. Together, these complete the proof of (3.18) for $u>0$. Note that the left-hand side of (3.18) is analytic for $\text{Re}(u)>0$ as can be seen using [Reference Temme61, Theorem 2.3, page 30]. Hence by analytic continuation, the result holds for $\text{Re}(u)>0$.◻

Proof of Theorem 2.2.

Let

(3.24)$$\begin{eqnarray}g(x)=\int _{0}^{\infty }\frac{v\cos (2\unicode[STIX]{x1D70B}xv)}{u^{2}+v^{2}}\,\text{d}v.\end{eqnarray}$$

That the above integral exists is clear from the fact that $f(v)=\frac{v}{2(u^{2}+v^{2})}$ satisfies the hypotheses of Theorem 3.1. To say that the two limits in Theorem 3.1 are equal implies, in particular, that they exist. The fact that $\int _{0}^{\infty }g(x)\,dx$ exists can be seen, in particular, from Lemma 3.4. Together, we conclude that $\sum _{m=1}^{\infty }g(m)$ is convergent. Employing Theorem 3.1 with $f(v)=\frac{v}{2(u^{2}+v^{2})}$ and $g(x)$ as in (3.24) and invoking Lemma 3.4, we see that

(3.25)$$\begin{eqnarray}\displaystyle \mathop{\sum }_{m=1}^{\infty }g(m) & = & \displaystyle \frac{1}{2}\lim _{M\rightarrow \infty }\biggl(\mathop{\sum }_{n=1}^{M}\frac{n}{u^{2}+n^{2}}-\int _{0}^{M}\frac{v}{u^{2}+v^{2}}\,dv\biggr)\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{2}\lim _{M\rightarrow \infty }\bigg\{\biggl(\mathop{\sum }_{n=1}^{M}\frac{n}{u^{2}+n^{2}}-\log M\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,\biggl(\log M-\int _{0}^{M}\frac{v}{u^{2}+v^{2}}\,dv\biggr)\bigg\}\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{2}\biggl[-\frac{1}{2}(\unicode[STIX]{x1D713}(iu)+\unicode[STIX]{x1D713}(-iu))\biggr]\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{2}\lim _{M\rightarrow \infty }\biggl(\log M-\int _{0}^{M}\frac{v}{u^{2}+v^{2}}\,dv\biggr)\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{2}\biggl[-\frac{1}{2}(\unicode[STIX]{x1D713}(iu)+\unicode[STIX]{x1D713}(-iu))\biggr]+\frac{1}{2}\lim _{M\rightarrow \infty }\biggl(\log \frac{M}{\sqrt{u^{2}+M^{2}}}+\log u\biggr)\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{2}\biggl(\log u-\frac{1}{2}(\unicode[STIX]{x1D713}(iu)+\unicode[STIX]{x1D713}(-iu))\biggr),\end{eqnarray}$$

where in the third step, we used [Reference Dixit20, Equation (3.8)]. Theorem 2.2 now follows from (3.25) and by employing the change of variable $v=t/(2\unicode[STIX]{x1D70B}m)$ and replacing $x$ by $m$ and $u$ by $u/(2\unicode[STIX]{x1D70B})$.◻

Proof. For $0<\text{Re}(s)<1$, we have [Reference Gradshteyn, Ryzhik, Zwillinger and Moll26, page 1101, Formula (3)]

$$\begin{eqnarray}\int _{0}^{\infty }t^{s-1}\cos t\,\text{d}t=\unicode[STIX]{x1D6E4}(s)\cos \biggl(\frac{\unicode[STIX]{x1D70B}s}{2}\biggr),\end{eqnarray}$$

and for $0<\text{Re}(s)<2$, we have [Reference Gradshteyn, Ryzhik, Zwillinger and Moll26, page 1101, Formula (6)]

$$\begin{eqnarray}\int _{0}^{\infty }t^{s-1}\frac{1}{1+t^{2}}\,\text{d}t=\frac{\unicode[STIX]{x1D70B}}{2}\text{cosec}\biggl(\frac{\unicode[STIX]{x1D70B}s}{2}\biggr).\end{eqnarray}$$

Using Parseval’s formula [Reference Paris and Kaminski52, page 83, Equation (3.1.13)], for $0<\text{Re}(s)<1$, one can obtain

$$\begin{eqnarray}\displaystyle \frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(c_{1})}\frac{\unicode[STIX]{x1D6E4}(s)}{\tan (\frac{\unicode[STIX]{x1D70B}s}{2})}u^{-s}\,\text{d}s & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(c_{1})}\frac{\unicode[STIX]{x1D6E4}(s)\cos (\frac{\unicode[STIX]{x1D70B}s}{2})}{\sin (\frac{\unicode[STIX]{x1D70B}s}{2})}u^{-s}\,\text{d}s\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{\unicode[STIX]{x1D70B}}\int _{0}^{\infty }\frac{2\cos t}{1+\frac{u^{2}}{t^{2}}}\,\frac{\text{d}t}{t}\nonumber\\ \displaystyle & = & \displaystyle \frac{2}{\unicode[STIX]{x1D70B}}\int _{0}^{\infty }\frac{t\cos t}{u^{2}+t^{2}}\,\text{d}t.\nonumber\end{eqnarray}$$

Now one can easily extend the region of validity of the above result to $0<\text{Re}(s)<2$ by noting that when we shift the line of integration $\text{Re}(s)=c_{1}$ to, say, $\text{Re}(s)=c_{2},1\leqslant c_{2}<2,$ one does not encounter any poles of the integrand, and also that the integrals over the horizontal segments tend to zero as the height $T$ tends to $\infty$.◻

Proof. By hypothesis, $j+\frac{1}{2N}=\frac{h}{N}-\frac{1}{2}$. Since $j$ is an integer and $\lfloor \frac{1}{2N}\rfloor =0$ for $N\geqslant 1$, we have $j=\lfloor j+\frac{1}{2N}\rfloor =\lfloor \frac{h}{N}-\frac{1}{2}\rfloor$.◻

Proof of Theorem 2.1.

The hypothesis $h\geqslant N/2$, $N\in \mathbb{N}$, will be used several times, without mention, in the proof. It is easy to see that the series $\sum _{n=0}^{\infty }n^{N-2h}\frac{\exp (-an^{N}x)}{1-\exp (-n^{N}x)}$ is absolutely and uniformly convergent for any $x>0,N\in \mathbb{N}$. Thus, interchanging the order of summation and integration in the first step below, we see that for $\text{Re}(s)>\max (\frac{N-2h+1}{N},1)=1$,

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\infty }x^{s-1}\mathop{\sum }_{n=1}^{\infty }n^{N-2h}\frac{\exp (-an^{N}x)}{1-\exp (-n^{N}x)}\,\text{d}x\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{n=1}^{\infty }n^{N-2h}\int _{0}^{\infty }x^{s-1}\frac{\exp (-an^{N}x)}{1-\exp (-n^{N}x)}\,\text{d}x\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{n=1}^{\infty }n^{N-2h-Ns}\int _{0}^{\infty }\frac{y^{s-1}e^{-ay}}{1-e^{-y}}\,\text{d}y\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D6E4}(s)\unicode[STIX]{x1D701}(s,a)\unicode[STIX]{x1D701}(Ns-N+2h),\nonumber\end{eqnarray}$$

where, in the second step, we employed the change of variable $y=n^{N}x$ and in the third, we used the fact [Reference Apostol5, page 251, Theorem 12.2] that for $\text{Re}(s)>1$,

$$\begin{eqnarray}\int _{0}^{\infty }\frac{y^{s-1}e^{-ay}}{1-e^{-y}}\,\text{d}y=\unicode[STIX]{x1D6E4}(s)\unicode[STIX]{x1D701}(s,a).\end{eqnarray}$$

Thus, for $\unicode[STIX]{x1D706}=\text{Re}(s)>1$,

(4.4)$$\begin{eqnarray}\mathop{\sum }_{n=1}^{\infty }n^{N-2h}\frac{\exp (-an^{N}x)}{1-\exp (-n^{N}x)}=\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(\unicode[STIX]{x1D706})}\unicode[STIX]{x1D6E4}(s)\unicode[STIX]{x1D701}(s,a)\unicode[STIX]{x1D701}(Ns-(N-2h))x^{-s}\,\text{d}s.\end{eqnarray}$$

We now obtain an alternate evaluation of the above integral by shifting the line of integration and then using Cauchy’s residue theorem. Consider the contour ${\mathcal{C}}$ determined by the line segments $[\unicode[STIX]{x1D706}-iT,\unicode[STIX]{x1D706}+iT],[\unicode[STIX]{x1D706}+iT,-r+iT],[-r+iT,-r-iT]$ and $[-r-iT,\unicode[STIX]{x1D706}-iT]$, where, $r$ is a sufficiently large positive real number which is not an integer and $\frac{2h}{N}-1<r<\frac{2h+1}{N}-1$. The reason for choosing the lower and upper bounds for $r$ will be explained soon. Let

(4.5)$$\begin{eqnarray}F(s):=\unicode[STIX]{x1D6E4}(s)\unicode[STIX]{x1D701}(s,a)\unicode[STIX]{x1D701}(Ns-(N-2h))x^{-s},\end{eqnarray}$$

and let $R_{a}$ denote the residue of $F(s)$ at the pole $s=a$. We first find the poles of $F(s)$ and residues at those poles.

(1) $F(s)$ has a simple pole at $s=0$ since $\unicode[STIX]{x1D6E4}(s)$ has a simple pole at $s=0$ with the residue

(4.6)$$\begin{eqnarray}R_{0}=\lim _{s\rightarrow 0}sF(s)=\unicode[STIX]{x1D701}(0,a)\unicode[STIX]{x1D701}(-N+2h)=-(a-\frac{1}{2})\unicode[STIX]{x1D701}(2h-N),\end{eqnarray}$$

since from [Reference Apostol5, page 264, Theorem 12.13], $\unicode[STIX]{x1D701}(-n,a)=-\frac{B_{n+1}(a)}{n+1},n\geqslant 0$ and $B_{1}(a)=a-\frac{1}{2}$.

(2) Since $\unicode[STIX]{x1D701}(s,a)$ has a simple pole at $s=1$, $F(s)$ has a simple pole at $s=1$ with residue

(4.7)$$\begin{eqnarray}R_{1}=\lim _{s\rightarrow 1}(s-1)F(s)=\frac{\unicode[STIX]{x1D701}(2h)}{x}.\end{eqnarray}$$

(3) Since $\frac{N-2h+1}{N}\neq -2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor$, Lemma 4.2 implies $\frac{N-2h+1}{N}\neq -2j$ for any $j\geqslant 1$. Thus, $F(s)$ has a simple pole at $s=\frac{N-2h+1}{N}$, owing to the pole of $\unicode[STIX]{x1D701}(s)$ at $s=1$, with the residue

(4.8)$$\begin{eqnarray}R_{\frac{N-2h+1}{N}}=\frac{1}{N}\unicode[STIX]{x1D6E4}\biggl(\frac{N-2h+1}{N}\biggr)\unicode[STIX]{x1D701}\biggl(\frac{N-2h+1}{N},a\biggr)x^{-\frac{(N-2h+1)}{N}}.\end{eqnarray}$$

(4) Consider the simple poles of $\unicode[STIX]{x1D6E4}(s)$ at $s=-2j,j\in \mathbb{N}$, and the trivial zeros of $\unicode[STIX]{x1D701}(Ns-N+2h)$ at $-2k,k\in \mathbb{N}$. It is important to see if some of these poles of $\unicode[STIX]{x1D6E4}(s)$ get canceled by the trivial zeros of $\unicode[STIX]{x1D701}(Ns-N+2h)$. To that end, suppose for some positive integers $j^{\prime }$ and $k^{\prime }$ we have $N(-2j^{\prime })-N+2h=-2k^{\prime }$. Then $k^{\prime }=Nj^{\prime }+\frac{N}{2}-h$. This implies that if $N$ is an odd positive integer, no pole of $\unicode[STIX]{x1D6E4}(s)$ at a negative even integer will get canceled by a trivial zero of $\unicode[STIX]{x1D701}(Ns-N+2h)$ since $k^{\prime }$ is not an integer. However, if $N$ is an even positive integer, then $k^{\prime }$ can equal $Nj^{\prime }+\frac{N}{2}-h$, while being a positive integer, implying $j^{\prime }>\frac{h}{N}-\frac{1}{2}$, that is, among the poles of $\unicode[STIX]{x1D6E4}(s)$ at negative even integers, only the poles $-2j$, $1\leqslant j\leqslant \lfloor \frac{h}{N}-\frac{1}{2}\rfloor$, contribute toward the evaluation of the line integral. To sum up, when $N$ is odd integer, $F(s)$ has simple poles at all negative even integers $-2j,j\geqslant 1$, and when $N$ is an even integer, $F(s)$ has simple poles at $-2j$, where $1\leqslant j\leqslant \lfloor \frac{h}{N}-\frac{1}{2}\rfloor$. The residue at such a pole is

(4.9)$$\begin{eqnarray}\displaystyle R_{-2j} & = & \displaystyle \lim _{s\rightarrow -2j}(s+2j)\unicode[STIX]{x1D6E4}(s)\unicode[STIX]{x1D701}(s,a)\unicode[STIX]{x1D701}(Ns-(N-2h))x^{-s}\nonumber\\ \displaystyle & = & \displaystyle -\frac{B_{2j+1}(a)}{(2j+1)!}\unicode[STIX]{x1D701}(-2jN-N+2h)x^{2j}.\end{eqnarray}$$

At this juncture, it deems necessary to explain why we choose the real part of the shifted line of integration to be $-r$ with $r>\frac{2h}{N}-1$. The reason is, this implies $-r<-2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor$, and thus all poles of $\unicode[STIX]{x1D6E4}(s)$ at negative even integers $-2j$, where $1\leqslant j\leqslant \lfloor \frac{h}{N}-\frac{1}{2}\rfloor$, lie inside the contour, thus contributing toward the evaluation of the line integral.

(5) Arguing as in (4), it can be seen that $F(s)$ has simple poles at $s=-(2j-1)$, $1\leqslant j\leqslant \lfloor \frac{h}{N}\rfloor$, and the residue at such a pole is

(4.10)$$\begin{eqnarray}\displaystyle R_{-(2j-1)} & = & \displaystyle \lim _{s\rightarrow -(2j-1)}(s+(2j-1))F(s)\nonumber\\ \displaystyle & = & \displaystyle (-1)^{h+1}2^{2h-1}\unicode[STIX]{x1D70B}^{2h}\biggl(\frac{-1}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{jN}\frac{B_{2j}(a)B_{2h-2jN}}{(2j)!(2h-2jN)!}x^{2j-1}.\end{eqnarray}$$

Now applying Cauchy’s residue theorem, we observe that

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}i}\biggl[\int _{\unicode[STIX]{x1D706}-iT}^{\unicode[STIX]{x1D706}+iT}+\int _{\unicode[STIX]{x1D706}+iT}^{-r+iT}+\int _{-r+iT}^{-r-iT}+\int _{-r-iT}^{\unicode[STIX]{x1D706}-iT}\biggr]\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\frac{\unicode[STIX]{x1D6E4}(s)}{x^{s}}\unicode[STIX]{x1D701}(s,a)\unicode[STIX]{x1D701}(Ns-(N-2h))\,\text{d}s\nonumber\\ \displaystyle & & \displaystyle \quad =R_{0}+R_{1}+R_{\frac{N-2h+1}{N}}+\mathop{\sum }_{j=1}^{\lfloor \frac{h}{N}\rfloor }R_{-(2j-1)}+\mathop{\sum }_{j=1}^{\lfloor \frac{h}{N}-\frac{1}{2}\rfloor }R_{-2j}.\nonumber\end{eqnarray}$$

Now let $T\rightarrow \infty$. Using Stirling’s formula for $\unicode[STIX]{x1D6E4}(s)$ and elementary bounds on the Riemann zeta function and the Hurwitz zeta function, it can be seen that the integrals along the horizontal segments $[\unicode[STIX]{x1D706}+iT,-r+iT]$, $[-r-iT,\unicode[STIX]{x1D706}-iT]$ approach zero as $T\rightarrow \infty$. Hence from (4.6)–(4.10),

(4.11)$$\begin{eqnarray}\mathop{\sum }_{n=1}^{\infty }n^{N-2h}\frac{\exp (-an^{N}x)}{1-\exp (-n^{N}x)}=P(x,a)+J(x,a),\end{eqnarray}$$

where $P(x,a)$ is the sum of all residues of $F(s)$, defined in (2.2), and

(4.12)$$\begin{eqnarray}J(x,a):=\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(-r)}\unicode[STIX]{x1D6E4}(s)\unicode[STIX]{x1D701}(s,a)\unicode[STIX]{x1D701}(Ns-(N-2h))x^{-s}\,\text{d}s.\end{eqnarray}$$

It remains to show that $J(x,a)$ agrees with $S(x,a)$ defined in (2.3) and (2.4) respectively when $N$ is odd and even. To evaluate $J(x,a)$, we first make a change of variable $s\leftrightarrow 1-s$ in (4.12) so that

(4.13)$$\begin{eqnarray}J(x,a)=\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(1+r)}\unicode[STIX]{x1D6E4}(1-s)\unicode[STIX]{x1D701}(1-s,a)\unicode[STIX]{x1D701}(2h-Ns)x^{s-1}\,\text{d}s.\end{eqnarray}$$

Now replace $s$ by $1-s$ in (1.4), then multiply both sides of the resulting identity by $\unicode[STIX]{x1D6E4}(1-s)$ to obtain, for $\text{Re}(s)>1$,

(4.14)$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6E4}(1-s)\unicode[STIX]{x1D701}(1-s,a)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{2\unicode[STIX]{x1D6E4}(1-s)\unicode[STIX]{x1D6E4}(s)}{(2\unicode[STIX]{x1D70B})^{s}}\bigg\{\cos \biggl(\frac{\unicode[STIX]{x1D70B}s}{2}\biggr)\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n^{s}}+\sin \biggl(\frac{\unicode[STIX]{x1D70B}s}{2}\biggr)\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{s}}\bigg\}\nonumber\\ \displaystyle & & \displaystyle \quad =(2\unicode[STIX]{x1D70B})^{1-s}\bigg\{\frac{1}{2\sin (\frac{\unicode[STIX]{x1D70B}s}{2})}\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n^{s}}+\frac{1}{2\cos (\frac{\unicode[STIX]{x1D70B}s}{2})}\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{s}}\bigg\},\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where in the last step, we used the reflection formula for the gamma function.

We note here that Kanemitsu et al. [Reference Kanemitsu, Tanigawa and Yoshimoto37, page 51] use a formula equivalent to (1.4), namely, for $\text{Re}(s)<0$,

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D701}(s,a) & = & \displaystyle \frac{\unicode[STIX]{x1D6E4}(1-s)}{(2\unicode[STIX]{x1D70B})^{1-s}}\biggl(\exp \biggl(\frac{-\unicode[STIX]{x1D70B}i(1-s)}{2}\biggr)\mathop{\sum }_{n=1}^{\infty }\frac{e^{2\unicode[STIX]{x1D70B}ian}}{n^{1-s}}\nonumber\\ \displaystyle & & \displaystyle +\,\exp \biggl(\frac{\unicode[STIX]{x1D70B}i(1-s)}{2}\biggr)\mathop{\sum }_{n=1}^{\infty }\frac{e^{-2\unicode[STIX]{x1D70B}ian}}{n^{1-s}}\biggr).\nonumber\end{eqnarray}$$

However, one can see that while the above formula is useful when $N$ is an even positive integer, it is not when $N$ is odd. In fact, employing it leads to very complicated integrals which do not seem to lead us to any concrete result. On the other hand, (1.4) works for any positive integer $N$, irrespective of its parity, as will be seen in the remainder of the proof.

Now substitute (4.14) in (4.13) and invoke the functional equation [Reference Apostol5, page 259] $\unicode[STIX]{x1D701}(w)=2^{w}\unicode[STIX]{x1D70B}^{w-1}\unicode[STIX]{x1D6E4}(1-w)\unicode[STIX]{x1D701}(1-w)\sin (\frac{1}{2}\unicode[STIX]{x1D70B}w)$ for $\unicode[STIX]{x1D701}(2h-Ns)$ to obtain after simplification

(4.15)$$\begin{eqnarray}J(x,a)=J_{1}(x,a)+J_{2}(x,a),\end{eqnarray}$$

where

(4.16)$$\begin{eqnarray}\displaystyle J_{1}(x,a) & := & \displaystyle \frac{(-1)^{h+1}2^{2h+1}\unicode[STIX]{x1D70B}^{2h}}{x}\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(1+r)}\biggl(\frac{(2\unicode[STIX]{x1D70B})^{N+1}}{x}\biggr)^{-s}\unicode[STIX]{x1D6E4}(1-2h+Ns)\nonumber\\ \displaystyle & & \displaystyle \times \,\unicode[STIX]{x1D701}(1-2h+Ns)\bigg\{\frac{\sin (\frac{N\unicode[STIX]{x1D70B}s}{2})}{2\sin (\frac{\unicode[STIX]{x1D70B}s}{2})}\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n^{s}}\bigg\}\,\text{d}s,\end{eqnarray}$$

(4.17)$$\begin{eqnarray}\displaystyle J_{2}(x,a) & := & \displaystyle \frac{(-1)^{h+1}2^{2h+1}\unicode[STIX]{x1D70B}^{2h}}{x}\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(1+r)}\biggl(\frac{(2\unicode[STIX]{x1D70B})^{N+1}}{x}\biggr)^{-s}\unicode[STIX]{x1D6E4}(1-2h+Ns)\nonumber\\ \displaystyle & & \displaystyle \times \,\unicode[STIX]{x1D701}(1-2h+Ns)\bigg\{\frac{\sin (\frac{N\unicode[STIX]{x1D70B}s}{2})}{2\cos (\frac{\unicode[STIX]{x1D70B}s}{2})}\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{s}}\bigg\}\,\text{d}s.\end{eqnarray}$$

We first evaluate $J_{2}(x,a)$. Its evaluation depends on the parity of $N$. Assume first that $N$ is odd. Employ the change of variable

(4.18)$$\begin{eqnarray}s_{1}=Ns-2h+1\end{eqnarray}$$

in (4.17) so that $c_{1}:=\text{Re}(s_{1})>1$ (since $r>\frac{2h}{N}-1$), write $\unicode[STIX]{x1D701}(s_{1})=\sum _{m=1}^{\infty }m^{-s_{1}}$, and then interchange the order of double sum and the integral, permitted because of absolute convergence, to arrive at

(4.19)$$\begin{eqnarray}\displaystyle J_{2}(x,a) & = & \displaystyle \frac{(-1)^{h+1}2^{2h+1}\unicode[STIX]{x1D70B}^{2h}}{Nx}\biggl(\frac{(2\unicode[STIX]{x1D70B})^{N+1}}{x}\biggr)^{\frac{1-2h}{N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{m,n=1}^{\infty }n^{\frac{1-2h}{N}}\sin (2\unicode[STIX]{x1D70B}na)E(X_{m,n}),\end{eqnarray}$$

where

(4.20)$$\begin{eqnarray}E(X_{m,n}):=\frac{(-1)^{h-1}}{2\unicode[STIX]{x1D70B}i}\int _{(c_{1})}\frac{\unicode[STIX]{x1D6E4}(s_{1})\cos (\frac{\unicode[STIX]{x1D70B}s_{1}}{2})}{2\cos (\frac{\unicode[STIX]{x1D70B}}{2}(\frac{s_{1}+2h-1}{N}))}X_{m,n}^{-s_{1}}\,\text{d}s_{1},\end{eqnarray}$$

with

(4.21)$$\begin{eqnarray}X_{m,n}:=2\unicode[STIX]{x1D70B}m\biggl(\frac{2\unicode[STIX]{x1D70B}n}{x}\biggr)^{1/N}.\end{eqnarray}$$

Using (4.2) in the second step below, we find that

(4.22)$$\begin{eqnarray}\displaystyle E(X_{m,n}) & = & \displaystyle \frac{-1}{2\unicode[STIX]{x1D70B}i}\int _{(c_{1})}\frac{\unicode[STIX]{x1D6E4}(s_{1})}{2\tan (\frac{\unicode[STIX]{x1D70B}s_{1}}{2})}\frac{\cos (\frac{\unicode[STIX]{x1D70B}}{2}(s_{1}+2h-1))}{\cos (\frac{\unicode[STIX]{x1D70B}}{2}(\frac{s_{1}+2h-1}{N}))}X_{m,n}^{-s_{1}}\,\text{d}s_{1}\nonumber\\ \displaystyle & = & \displaystyle \frac{(-1)^{\frac{N+1}{2}}}{2}\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(c_{1})}\frac{\unicode[STIX]{x1D6E4}(s_{1})}{\tan (\frac{\unicode[STIX]{x1D70B}s_{1}}{2})}\nonumber\\ \displaystyle & & \displaystyle \times \,\sum _{j=-(N-1)}^{N-1}i^{j}e^{-\frac{ij\unicode[STIX]{x1D70B}}{2}(\frac{s_{1}+2h-1}{N})}X_{m,n}^{-s_{1}}\,\text{d}s_{1}\nonumber\\ \displaystyle & = & \displaystyle \frac{(-1)^{\frac{N+1}{2}}}{2}\sum _{j=-(N-1)}^{N-1}i^{j}e^{\frac{-ij\unicode[STIX]{x1D70B}(2h-1)}{2N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(c_{1})}\frac{\unicode[STIX]{x1D6E4}(s_{1})}{\tan (\frac{\unicode[STIX]{x1D70B}s_{1}}{2})}{X_{m,n,j}^{\ast }}^{-s_{1}}\,\text{d}s_{1},\end{eqnarray}$$

where

(4.23)$$\begin{eqnarray}X_{m,n,j}^{\ast }:=X_{m,n}e^{\frac{ij\unicode[STIX]{x1D70B}}{2N}}.\end{eqnarray}$$

Now (4.18) and the inequality $\frac{2h}{N}-1<r<\frac{2h+1}{N}-1$ along with the fact that $\text{Re}(s)=1+r$ imply that $1<\text{Re}(s_{1})<2$. The reason why we initially chose $r<\frac{2h+1}{N}-1$ is because, we need $\text{Re}(s_{1})<2$ in order to use Lemma 4.1. Hence invoking Lemma 4.1 to simplify the above representation for $E(X_{m,n})$ and then substituting the resultant in (4.19) gives

$$\begin{eqnarray}\displaystyle J_{2}(x,a) & = & \displaystyle \frac{2}{\unicode[STIX]{x1D70B}Nx}(-1)^{h+\frac{N+3}{2}}(2\unicode[STIX]{x1D70B})^{2h}\biggl(\frac{(2\unicode[STIX]{x1D70B})^{N+1}}{x}\biggr)^{\frac{1-2h}{N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\sum _{j=-(N-1)}^{N-1}i^{j}e^{\frac{-ij\unicode[STIX]{x1D70B}(2h-1)}{2N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n=1}^{\infty }n^{\frac{1-2h}{N}}\sin (2\unicode[STIX]{x1D70B}na)\mathop{\sum }_{m=1}^{\infty }\int _{0}^{\infty }\frac{t\cos (t)}{{X_{m,n,j}^{\ast }}^{2}+t^{2}}\,\text{d}t.\nonumber\end{eqnarray}$$

Now note that $\text{Re}(X_{m,n,j}^{\ast })=2\unicode[STIX]{x1D70B}m(\frac{2\unicode[STIX]{x1D70B}n}{x})^{\frac{1}{N}}\cos (\frac{\unicode[STIX]{x1D70B}j}{2N})>0$ as

$$\begin{eqnarray}-\!\frac{\unicode[STIX]{x1D70B}}{2}<-\frac{\unicode[STIX]{x1D70B}(N-1)}{2N}\leqslant \frac{\unicode[STIX]{x1D70B}j}{2N}\leqslant \frac{\unicode[STIX]{x1D70B}(N-1)}{2N}<\frac{\unicode[STIX]{x1D70B}}{2}.\end{eqnarray}$$

Hence apply Theorem 2.2 and then replace $j$ by $2j$ in the second step below to deduce that

(4.24)$$\begin{eqnarray}\displaystyle J_{2}(x,a) & = & \displaystyle (-1)^{h+\frac{N+3}{2}}\frac{(2\unicode[STIX]{x1D70B})^{2h}}{\unicode[STIX]{x1D70B}Nx}\biggl(\frac{(2\unicode[STIX]{x1D70B})^{N+1}}{x}\biggr)^{\frac{1-2h}{N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\sum _{j=-(N-1)}^{N-1}i^{j}e^{\frac{-ij\unicode[STIX]{x1D70B}(2h-1)}{2N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n=1}^{\infty }n^{\frac{1-2h}{N}}\sin (2\unicode[STIX]{x1D70B}na)\bigg\{\log \biggl(\biggl(\frac{2\unicode[STIX]{x1D70B}n}{x}\biggr)^{\frac{1}{N}}e^{\frac{i\unicode[STIX]{x1D70B}j}{2N}}\biggr)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{1}{2}\biggl(\unicode[STIX]{x1D713}\biggl(i\biggl(\frac{2\unicode[STIX]{x1D70B}n}{x}\biggr)^{\frac{1}{N}}e^{\frac{i\unicode[STIX]{x1D70B}j}{2N}}\biggr)+\unicode[STIX]{x1D713}\biggl(-i\biggl(\frac{2\unicode[STIX]{x1D70B}n}{x}\biggr)^{\frac{1}{N}}e^{\frac{i\unicode[STIX]{x1D70B}j}{2N}}\biggr)\biggr)\bigg\}\nonumber\\ \displaystyle & & \displaystyle =\frac{(-1)^{h+\frac{N+3}{2}}}{\unicode[STIX]{x1D70B}N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{\frac{N-2h+1}{N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}(-1)^{j}\text{exp}\biggl(\frac{i\unicode[STIX]{x1D70B}(1-2h)j}{N}\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h-1}{N}}}\bigg\{\log \biggl(\frac{1}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{1}{2}\biggl(\unicode[STIX]{x1D713}\biggl(\frac{i}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)+\unicode[STIX]{x1D713}\biggl(-\frac{i}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)\biggr)\bigg\},\end{eqnarray}$$

where $A_{N,j}(y)=\unicode[STIX]{x1D70B}(2\unicode[STIX]{x1D70B}y)^{\frac{1}{N}}e^{\frac{i\unicode[STIX]{x1D70B}j}{N}}$. This completes the evaluation of $J_{2}(x,a)$ when $N$ is odd.

Now consider the case when $N$ is even. Note that (4.19) still holds with $E(X_{m,n})$ and $X_{m,n}$ defined in (4.20) and (4.21). Use (4.3) and (4.23) in the second step below to simplify $E(X_{m,n})$ as

(4.25)$$\begin{eqnarray}\displaystyle E(X_{m,n}) & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(c_{1})}\frac{\unicode[STIX]{x1D6E4}(s_{1})\sin (\frac{\unicode[STIX]{x1D70B}}{2}(s_{1}+2h-1))}{2\cos (\frac{\unicode[STIX]{x1D70B}}{2}(\frac{s_{1}+2h-1}{N}))}X_{m,n}^{-s_{1}}\,\text{d}s_{1}\nonumber\\ \displaystyle & = & \displaystyle \frac{(-1)^{\frac{N}{2}}}{2\unicode[STIX]{x1D70B}i}\sum _{j=-(N-1)}^{N-1}i^{j}\exp \biggl(\frac{i\unicode[STIX]{x1D70B}j(2h-1)}{2N}\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,\int _{(c_{1})}\unicode[STIX]{x1D6E4}(s_{1}){X_{m,n,-j}^{\ast }}^{-s_{1}}\,\text{d}s_{1}\nonumber\\ \displaystyle & = & \displaystyle (-1)^{\frac{N}{2}}\sum _{j=-(N-1)}^{N-1}i^{j}\exp \biggl(\frac{i\unicode[STIX]{x1D70B}j(2h-1)}{2N}\biggr)e^{-X_{m,n,-j}^{\ast }},\end{eqnarray}$$

since $\text{Re}(X_{m,n,-j}^{\ast })>0$, where in the last step, we used the inverse Mellin transform of $\unicode[STIX]{x1D6E4}(s_{1})$. Replace $j$ by $-2j-1$ in (4.25), then substitute the resultant in (4.19) to get

(4.26)$$\begin{eqnarray}\displaystyle J_{2}(x,a) & = & \displaystyle \frac{i(-1)^{h+\frac{N}{2}}}{N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{\frac{N-2h+1}{N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{j=-\frac{N}{2}}^{\frac{N}{2}-1}(-1)^{j}e^{\frac{i\unicode[STIX]{x1D70B}(2j+1)(1-2h)}{2N}}\mathop{\sum }_{n=1}^{\infty }\frac{n^{\frac{1-2h}{N}}\sin (2\unicode[STIX]{x1D70B}na)}{\exp (2A_{N,j+\frac{1}{2}}(\frac{n}{x}))-1}.\end{eqnarray}$$

From (4.24) and (4.26), we obtain an expression for $J_{2}(x,a)$ for all positive integers $N$.

Now $J_{1}(x,a)$ from (4.16) can be evaluated in a similar way to obtain

(4.27)$$\begin{eqnarray}\displaystyle J_{1}(x,a) & = & \displaystyle \frac{(-1)^{h+1}}{N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{\frac{N-2h+1}{N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}e^{\frac{i\unicode[STIX]{x1D70B}(1-2h)j}{N}}\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h-1}{N}}(\text{exp}(2A_{N,j}(\frac{n}{x}))-1)}\end{eqnarray}$$

for $N$ odd, whereas, for $N$ even,

(4.28)$$\begin{eqnarray}\displaystyle J_{1}(x,a) & = & \displaystyle \frac{(-1)^{h+1}}{N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{\frac{N-2h+1}{N}}\mathop{\sum }_{j=-\frac{N}{2}}^{\frac{N}{2}-1}e^{\frac{i\unicode[STIX]{x1D70B}(1-2h)(j+\frac{1}{2})}{N}}\nonumber\\ \displaystyle & & \displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h-1}{N}}(\text{exp}(2A_{N,j+\frac{1}{2}}(\frac{n}{x}))-1)}.\end{eqnarray}$$

In fact, the above expressions for $J_{1}(x,a)$ differ from the expression in the first equality in [Reference Kanemitsu, Tanigawa and Yoshimoto35, Equation (2.18)] only in that the numerator of the summand of the infinite series in them involves $\cos (2\unicode[STIX]{x1D70B}na)$, which is absent in the latter.

Finally, adding the corresponding sides of (4.27) and (4.24) when $N$ is odd, and respectively of (4.28) and (4.26) when $N$ is even gives expressions for $J(x,a)$ (see (4.15)). These are nothing but the expressions for $S(x,a)$ claimed in the statement of Theorem 2.1. Along with (4.11), this completes the proof of Theorem 2.1.◻

Proof. Substitute $N=2M$ and $h=M-h^{\prime }+M\ell$ on both sides of (2.1). Then the resulting left-hand side is the same as the Lambert series in (4.31). With the above substitutions,

(4.34)$$\begin{eqnarray}\displaystyle P(x,a) & = & \displaystyle \frac{1}{2M}\unicode[STIX]{x1D6E4}\biggl(-\ell +\frac{2h^{\prime }+1}{2M}\biggr)\unicode[STIX]{x1D701}\biggl(-\ell +\frac{2h^{\prime }+1}{2M},a\biggr)x^{\ell -\frac{2h^{\prime }+1}{2M}}\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{j=0}^{\lfloor \frac{\ell }{2}-\frac{h^{\prime }}{2M}\rfloor }\frac{\unicode[STIX]{x1D701}(-2j,a)}{(2j)!}\unicode[STIX]{x1D701}(2M(\ell -2j)-2h^{\prime })x^{2j}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{\unicode[STIX]{x1D701}(2M(\ell +1)-2h^{\prime })}{x}\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{j=1}^{\lfloor \frac{\ell }{2}-\frac{h^{\prime }}{2M}+\frac{1}{2}\rfloor }\frac{(-1)^{2j-1}}{(2j-1)!}\frac{\unicode[STIX]{x1D701}(-(2j-1),a)}{x^{1-2j}}\nonumber\\ \displaystyle & & \displaystyle \times \,\unicode[STIX]{x1D701}(-4Mj+2M-2h^{\prime }+2M\ell ).\end{eqnarray}$$

Note that $0<h^{\prime }<M\Rightarrow 0<\frac{1}{2}-\frac{h^{\prime }}{2m}<\frac{1}{2}\Rightarrow \lfloor \frac{1}{2}-\frac{h^{\prime }}{2m}\rfloor =0$, and $0<h^{\prime }<M\Rightarrow -\frac{1}{2}<-\frac{h^{\prime }}{2M}<0\Rightarrow \lfloor -\frac{h^{\prime }}{2M}\rfloor =-1$. Thus,

(4.35)$$\begin{eqnarray}\displaystyle \biggl\lfloor\frac{\ell }{2}+\frac{1}{2}-\frac{h^{\prime }}{2m}\biggr\rfloor & = & \displaystyle \left\{\begin{array}{@{}ll@{}}\displaystyle \frac{\ell }{2}\quad & \ell \text{ is even},\\ \displaystyle \frac{\ell -1}{2}\quad & \ell \text{ is odd},\end{array}\right.\end{eqnarray}$$

(4.36)$$\begin{eqnarray}\displaystyle \biggl\lfloor\frac{\ell }{2}-\frac{h^{\prime }}{2M}\biggr\rfloor & = & \displaystyle \left\{\begin{array}{@{}ll@{}}\displaystyle \frac{\ell }{2}-1\quad & \ell \text{ is even},\\ \displaystyle \frac{\ell -1}{2}\quad & \ell \text{ is odd}.\end{array}\right.\end{eqnarray}$$

Using (4.35) and (4.36), we see that, irrespective of the parity of $\ell$, the two finite sums over $j$ in (4.34) combine together to give

$$\begin{eqnarray}\mathop{\sum }_{j=0}^{\ell }\frac{(-1)^{j}}{(j)!}\unicode[STIX]{x1D701}(-j,a)\unicode[STIX]{x1D701}(2M(\ell -j)-2h^{\prime })x^{j},\end{eqnarray}$$

which, when combined with the other expression in (4.34), shows that our $P(x,a)$ equals $P(x)$, which is defined in (4.30).

Next, we have to show that our $S(x,a)$ from (2.4) matches with the expressions for $U(x,a)$ in (4.32) and (4.33) corresponding to $M$ even and $M$ odd, respectively. We only prove this in the case when $M$ is even. That for $M$ odd can be similarly proved.

Substitute $N=2M$, $h=M-h^{\prime }+M\ell$, with $M$ even, say $M=2k$, in (2.4) and simplify to get

$$\begin{eqnarray}\displaystyle S(x,a) & = & \displaystyle \frac{(-1)^{h^{\prime }+1}}{4k}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{-\ell +\frac{2h^{\prime }+1}{4k}}\mathop{\sum }_{n=1}^{\infty }n^{-1-\ell +\frac{1+2h^{\prime }}{4k}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{j=-2k}^{2k-1}\exp \biggl(\frac{i\unicode[STIX]{x1D70B}(1-4k+2h^{\prime }-4k\ell )(j+\frac{1}{2})}{4k}\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{(\cos (2\unicode[STIX]{x1D70B}na)+i(-1)^{j+1}\sin (2\unicode[STIX]{x1D70B}na))}{\exp (2\unicode[STIX]{x1D70B}(\frac{2\unicode[STIX]{x1D70B}n}{x})^{\frac{1}{4k}}e^{\frac{i\unicode[STIX]{x1D70B}}{4k}(j+\frac{1}{2})})-1}.\nonumber\end{eqnarray}$$

Now split the sum over $j$ according to the parity of $j$ and simplify so as to obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle S(x,a)=\frac{(-1)^{h^{\prime }+1}}{4k}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{-\ell +\frac{2h^{\prime }+1}{4k}}\mathop{\sum }_{n=1}^{\infty }n^{-1-\ell +\frac{1+2h^{\prime }}{4k}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\bigg\{\mathop{\sum }_{j=-k}^{k-1}\frac{\exp (-i(2\unicode[STIX]{x1D70B}na-\frac{\unicode[STIX]{x1D70B}(2h^{\prime }+1)}{4k}(\frac{4j+1}{2})+\frac{(\ell +1)}{2}(4j+1)\unicode[STIX]{x1D70B}))}{\exp (2\unicode[STIX]{x1D70B}(\frac{2\unicode[STIX]{x1D70B}n}{x})^{\frac{1}{4k}}e^{\frac{i\unicode[STIX]{x1D70B}}{4k}(\frac{4j+1}{2})})-1}\nonumber\\ \displaystyle & & \displaystyle \quad +\,\mathop{\sum }_{j=-k}^{k-1}\frac{\exp (i(2\unicode[STIX]{x1D70B}na+\frac{\unicode[STIX]{x1D70B}(2h^{\prime }+1)}{4k}(\frac{4j+3}{2})-\frac{(\ell +1)}{2}(4j+3)\unicode[STIX]{x1D70B}))}{\exp (2\unicode[STIX]{x1D70B}(\frac{2\unicode[STIX]{x1D70B}n}{x})^{\frac{1}{4k}}e^{\frac{i\unicode[STIX]{x1D70B}}{4k}(\frac{4j+3}{2})})-1}\bigg\}.\nonumber\end{eqnarray}$$

Replace $j$ by $-j-1$ in the second sum and then observe that the resulting corresponding summands of the two sums are complex conjugates of each other so that

$$\begin{eqnarray}\displaystyle S(x,a) & = & \displaystyle \frac{(-1)^{h^{\prime }+1}}{4k}\biggl(\frac{x}{2\unicode[STIX]{x1D70B}}\biggr)^{\ell -\frac{(2h^{\prime }+1)}{4k}}\mathop{\sum }_{n=1}^{\infty }n^{-1-\ell +\frac{1+2h^{\prime }}{4k}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{j=-k}^{k-1}2\text{Re}\biggl(\frac{e^{iuv}}{\exp (ae^{-iu})-1}\biggr),\nonumber\end{eqnarray}$$

where $a=2A(\frac{n}{x}),u=-\frac{\unicode[STIX]{x1D70B}}{4k}(\frac{4j+1}{2})$, and $uv=-2\unicode[STIX]{x1D70B}an+\frac{\unicode[STIX]{x1D70B}(2h^{\prime }+1)}{4k}(\frac{4j+1}{2})-\frac{\unicode[STIX]{x1D70B}(\ell +1)(4j+1)}{2}$. Using Lemma 4.4, the hypotheses of Theorem 4.5, (4.29) and the fact that $k=M/2$, we deduce that

$$\begin{eqnarray}S(x,a)=\frac{(-1)^{h^{\prime }}}{4k}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{-\ell +\frac{2h^{\prime }+1}{4k}}\mathop{\sum }_{j=-\frac{M}{2}}^{\frac{M}{2}-1}\mathop{\sum }_{n=1}^{\infty }f_{2j+1}(n,h^{\prime },\ell ,x)n^{-1-\ell +\frac{1+2h^{\prime }}{4k}},\end{eqnarray}$$

which is nothing but (4.32). Thus we derive (4.31) from (2.1). As remarked before, (4.33) can be proved by a similar argument.◻

Proof. Let $\lfloor \frac{h}{N}-\frac{1}{2}\rfloor =m$. Since $h>N/2$, we have $m\in \mathbb{N}\cup \{0\}$. Let $\frac{h}{N}-\frac{1}{2}=m+r$, where $0\leqslant r<1$. Then $h=\frac{N}{2}+N(m+r)$. From the hypothesis, $N-2h+1=-2Nm$. Since the last two equations imply $r=\frac{1}{2N}$, we get $h=\frac{N+1}{2}+Nm$. The other direction is trivial.◻

Proof of Theorem 2.3.

The setup for the proof of this theorem is exactly similar to that of Theorem 2.1. Hence we only give details where they differ from those of the latter.

Note that $a\in (0,1]$ is fixed, and our integrand $F(s)$, defined in (4.5), is $F(s):=\unicode[STIX]{x1D6E4}(s)\unicode[STIX]{x1D701}(s,a)\unicode[STIX]{x1D701}(Ns-(N-2h))x^{-s}$. The poles of $\unicode[STIX]{x1D6E4}(s)$ include negative even integers, whereas $\unicode[STIX]{x1D701}(Ns-(N-2h))$ has a simple pole at $s=\frac{N-2h+1}{N}$. Since $N$ is odd, it may happen that $\frac{N-2h+1}{N}=-2j$ for some positive integers $N$ and $j$. As explained in the Introduction, Lemma 4.2 then implies that $j=\lfloor \frac{h}{N}-\frac{1}{2}\rfloor$. This may imply a double order pole of $F(s)$ at $s=\frac{N-2h+1}{N}=-2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor$ if $\unicode[STIX]{x1D701}(\frac{N-2h+1}{N},a)\neq 0$, or a simple pole (or a removable singularity) if $\unicode[STIX]{x1D701}(\frac{N-2h+1}{N},a)=0$. However, even if $\unicode[STIX]{x1D701}(\frac{N-2h+1}{N},a)=0$, one may first calculate the residue assuming a double pole and then apply this fact, and the answer obtained would be same as that deduced by first applying $\unicode[STIX]{x1D701}(\frac{N-2h+1}{N},a)=0$ and then accordingly calculating the residue.

Thus the residue at $\frac{N-2h+1}{N}$ is given by

(5.1)$$\begin{eqnarray}\displaystyle & & \displaystyle R_{\frac{N-2h+1}{N}}\nonumber\\ \displaystyle & & \displaystyle \quad =\lim _{s\rightarrow \frac{N-2h+1}{N}}\biggl(\frac{\text{d}}{\text{d}s}\biggl(s-\frac{N-2h+1}{N}\biggr)^{2}\frac{\unicode[STIX]{x1D6E4}(s)}{x^{s}}\unicode[STIX]{x1D701}(s,a)\unicode[STIX]{x1D701}(Ns-(N-2h))\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{x^{2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor }}{N(2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor )!}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\bigg\{-\frac{B_{2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor +1}(a)}{2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor +1}\biggl(\unicode[STIX]{x1D713}\biggl(2\biggl\lfloor\frac{h}{N}-\frac{1}{2}\biggr\rfloor+1\biggr)+N\unicode[STIX]{x1D6FE}-\log x\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\unicode[STIX]{x1D701}^{\prime }\biggl(-2\biggl\lfloor\frac{h}{N}-\frac{1}{2}\biggr\rfloor,a\biggr)\bigg\}.\end{eqnarray}$$

Thus, from (4.6), (4.7), (4.9), (4.10), (5.1) and (2.3), we see that

(5.2)$$\begin{eqnarray}\mathop{\sum }_{n=1}^{\infty }n^{N-2h}\frac{\text{exp}(-an^{N}x)}{1-\text{exp}(-n^{N}x)}=P^{\ast }(x,a)+S(x,a),\end{eqnarray}$$

where

(5.3)$$\begin{eqnarray}\displaystyle P^{\ast }(x,a) & := & \displaystyle -\biggl(a-\frac{1}{2}\biggr)\unicode[STIX]{x1D701}(-N+2h)+\frac{\unicode[STIX]{x1D701}(2h)}{x}\nonumber\\ \displaystyle & & \displaystyle -\,\mathop{\sum }_{j=1}^{\lfloor \frac{h}{N}-\frac{1}{2}\rfloor -1}\frac{B_{2j+1}(a)}{(2j+1)!}\frac{\unicode[STIX]{x1D701}(2h-(2j+1)N)}{x^{-2j}}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{x^{2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor }}{N(2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor )!}\nonumber\\ \displaystyle & & \displaystyle \times \,\bigg\{\frac{-B_{2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor +1}(a)}{2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor +1}\biggl(\unicode[STIX]{x1D713}\biggl(2\biggl\lfloor\frac{h}{N}-\frac{1}{2}\biggr\rfloor+1\biggr)+N\unicode[STIX]{x1D6FE}-\log x\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D701}^{\prime }\biggl(-2\biggl\lfloor\frac{h}{N}-\frac{1}{2}\biggr\rfloor,a\biggr)\bigg\}\nonumber\\ \displaystyle & & \displaystyle +\,(-1)^{h+1}2^{2h-1}\unicode[STIX]{x1D70B}^{2h}\mathop{\sum }_{j=1}^{\lfloor \frac{h}{N}\rfloor }\biggl(\frac{-1}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{jN}\frac{B_{2j}(a)B_{2h-2jN}}{(2j)!(2h-2jN)!}x^{2j-1},\end{eqnarray}$$

and the calculation for $S(x,a)$ remains exactly the same as in the proof of Theorem 2.1.

As we now show, (5.2) can be simplified to a great extent using the following result of Koyama and Kurokawa [Reference Koyama and Kurokawa39, page 7] for an even positive integer $k$ and $0<a\leqslant 1$:

(5.4)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D701}^{\prime }(-k,a) & = & \displaystyle \frac{2(-1)^{\frac{k}{2}}k!}{(2\unicode[STIX]{x1D70B})^{k+1}}\bigg\{\mathop{\sum }_{n=1}^{\infty }\frac{(\log n)\sin (2\unicode[STIX]{x1D70B}na)}{n^{k+1}}\nonumber\\ \displaystyle & & \displaystyle +\,(\log (2\unicode[STIX]{x1D70B})-\unicode[STIX]{x1D713}(k+1))\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{k+1}}+\frac{\unicode[STIX]{x1D70B}}{2}\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n^{k+1}}\bigg\}.\end{eqnarray}$$

Though Koyama and Kurokawa write $\log (2\unicode[STIX]{x1D70B})+\unicode[STIX]{x1D6FE}-\sum _{i=1}^{k}\frac{1}{i}$ in place of $(\log (2\unicode[STIX]{x1D70B})-\unicode[STIX]{x1D713}(k+1))$, using the functional equation $\unicode[STIX]{x1D713}(z+1)=\unicode[STIX]{x1D713}(z)+\frac{1}{z}$, it is seen that they are equal. Also, even though they work with $0<a<1$, the formula holds for $a=1$ as long as $k$ is even, $k>0$, and is then a well-known result $\unicode[STIX]{x1D701}^{\prime }(-k)=\frac{1}{2}(-1)^{\frac{k}{2}}(2\unicode[STIX]{x1D70B})^{-k}(k!)\unicode[STIX]{x1D701}(k+1)$ [Reference Kanemitsu, Tanigawa and Yoshimoto36, Equation (1)].

It is important to note that (5.4) also holds for $k=0$ but only for $0<a<1$, and is then an equivalent form of the well-known Kummer formula for $\log \unicode[STIX]{x1D6E4}(a)$ [Reference Kummer40, page 4].

By Lemma 5.1, we know that for $h>N/2$, we have $\frac{N-2h+1}{N}=-2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor$ if and only if $h=\frac{N+1}{2}+Nm$, where $m\in \mathbb{N}$. Thus, we let $h=\frac{N+1}{2}+Nm$ in (5.2), employ (5.4) with $k=2m$ in the expression for the residue in (5.3) arising due to double pole to simplify it as

(5.5)$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{x^{2m}}{N(2m)!}\bigg\{-\frac{B_{2m+1}(a)}{2m+1}(\unicode[STIX]{x1D713}(2m+1)+N\unicode[STIX]{x1D6FE}-\log x)+\unicode[STIX]{x1D701}^{\prime }(-2m,a)\bigg\}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{x^{2m}B_{2m+1}(a)}{N(2m+1)!}\biggl(-N\unicode[STIX]{x1D6FE}+\log \biggl(\frac{x}{2\unicode[STIX]{x1D70B}}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{2(-1)^{m}x^{2m}}{N(2\unicode[STIX]{x1D70B})^{2m+1}}\biggl(\mathop{\sum }_{n=1}^{\infty }\frac{(\log n)\sin (2\unicode[STIX]{x1D70B}na)}{n^{2m+1}}+\frac{\unicode[STIX]{x1D70B}}{2}\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n^{2m+1}}\biggr),\end{eqnarray}$$

where, in the course of simplification, the series $\sum _{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{2m+1}}$ is expressed in terms of Bernoulli polynomials using their Fourier expansion [Reference Abramowitz and Stegun1, page 805]:

(5.6)$$\begin{eqnarray}B_{2m+1}(a)=\frac{2(-1)^{m+1}(2m+1)!}{(2\unicode[STIX]{x1D70B})^{2m+1}}\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{2m+1}}.\end{eqnarray}$$

Moreover, part of the expression for $S(x,a)$ in (2.3) can be simplified:

(5.7)$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{(-1)^{h+1}}{N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{\frac{N-2h+1}{N}}\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}\text{exp}\biggl(\frac{i\unicode[STIX]{x1D70B}(1-2h)j}{N}\biggr)\frac{(-1)^{j+\frac{N+1}{2}}}{\unicode[STIX]{x1D70B}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h-1}{N}}}\log \biggl(\frac{1}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{(-1)^{m+\frac{N+3}{2}}}{N}\biggl(\frac{x}{2\unicode[STIX]{x1D70B}}\biggr)^{2m}\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}\frac{(-1)^{2j+\frac{N+1}{2}}}{\unicode[STIX]{x1D70B}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{2m+1}}\log \biggl(\biggl(\frac{2\unicode[STIX]{x1D70B}n}{x}\biggr)^{\frac{1}{N}}e^{\frac{i\unicode[STIX]{x1D70B}j}{N}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{(-1)^{m+1}}{\unicode[STIX]{x1D70B}N}\biggl(\frac{x}{2\unicode[STIX]{x1D70B}}\biggr)^{2m}\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}\bigg\{\biggl(\frac{1}{N}\log \biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)+\frac{i\unicode[STIX]{x1D70B}j}{N}\biggr)\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{2m+1}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{1}{N}\mathop{\sum }_{n=1}^{\infty }\frac{(\log n)\sin (2\unicode[STIX]{x1D70B}na)}{n^{2m+1}}\bigg\}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{x^{2m}}{N(2m+1)!}\log \biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)B_{2m+1}(a)-\frac{(-1)^{m}}{\unicode[STIX]{x1D70B}N}\biggl(\frac{x}{2\unicode[STIX]{x1D70B}}\biggr)^{2m}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{n=1}^{\infty }\frac{(\log n)\sin (2\unicode[STIX]{x1D70B}na)}{n^{2m+1}},\end{eqnarray}$$

where in the last step, we again used (5.6). Now combine (5.5) and (5.7) to deduce that

(5.8)$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{x^{2m}}{N(2m)!}\bigg\{-\frac{B_{2m+1}(a)}{2m+1}(\unicode[STIX]{x1D713}(2m+1)+N\unicode[STIX]{x1D6FE}-\log x)+\unicode[STIX]{x1D701}^{\prime }(-2m,a)\bigg\}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{(-1)^{h+1}}{N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{\frac{N-2h+1}{N}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}\text{exp}\biggl(\frac{i\unicode[STIX]{x1D70B}(1-2h)j}{N}\biggr)\frac{(-1)^{j+\frac{N+1}{2}}}{\unicode[STIX]{x1D70B}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h-1}{N}}}\log \biggl(\frac{1}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =-\unicode[STIX]{x1D6FE}B_{2m+1}(a)\frac{x^{2m}}{(2m+1)!}+\frac{(-1)^{m}x^{2m}\unicode[STIX]{x1D70B}}{N(2\unicode[STIX]{x1D70B})^{2m+1}}\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n^{2m+1}}.\end{eqnarray}$$

Substituting (5.8) in (5.2) and noting that $m=\lfloor \frac{h}{N}-\frac{1}{2}\rfloor =\frac{2h-1-N}{2N}$ leads us to (2.7).◻

Proof of Theorem 2.4.

Let $h=\frac{N+1}{2}+Nm,m>0$, $x=2^{N}\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}^{N}=\unicode[STIX]{x1D70B}^{N+1}$ in Theorem 2.3. To write the sum over $j$ going from $0$ to $\lfloor \frac{h}{N}\rfloor$ in terms of $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}$, we use the fact that

(5.9)$$\begin{eqnarray}\unicode[STIX]{x1D70B}(2\unicode[STIX]{x1D70B})^{(2m+1)N-2jN}x^{2j-1}=2^{2Nm}\unicode[STIX]{x1D6FC}^{2j+\frac{2N}{N+1}(m-j)}\unicode[STIX]{x1D6FD}^{N+\frac{2N^{2}}{N+1}(m-j)}.\end{eqnarray}$$

Now rearrange the terms of the resulting identity upon the aforementioned substitutions, multiply both sides of the rearranged identity by $\unicode[STIX]{x1D6FC}^{-2Nm/(N+1)}$, and then simplify to arrive at (2.8).◻

Proof of Theorem 2.7.

Before we prove Theorem 2.7, it is important to know how it differs from Theorem 2.3. In Theorem 2.3, the condition $\frac{N-2h+1}{N}=-2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor \neq 0$ suggested that we separately consider the contribution $-(a-\frac{1}{2})\unicode[STIX]{x1D701}(-N+2h)$ arising due to the pole of $\unicode[STIX]{x1D6E4}(s)$ at $s=0$.

However, in Theorem 2.7, we have the condition $\frac{N-2h+1}{N}=-2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor =0$, that is, $h=\frac{N+1}{2}$, which means that the integrand $F(s)$, defined in (4.5), has a double order pole at $s=0$ except when $a=1/2$ as will be explained below. So we can as well use the same formula that we used in the proof of Theorem 2.3 to calculate the residue at the double order pole at $s=\frac{N-2h+1}{N}=-2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor \neq 0$, that is (5.1), to calculate the residue at the double order pole at $s=\frac{N-2h+1}{N}=-2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor =0$ in Theorem 2.7. But then the term $-(a-\frac{1}{2})\unicode[STIX]{x1D701}(-N+2h)$ appearing in Theorem 2.3 does not appear in this context. Note also that [Reference Apostol5, page 264, Equation (17)] $\unicode[STIX]{x1D701}(0,a)=\frac{1}{2}-a\neq 0$, except when $a=\frac{1}{2}$, which indeed means that we have a double order pole when $a\neq \frac{1}{2}$. Also when $a=\frac{1}{2}$, even though we get a simple pole at $s=0$, one can always apply (5.1) in this case too and get the correct residue contribution.

Taking the above thing into account, we let $h=\frac{N+1}{2}$ in (5.2) and simplify the resultant using the facts [Reference Temme61, page 3] $B_{1}(a)=(a-\frac{1}{2})$, [Reference Temme61, page 54] $\unicode[STIX]{x1D713}(1)=-\unicode[STIX]{x1D6FE}$ and [Reference Lerch43, Equations (9), $(22^{a})$] $\unicode[STIX]{x1D701}^{\prime }(0,a)=\log \unicode[STIX]{x1D6E4}(a)-\frac{1}{2}\log (2\unicode[STIX]{x1D70B})$. This results in

(6.1)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{\text{exp}(-an^{N}x)}{n(1-\text{exp}(-n^{N}x))}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{\unicode[STIX]{x1D701}(N+1)}{x}+\frac{1}{N}\biggl(\biggl(\frac{1}{2}-a\biggr)((N-1)\unicode[STIX]{x1D6FE}-\log x)+\log \unicode[STIX]{x1D6E4}(a)-\frac{1}{2}\log 2\unicode[STIX]{x1D70B}\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,(-1)^{\frac{N+3}{2}}2^{N}\unicode[STIX]{x1D70B}^{N+1}\mathop{\sum }_{j=1}^{\lfloor \frac{N+1}{2N}\rfloor }\biggl(\frac{-1}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{jN}\frac{B_{2j}(a)B_{N+1-2jN}}{(2j)!(N+1-2jN)!}x^{2j-1}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{(-1)^{\frac{N+3}{2}}}{N}\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}(-1)^{j}\bigg\{\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n(\text{exp}(2A_{N,j}(\frac{n}{x}))-1)}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{(-1)^{j+\frac{N+1}{2}}}{\unicode[STIX]{x1D70B}}\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n}\times \bigg\{\log \biggl(\frac{1}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{1}{2}\biggl(\unicode[STIX]{x1D713}\biggl(\frac{i}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)+\unicode[STIX]{x1D713}\biggl(-\frac{i}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)\biggr)\bigg\}\bigg\},\end{eqnarray}$$

where $A_{N,j}(y)$ is defined in Theorem 2.1. Now (2.12) is proved by letting $x=2^{N}\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}^{N}=\unicode[STIX]{x1D70B}^{N+1}$ in (6.1), making use of (5.9) with $m=0$ and then by simplifying the resultant.◻

Proof of Corollary 2.8.

As mentioned in the proof of Theorem 2.7, the term $-(a-\frac{1}{2})\unicode[STIX]{x1D701}(-N+2h)$ does not appear when $\frac{N-2h+1}{N}=-2\lfloor \frac{h}{N}-\frac{1}{2}\rfloor =0$. With this understanding, we let $h=\frac{N+1}{2}$ in Theorem 2.3 and while simplifying, we use the following formula valid for $0<a<1$ [Reference Gradshteyn, Ryzhik, Zwillinger and Moll26, page 45, 1.441.2]:

$$\begin{eqnarray}\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n}=-\frac{1}{2}\log (2(1-\cos (2\unicode[STIX]{x1D70B}a))).\end{eqnarray}$$

This results in

(6.2)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{\text{exp}(-an^{N}x)}{n(1-\text{exp}(-n^{N}x))}\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D6FE}\biggl(\frac{1}{2}-a\biggr)-\frac{\log (2\sin (\unicode[STIX]{x1D70B}a))}{2N}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,(-1)^{\frac{N+3}{2}}2^{N}\unicode[STIX]{x1D70B}^{N+1}\mathop{\sum }_{j=0}^{\lfloor \frac{N+1}{2N}\rfloor }\biggl(\frac{-1}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{jN}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\frac{B_{2j}(a)B_{N+1-2jN}}{(2j)!(N+1-2jN)!}x^{2j-1}+\frac{(-1)^{\frac{N+3}{2}}}{N}\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}(-1)^{j}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\bigg\{\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n(\text{exp}(2\unicode[STIX]{x1D70B}(\frac{2\unicode[STIX]{x1D70B}n}{x})^{\frac{1}{N}}e^{\frac{i\unicode[STIX]{x1D70B}j}{N}})-1)}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{(-1)^{j+\frac{N+3}{2}}}{2\unicode[STIX]{x1D70B}}\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n}\biggl(\unicode[STIX]{x1D713}\biggl(i\biggl(\frac{2\unicode[STIX]{x1D70B}n}{x}\biggr)^{\frac{1}{N}}e^{\frac{i\unicode[STIX]{x1D70B}j}{N}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\unicode[STIX]{x1D713}\biggl(-i\biggl(\frac{2\unicode[STIX]{x1D70B}n}{x}\biggr)^{\frac{1}{N}}e^{\frac{i\unicode[STIX]{x1D70B}j}{N}}\biggr)\biggr)\bigg\}.\end{eqnarray}$$

Now let $x=2^{N}\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}^{N}=\unicode[STIX]{x1D70B}^{N+1}$ in (6.2) and simplify so as to obtain (2.14).◻

Proof of Corollary 2.9.

Let $a=1/2,N=1$ in (2.14) and simplify.◻

Proof. Let $N=1,a=1/2$, $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FD}=\unicode[STIX]{x1D70B}$ in (2.14) and simplify.

Proof of Theorem 2.10.

Since the proof is similar to that of Theorem 2.1, we will be brief.

As before, (4.4) holds, but now for $\text{Re}(s)=l>\max (\frac{N-2h+1}{N},1)$. We choose the contour $[\unicode[STIX]{x1D706}-iT,\unicode[STIX]{x1D706}+iT],[\unicode[STIX]{x1D706}+iT,-r+iT],[-r+iT,-r-iT]$ and $[-r-iT,\unicode[STIX]{x1D706}-iT]$, where, $r$ is a positive real number such that $0<r<\frac{1}{N}$, the reason for which will be clear soon. The poles of the integrand $F(s)$, defined in (4.5), that are enclosed in the contour are the simple poles at $s=0,1$ and $\frac{N-2h+1}{N}$, the residues of whom are same as those calculated in (4.6), (4.7) and (4.8), respectively. Thus, using Cauchy’s residue theorem, letting $T\rightarrow \infty$ and noting that the integrals along the horizontal segments approach zero, and invoking (4.4), we see that

(7.1)$$\begin{eqnarray}\mathop{\sum }_{n=1}^{\infty }n^{N-2h}\frac{\exp (-an^{N}x)}{1-\exp (-n^{N}x)}=R_{0}+R_{1}+R_{\frac{N-2h+1}{N}}+J(x,a),\end{eqnarray}$$

where $J(x,a)$ is defined in (4.12). We first prove part (i), that is, when $N$ is an odd positive integer. From (4.12) to (4.23), the calculations for evaluating $J(x,a)$ remain exactly the same. Now (4.18) and the inequality $0<r<1/N$ along with the fact that $\text{Re}(s)=1+r$ imply $N-2h+1<c_{1}:=\text{Re}(s_{1})<N-2h+2$. In order to apply Lemma 4.1, we need to again shift the line of integration from $\text{Re}(s_{1})=c_{1}$ to $\text{Re}(s_{1})=c_{2}$, where $0<c_{2}<2$. In doing so, we encounter poles of the integrand $\frac{\unicode[STIX]{x1D6E4}(s_{1})}{\tan (\frac{\unicode[STIX]{x1D70B}s_{1}}{2})}{X_{m,n,j}^{\ast }}^{-s_{1}}$ at $s=2,4,\ldots ,N-2h+1$. Again, the integrals along the horizontal segments approach zero as the height of the contour tends to $\infty$. Now

$$\begin{eqnarray}\lim _{s_{1}\rightarrow 2k}\frac{(s_{1}-2k)\unicode[STIX]{x1D6E4}(s_{1})}{\tan (\frac{\unicode[STIX]{x1D70B}s_{1}}{2})}{X_{m,n,j}^{\ast }}^{-s_{1}}=\frac{2}{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D6E4}(2k){X_{m,n,j}^{\ast }}^{-2k}\end{eqnarray}$$

along with Lemma 4.1 and (4.22) imply that

$$\begin{eqnarray}\displaystyle E(X_{m,n}) & = & \displaystyle \frac{(-1)^{\frac{N+1}{2}}}{\unicode[STIX]{x1D70B}}\sum _{j=-(N-1)}^{N-1}i^{j}e^{\frac{-ij\unicode[STIX]{x1D70B}(2h-1)}{2N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl(\int _{0}^{\infty }\frac{t\cos t}{{X_{m,n,j}^{\ast }}^{2}+t^{2}}\,\text{d}t+\mathop{\sum }_{k=1}^{\frac{N-2h+1}{2}}\unicode[STIX]{x1D6E4}(2k){X_{m,n,j}^{\ast }}^{-2k}\biggr)\nonumber\end{eqnarray}$$

so that along with (4.19), we have

(7.2)$$\begin{eqnarray}\displaystyle J_{2}(x,a) & = & \displaystyle \frac{2}{\unicode[STIX]{x1D70B}Nx}(-1)^{h+\frac{N+3}{2}}(2\unicode[STIX]{x1D70B})^{2h}\biggl(\frac{(2\unicode[STIX]{x1D70B})^{N+1}}{x}\biggr)^{\frac{1-2h}{N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\sum _{j=-(N-1)}^{N-1}i^{j}e^{\frac{-ij\unicode[STIX]{x1D70B}(2h-1)}{2N}}\mathop{\sum }_{n=1}^{\infty }n^{\frac{1-2h}{N}}\sin (2\unicode[STIX]{x1D70B}na)\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{m=1}^{\infty }\biggl(\int _{0}^{\infty }\frac{t\cos (t)}{{X_{m,n,j}^{\ast }}^{2}+t^{2}}\,dt+\mathop{\sum }_{k=1}^{\frac{N-2h+1}{2}}\unicode[STIX]{x1D6E4}(2k){X_{m,n,j}^{\ast }}^{-2k}\biggr).\end{eqnarray}$$

Employ Theorem 2.2 using (4.23) and (4.21) to see that

(7.3)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{m=1}^{\infty }\biggl(\int _{0}^{\infty }\frac{t\cos (t)}{{X_{m,n,j}^{\ast }}^{2}+t^{2}}\,dt+\mathop{\sum }_{k=1}^{\frac{N-2h+1}{2}}\unicode[STIX]{x1D6E4}(2k){X_{m,n,j}^{\ast }}^{-2k}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2}\biggl(\log \biggl(\frac{1}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)-\frac{1}{2}\biggl(\unicode[STIX]{x1D713}\biggl(\frac{i}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)+\unicode[STIX]{x1D713}\biggl(-\frac{i}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,T(N,h,x,j)\biggr),\end{eqnarray}$$

where

$$\begin{eqnarray}T(N,h,x,j):=2\mathop{\sum }_{k=1}^{\frac{N-2h+1}{2}}\frac{\unicode[STIX]{x1D6E4}(2k)\unicode[STIX]{x1D701}(2k)}{(2\unicode[STIX]{x1D70B}(\frac{2\unicode[STIX]{x1D70B}n}{x})^{1/N}e^{\frac{i\unicode[STIX]{x1D70B}j}{N}})^{2k}}.\end{eqnarray}$$

From [Reference Abramowitz and Stegun1, page 259, formula 6.3.18], for $|\text{arg}\,z|<\unicode[STIX]{x1D70B}$, as $z\rightarrow \infty$,

(7.4)$$\begin{eqnarray}\unicode[STIX]{x1D713}(z)\sim \log z-\frac{1}{2z}-\frac{1}{12z^{2}}+\frac{1}{120z^{4}}-\frac{1}{252z^{6}}+\cdots \,.\end{eqnarray}$$

This implies

$$\begin{eqnarray}\displaystyle & & \displaystyle \log \biggl(\frac{1}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)-\frac{1}{2}\biggl(\unicode[STIX]{x1D713}\biggl(\frac{i}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)+\unicode[STIX]{x1D713}\biggl(-\frac{i}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =O_{N,x}(n^{-2/N}),\nonumber\end{eqnarray}$$

and since $1\leqslant k\leqslant \frac{N-2h+1}{2}$, $T(N,h,x,j)=O_{N,x}(n^{-2/N})$. Thus if we multiply both sides of (7.3) by $n^{\frac{1-2h}{N}}\sin (2\unicode[STIX]{x1D70B}na)$ and then sum over $n$, we can write the sum as

(7.5)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{n=1}^{\infty }n^{\frac{1-2h}{N}}\sin (2\unicode[STIX]{x1D70B}na)\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{m=1}^{\infty }\biggl(\int _{0}^{\infty }\frac{t\cos (t)}{{X_{m,n,j}^{\ast }}^{2}+t^{2}}\,dt+\mathop{\sum }_{k=1}^{\frac{N-2h+1}{2}}\unicode[STIX]{x1D6E4}(2k){X_{m,n,j}^{\ast }}^{-2k}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2}\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h-1}{N}}}\bigg\{\log \biggl(\frac{1}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{1}{2}\biggl(\unicode[STIX]{x1D713}\biggl(\frac{i}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)+\unicode[STIX]{x1D713}\biggl(-\frac{i}{\unicode[STIX]{x1D70B}}A_{N,j}\biggl(\frac{n}{x}\biggr)\biggr)\biggr)\bigg\}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{1}{2}\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h-1}{N}}}T(N,h,x,j),\end{eqnarray}$$

since the series $\sum _{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h+1}{N}}}$ converges for $0<a\leqslant 1$ as long as $(2h+1)/N>0$, that is, $h\geqslant 0$, which is what we have in our hypotheses. (This also explains why we fail to obtain a transformation for our series when $h<0$.)

Now substituting (7.5) in (7.2), noting that the expression for $J_{1}(x,a)$ remains exactly as in (4.27), we deduce along with (7.1), (4.6), (4.7), (4.8) and (4.15) that for $0<a\leqslant 1$,

(7.6)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{n=1}^{\infty }n^{N-2h}\frac{\text{exp}(-an^{N}x)}{1-\text{exp}(-n^{N}x)}\nonumber\\ \displaystyle & & \displaystyle \quad =-\biggl(a-\frac{1}{2}\biggr)\unicode[STIX]{x1D701}(-N+2h)+\frac{\unicode[STIX]{x1D701}(2h)}{x}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{1}{N}\unicode[STIX]{x1D6E4}\biggl(\frac{N-2h+1}{N}\biggr)\unicode[STIX]{x1D701}\biggl(\frac{N-2h+1}{N},a\biggr)x^{-\frac{(N-2h+1)}{N}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,S(x,a)+\frac{(-1)^{h+\frac{N+3}{2}}}{\unicode[STIX]{x1D70B}N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{\frac{N-2h+1}{N}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}(-1)^{j}\text{exp}\biggl(\frac{i\unicode[STIX]{x1D70B}(1-2h)j}{N}\biggr)\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h-1}{N}}}T(N,h,x,j).\end{eqnarray}$$

Thus (7.6) leads to (2.16) for $a=1$ with $g(N,h,1)=-\frac{1}{2}\unicode[STIX]{x1D701}(-N+2h)$.

When $0<a<1$, in view of (2.15), (2.16) and (7.6), it suffices to show that

(7.7)$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{(-1)^{h+\frac{N+3}{2}}}{\unicode[STIX]{x1D70B}N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{\frac{N-2h+1}{N}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}(-1)^{j}e^{\frac{i\unicode[STIX]{x1D70B}(1-2h)j}{N}}\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h-1}{N}}}T(N,h,x,j)\nonumber\\ \displaystyle & & \displaystyle \quad =\biggl(a-\frac{1}{2}\biggr)\unicode[STIX]{x1D701}(-N+2h).\end{eqnarray}$$

Use Euler’s formula (2.9) along with $\unicode[STIX]{x1D701}(1-2k)=-B_{2k}/(2k)$, or equivalently, use the functional equation for $\unicode[STIX]{x1D701}(2k)$ to simplify $T(N,h,x,j)$ as

$$\begin{eqnarray}T(N,h,x,j)=\mathop{\sum }_{k=1}^{\frac{N-2h+1}{2}}\frac{(-1)^{k}\unicode[STIX]{x1D701}(1-2k)}{((\frac{2\unicode[STIX]{x1D70B}n}{x})^{1/N}e^{\frac{i\unicode[STIX]{x1D70B}j}{N}})^{2k}}.\end{eqnarray}$$

Now substitute the above representation of $T(N,h,x,j)$ in (7.7), separate the term corresponding to $k=\frac{N-2h+1}{2}$ on the left side and invoke [Reference Gradshteyn, Ryzhik, Zwillinger and Moll26, page 45, Formula 1.441.1] $\sum _{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n}=-\unicode[STIX]{x1D70B}(a-\frac{1}{2})$ to see that this term equals $(a-\frac{1}{2})\unicode[STIX]{x1D701}(-N+2h)$. Thus we need only show that

(7.8)$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{(-1)^{h+\frac{N+3}{2}}}{\unicode[STIX]{x1D70B}N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{\frac{N-2h+1}{N}}\mathop{\sum }_{k=1}^{\frac{N-2h-1}{2}}\frac{(-1)^{k}\unicode[STIX]{x1D701}(1-2k)}{(\frac{2\unicode[STIX]{x1D70B}}{x})^{2k/N}}\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{\frac{2h+2k-1}{N}}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}(-1)^{j}e^{\frac{i\unicode[STIX]{x1D70B}(1-2h-2k)j}{N}}=0.\end{eqnarray}$$

In the sum in (4.2), replace $j$ by $2j$, then let $z=\frac{\unicode[STIX]{x1D70B}}{2N}(2h+2k-1)$ so that

(7.9)$$\begin{eqnarray}\mathop{\sum }_{j=-\frac{(N-1)}{2}}^{\frac{(N-1)}{2}}(-1)^{j}e^{\frac{i\unicode[STIX]{x1D70B}(1-2h-2k)j}{N}}=\frac{\cos (\frac{\unicode[STIX]{x1D70B}(2h+2k-1)}{2})}{\cos (\frac{\unicode[STIX]{x1D70B}(2h+2k-1)}{2N})}.\end{eqnarray}$$

Now $\cos (\frac{\unicode[STIX]{x1D70B}(2h+2k-1)}{2})=0$, but we should also show $\cos (\frac{\unicode[STIX]{x1D70B}(2h+2k-1)}{2N})\neq 0$. To that end, note that $1\leqslant k<\frac{N-2h+1}{2}$ implies $\frac{2h+1}{N}\leqslant \frac{2h+2k-1}{N}<1$. Also, $h\geqslant 0$ implies $\frac{2h+1}{N}\geqslant \frac{1}{N}$. Combining, we see that $\frac{1}{N}\leqslant \frac{2h+2k-1}{N}<1$, so that $\cos (\frac{\unicode[STIX]{x1D70B}(2h+2k-1)}{2N})\neq 0$ for $N>1$. Thus, the sum over $j$ in (7.9) equals $0$ for $N>1$ which implies (7.8). For $N=1$, note that $h<N/2$ along with $h\geqslant 0$ implies $h=0$ so that the sum over $k$ in (7.8) is empty, and hence (7.8) holds again. Hence (7.7) holds and thus (2.16) is valid with $g(N,h,a)=0$ for $0<a<1$.

We omit the proof of (2.17) since it is exactly along the lines of the proof of Theorem 2.1. By a similar argument one sees that (2.17) holds also when $h<0$, unlike when $N$ is odd.◻

Proof of Corollary 2.11.

Let $N=1$ so that $h=0$ in part (i) of Theorem 2.10. Use the fact [Reference Olver, Lozier, Boisvert and Clark51, page 608, Formula 25.11.12] $\unicode[STIX]{x1D701}(\ell ,a)=\frac{(-1)^{\ell }}{(\ell -1)!}\unicode[STIX]{x1D713}^{(\ell -1)}(a)$, let $x=2\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}=\unicode[STIX]{x1D70B}^{2}$ and simplify.◻

Proof of Theorem 2.12.

Here $N$ is an even positive integer. We first prove the result for a nonnegative integer $m$ using Theorem 2.1. For $m<0$, it can be proved using Theorem 2.10.

Suppose $m$ is a nonnegative integer. Then let $h=\frac{N}{2}+Nm$, $x=2^{N}\unicode[STIX]{x1D6FC}$ in Theorem 2.1 and let $\unicode[STIX]{x1D6FD}>0$ be defined by $\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}^{N}=\unicode[STIX]{x1D70B}^{N+1}$. After rearranging some terms, we obtain

(8.1)$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl(a-\frac{1}{2}\biggr)\unicode[STIX]{x1D701}(2Nm)+\mathop{\sum }_{j=1}^{m}\frac{B_{2j+1}(a)}{(2j+1)!}\unicode[STIX]{x1D701}(2N(m-j))(2^{N}\unicode[STIX]{x1D6FC})^{2j}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\mathop{\sum }_{n=1}^{\infty }\frac{n^{-2Nm}\text{exp}(-a(2n)^{N}\unicode[STIX]{x1D6FC})}{1-\exp (-(2n)^{N}\unicode[STIX]{x1D6FC})}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{N}\unicode[STIX]{x1D6E4}\biggl(\frac{1-2Nm}{N}\biggr)\unicode[STIX]{x1D701}\biggl(\frac{1-2Nm}{N},a\biggr)(2^{N}\unicode[STIX]{x1D6FC})^{\frac{2Nm-1}{N}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{(-1)^{\frac{N}{2}+1}}{N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{2^{N}\unicode[STIX]{x1D6FC}}\biggr)^{\frac{1-2Nm}{N}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{j=-\frac{N}{2}}^{\frac{N}{2}-1}e^{\frac{i\unicode[STIX]{x1D70B}(2j+1)}{2N}}\exp \biggl(-\frac{i\unicode[STIX]{x1D70B}}{2}(2m+1)(2j+1)\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)+i(-1)^{j+\frac{N}{2}+1}\sin (2\unicode[STIX]{x1D70B}na)}{n^{2m+1-\frac{1}{N}}(\text{exp}((2n)^{\frac{1}{N}}\unicode[STIX]{x1D6FD}e^{\frac{i\unicode[STIX]{x1D70B}(2j+1)}{2N}})-1)}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{(-1)^{\frac{N}{2}+1}}{2}(2\unicode[STIX]{x1D70B})^{N+2Nm}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{j=0}^{m}\biggl(\frac{-1}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{jN}\frac{B_{2j}(a)B_{N+2N(m-j)}}{(2j)!(N+2N(m-j))!}(2^{N}\unicode[STIX]{x1D6FC})^{2j-1},\end{eqnarray}$$

where we have used the fact $2A_{N,j+\frac{1}{2}}(\frac{n}{x})=(2n)^{\frac{1}{N}}\unicode[STIX]{x1D6FD}e^{\frac{i\unicode[STIX]{x1D70B}(2j+1)}{2N}}$.

We now simplify some of the expressions on the right-hand side. Since $\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FD}^{N}=\unicode[STIX]{x1D70B}^{N+1}$,

(8.2)$$\begin{eqnarray}\displaystyle & \displaystyle \biggl(\frac{\unicode[STIX]{x1D70B}}{\unicode[STIX]{x1D6FC}}\biggr)^{\frac{1-2Nm}{N}}=\unicode[STIX]{x1D6FC}^{\frac{2Nm-1}{N+1}}\unicode[STIX]{x1D6FD}^{\frac{1-2Nm}{N+1}}, & \displaystyle\end{eqnarray}$$

(8.3)$$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D70B}^{N+2Nm-2Nj}\unicode[STIX]{x1D6FC}^{2j-1}=\unicode[STIX]{x1D6FC}^{\frac{2j+2Nm-1}{N+1}}\unicode[STIX]{x1D6FD}^{N+\frac{2N^{2}(m-j)-N}{N+1}}. & \displaystyle\end{eqnarray}$$

We now split the sum $\sum _{j=-\frac{N}{2}}^{\frac{N}{2}-1}=\sum _{j=0}^{\frac{N}{2}-1}+\sum _{j=-\frac{N}{2}}^{-1}$ and replace $j$ by $-1-j$ in the second sum. Then combining the corresponding terms in the resulting two finite sums on the above right-hand side, using the fact that $\exp (-\frac{1}{2}(i\unicode[STIX]{x1D70B}(2j+1)(2m+1)))=i(-1)^{j+m+1}$ and then simplifying, we get

(8.4)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j=-\frac{N}{2}}^{\frac{N}{2}-1}e^{\frac{i\unicode[STIX]{x1D70B}(2j+1)}{2N}}\exp \biggl(-\frac{i\unicode[STIX]{x1D70B}}{2}(2m+1)(2j+1)\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)+i(-1)^{j+\frac{N}{2}+1}\sin (2\unicode[STIX]{x1D70B}na)}{n^{2m+1-\frac{1}{N}}(\text{exp}((2n)^{\frac{1}{N}}\unicode[STIX]{x1D6FD}e^{\frac{i\unicode[STIX]{x1D70B}(2j+1)}{2N}})-1)}\nonumber\\ \displaystyle & & \displaystyle \quad =-2(-1)^{m+1}\mathop{\sum }_{j=0}^{\frac{N}{2}-1}(-1)^{j}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\biggl[\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{n^{2m+1-\frac{1}{N}}}\text{Im}\biggl(\frac{e^{\frac{i\unicode[STIX]{x1D70B}(2j+1)}{2N}}}{\exp ((2n)^{\frac{1}{N}}\unicode[STIX]{x1D6FD}e^{\frac{i\unicode[STIX]{x1D70B}(2j+1)}{2N}})-1}\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,(-1)^{j+\frac{N}{2}+1}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{n^{2m+1-\frac{1}{N}}}\text{Re}\biggl(\frac{e^{\frac{i\unicode[STIX]{x1D70B}(2j+1)}{2N}}}{\exp ((2n)^{\frac{1}{N}}\unicode[STIX]{x1D6FD}e^{\frac{i\unicode[STIX]{x1D70B}(2j+1)}{2N}})-1}\biggr)\biggr].\end{eqnarray}$$

Now substituting (8.4) in (8.1), using (8.2) and (8.3), and then multiplying both sides of the resulting identity by $\unicode[STIX]{x1D6FC}^{-(\frac{2Nm-1}{N+1})}$, we arrive at (2.19). This completes the proof for $m>0$. For $m<0$, the result follows from part (ii) of Theorem 2.10. The argument is exactly the same as above and is hence omitted.◻

Proof of Theorem 2.13.

Let $h=\frac{N}{2}$ in Theorem 2.1. This gives

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{\exp (-an^{N}x)}{1-\exp (-n^{N}x)} & = & \displaystyle \frac{1}{2}\biggl(a-\frac{1}{2}\biggr)+\frac{\unicode[STIX]{x1D701}(N)}{x}+\frac{1}{N}\unicode[STIX]{x1D6E4}\biggl(\frac{1}{N}\biggr)\unicode[STIX]{x1D701}\biggl(\frac{1}{N},a\biggr)x^{-\frac{1}{N}}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{(-1)^{\frac{N}{2}+1}}{N}\biggl(\frac{2\unicode[STIX]{x1D70B}}{x}\biggr)^{\frac{1}{N}}\mathop{\sum }_{j=-\frac{N}{2}}^{\frac{N}{2}-1}e^{\frac{i\unicode[STIX]{x1D70B}(1-2h)(j+\frac{1}{2})}{N}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)+i(-1)^{j+\frac{N}{2}+1}\sin (2\unicode[STIX]{x1D70B}na)}{n^{1-\frac{1}{N}}(\text{exp}(2A_{N,j+\frac{1}{2}}(\frac{n}{x}))-1)}.\nonumber\end{eqnarray}$$

Wigert’s formula [Reference Wigert67, pages 8–9, Equation (5)] (see also [Reference Dixit and Maji21, Equation (1.2)]) is a special case of the above formula when $a=1$.

Now let $N=2$ and simplify so as to obtain

(8.5)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{\exp (-an^{2}x)}{1-\exp (-n^{2}x)}=\frac{1}{2}\biggl(a-\frac{1}{2}\biggr)+\frac{\unicode[STIX]{x1D70B}^{2}}{6x}+\frac{1}{2}\sqrt{\frac{\unicode[STIX]{x1D70B}}{x}}\unicode[STIX]{x1D701}\biggl(\frac{1}{2},a\biggr)+\sqrt{\frac{2\unicode[STIX]{x1D70B}}{x}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\biggl[\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{\sqrt{n}}\text{Re}\biggl(\frac{\exp (i\unicode[STIX]{x1D70B}/4)}{\exp ((2\unicode[STIX]{x1D70B})^{\frac{3}{2}}\sqrt{\frac{n}{x}}e^{-\frac{i\unicode[STIX]{x1D70B}}{4}})-1}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad +\,\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{\sqrt{n}}\text{Im}\biggl(\frac{\exp (i\unicode[STIX]{x1D70B}/4)}{\exp ((2\unicode[STIX]{x1D70B})^{\frac{3}{2}}\sqrt{\frac{n}{x}}e^{-\frac{i\unicode[STIX]{x1D70B}}{4}})-1}\biggr)\biggr].\end{eqnarray}$$

Letting $a=1$ and using Lemmas 4.4 and 8.1, we obtain a formula of Ramanujan [Reference Ramanujan55], [Reference Berndt8, page 314], [Reference Ramanujan56, page 332]:

(8.6)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{1}{\exp (n^{2}x)-1}=\frac{1}{4}+\frac{\unicode[STIX]{x1D70B}^{2}}{6x}+\frac{1}{2}\sqrt{\frac{\unicode[STIX]{x1D70B}}{x}}\unicode[STIX]{x1D701}\biggl(\frac{1}{2}\biggr)+\sqrt{\frac{\unicode[STIX]{x1D70B}}{2x}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{n=1}^{\infty }\frac{1}{\sqrt{n}}\biggl(\frac{\cos (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}}+\frac{\unicode[STIX]{x1D70B}}{4})-e^{-2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}}}\cos (\frac{\unicode[STIX]{x1D70B}}{4})}{\cosh (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}})-\cos (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}})}\biggr).\end{eqnarray}$$

For $0<a<1$, one is able to further simplify (8.5). To that end, keeping in mind that (1.4) holds also for $\text{Re}(s)<1$, we let $s=1/2$ in it to obtain

(8.7)$$\begin{eqnarray}\unicode[STIX]{x1D701}\biggl(\frac{1}{2},a\biggr)=\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{\sqrt{n}}+\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{\sqrt{n}}.\end{eqnarray}$$

Invoking (8.7) in (8.5), then simplifying using Lemmas 4.4 and 8.1 leads to

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{\exp (-an^{2}x)}{1-\exp (-n^{2}x)}=\frac{1}{2}\biggl(a-\frac{1}{2}\biggr)+\frac{\unicode[STIX]{x1D70B}^{2}}{6x}+\frac{1}{2}\sqrt{\frac{\unicode[STIX]{x1D70B}}{x}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\biggl[\mathop{\sum }_{n=1}^{\infty }\frac{\cos (2\unicode[STIX]{x1D70B}na)}{\sqrt{n}}\biggl(\frac{\sinh (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}})-\sin (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}})}{\cosh (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}})-\cos (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}})}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad +\,\mathop{\sum }_{n=1}^{\infty }\frac{\sin (2\unicode[STIX]{x1D70B}na)}{\sqrt{n}}\biggl(\frac{\sinh (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}})+\sin (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}})}{\cosh (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}})-\cos (2\unicode[STIX]{x1D70B}^{\frac{3}{2}}\sqrt{\frac{n}{x}})}\biggr)\biggr].\nonumber\end{eqnarray}$$

Finally, let $x=\unicode[STIX]{x1D6FC}$ and let $\unicode[STIX]{x1D6FD}=4\unicode[STIX]{x1D70B}^{3}/\unicode[STIX]{x1D6FC}$ to arrive at (2.20). To derive (2.21), let $a=1/2$ in (2.20).◻

Proof. Let $a=1/4,N=1$ in (2.14) and simplify. This leads to (9.1).

Proof. Let $a=1/2$, $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FD}=\unicode[STIX]{x1D70B}$ in Corollary 2.11 and use the fact [Reference Olver, Lozier, Boisvert and Clark51, page 144, Formula 5.15.3] $\unicode[STIX]{x1D713}^{\prime }(1/2)=\unicode[STIX]{x1D70B}^{2}/2$.

Proof. Let $a=1/4,\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FD}=\unicode[STIX]{x1D70B}$ in Corollary 2.11 and use the fact [Reference Olver, Lozier, Boisvert and Clark51, page 144, Formula 5.15.1] $\unicode[STIX]{x1D713}^{\prime }(1/4)=8G+\unicode[STIX]{x1D70B}^{2}$.