2·1. Saddle point review

By [ Reference Manstavičius and Petuchovas13 , theorem 2 and corollary 5], uniformly for $n \ge m \ge 1$ we have

(2·1) \begin{equation} {}p_{n,m}= \frac{\exp\!\left( \sum_{i=1}^{m} \frac{x^i}{i}\right)}{x^n \sqrt{2\pi \lambda}} \left(1+O\left( u^{-1}\right)\right),\end{equation}

where $x=x_{n,m}$ is the saddle point defined as the unique positive solution to (1·11), and $\lambda=\lambda_{n,m}$ is defined as

(2·2) \begin{equation} {}\lambda = \sum_{i=1}^{m} ix^i.\end{equation}

Proof. We study x. Considering only the $i=m$ term in (1·11) gives $x \le n^{1/m}$ . If $x \lt 1$ then $\sum_{i=1}^m x^i \lt m \le n$ , a contradiction, hence $x \ge 1$ . Consequently, by the definition of x, $mx^m \ge n$ holds, which implies

\[ {}x \ge n^{\frac{1}{m}} m^{-\frac{1}{m}} \ge n^{\frac{1}{m}}e^{-\frac{1}{e}}.\]

We turn to $\lambda$ . We have $\lambda \le m \sum_{i=1}^{m} x^i = nm$ . By [ Reference Manstavičius and Petuchovas13 , lemma 9],

(2·3) \begin{equation} {} {}\lambda = nm(1+O(1/\log u)) {}\end{equation}

uniformly for $u \gt 1$ . We also have $x^m \le n=\sum_{i=1}^{m} x^i \lt x^m/(1-x^{-1})$ and so

\[ {}mn(1-x^{-1}) \lt mx^m \le \sum_{i=1}^{m}ix^i = \lambda \le nm \]

implying $\lambda=nm(1+O(x^{-1}))=nm(1+O(n^{-1/m}))$ . The estimate (2·3) already shows $\lambda \gg nm$ when $u \ge C$ . If $u \lt C$ then, since $x \ge 1$ , we find $\lambda \ge \sum_{i=1}^{m} i \gg nm$ .

2·2. Dickman function review

We define $\xi \colon [1,\infty) \to [0,\infty)$ , a function of variable $u\ge 1$ , by

\[e^{\xi(u)} = 1+u\xi(u).\]

Let

\[ {}I(s) = \int_{0}^{s} \frac{e^t-1}{t} {\textrm{d}}t\]

which defines an entire function. Observe that

\[ I'(\xi) = \frac{e^{\xi}-1}{\xi}=u,\]

and a short computation shows $I''(\xi)=1/\xi'$ . The following lemma is proved by induction.

It has the following direct consequence.

The function I arises when studying $\rho$ and its Laplace transform.

Lemma 2·6. [ Reference de Bruijn3 , equation (1·9)] [ Reference Alladi1 , equation (3·9)]. Let $\gamma$ be the Euler–Mascheroni constant. We have

\[ {}\hat{\rho}(s) \;:\!=\; \int_{0}^{\infty} e^{-sv}\rho(v)\, {\textrm{d}}v = \exp\!\left( \gamma + I(-s) \right)\]

for $s \in \mathbb{C}$ . Uniformly for $u \ge 1$ ,

(2·4) \begin{equation} {} {}\rho(u)=\frac{1}{\sqrt{2\pi I''(\xi)}}\exp\!\left( \gamma-u\xi + I(\xi)\right)(1 + O( u^{-1})). {}\end{equation}

We have the following bounds on $\hat{\rho}$ .

Lemma 2·7. [ Reference Hildebrand and Tenenbaum10 , lemma 2·7] The following bounds hold for $s=-\xi(u)+it$ :

\[ {} {}\hat{\rho}(s) = \begin{cases} O\left(\exp\!\left(I(\xi)-\frac{t^2u}{2\pi^2}\right)\right) & \text{if }|t| \le \pi,\\[5pt] {} {} {}O\left(\exp\!\left(I(\xi)-\frac{u}{\pi^2+\xi^2}\right)\right) & \text{if }|t| \ge \pi,\\[5pt] {} {} {}\frac{1}{s} + O\left( \frac{1+u \xi }{|s|^2} \right) & \text{if }1+u\xi =O(|t|).\end{cases}\]

2·3. T

We define the following function, holomorphic in the strip $|\Im s | \lt 2\pi m$ :

\[ {}T(s) \;:\!=\; \int_{0}^{s} \frac{e^t-1}{t}\left( \frac{\frac{t}{m}}{1-e^{-t/m}}-1\right){\textrm{d}}t.\]

The Bernoulli numbers $B_i$ are defined by

\[\frac{s}{1-e^{-s}} = 1+\frac{s}{2} +\frac{s^2}{12}-\frac{s^4}{720}+ \ldots = \sum_{i\ge 0} B_i \frac{s^i}{i!}.\]

This series converges for $|\Im s| \lt 2\pi$ . It is known that $B_i \ll i!/(2\pi)^i$ . It is then easy to see, from the estimate $\int_{0}^{s} (e^t-1)t^{i-1} {\textrm{d}}t \ll |s|^{i-1} (|e^s|+|s|)$ which holds for $\Re s \ge 0$ and $i \ge 1$ , that

(2·5) \begin{equation} {}T(s) = \sum_{i=1}^{k} \frac{B_i}{m^i i!} \int_{0}^{s} (e^t-1)t^{i-1} {\textrm{d}}t + O\bigg( \frac{|s|^k (|e^{s}|+|s|)}{(2\pi m)^{k+1}} \bigg)\end{equation}

holds for $k \ge 0$ and $s=\sigma+it$ with $\sigma \in [0,\pi m]$ and $t \in [\!-\pi m,\pi m]$ . Applying (2·5) with $k=1$ we obtain

Corollary 2·8. [ Reference Manstavičius and Petuchovas13 , lemma 11]. For $s=\sigma+it$ with $\sigma\in [0,\pi m]$ and $t\in [\!-\pi m,\pi m]$ ,

\[ {}\left|T(s)+\frac{s}{2m}\right| \ll \frac{e^{\sigma}}{m} + \frac{t^2}{m^2}.\]

Applying (2·5) with $k=2$ and $s=\xi=\xi(u)$ we obtain

(2·6) \begin{equation}T(\xi) = \frac{(u-1)\xi}{2m} + \frac{e^{\xi}(\xi-1)-\frac{\xi^2}{2}+1}{12m^2} + O\bigg( \frac{u\log^3 (u+1)}{m^3}\bigg)\end{equation}

when $\xi(u) \le \pi m$ holds.

Lemma 2·9. Suppose $\xi=\xi(u) \le \pi m$ . For any $k \ge 1$ we have

(2·7) \begin{equation} {}\begin{split} {} {}T^{(k)}(\xi) &= I^{(k)}(\xi)\left( \frac{\xi/m}{1-e^{-\xi/m}}-1+O_k\left( \frac{1}{m}\right)\right) \\[5pt] {} {}&=\frac{e^{\xi}}{2m}\left(1+O_k\left(\frac{1}{\log(u+1)}+\frac{\log(u+1)}{m}\right)\right). {}\end{split} {}\end{equation}

If additionally $t \in [\!-\pi m, \pi m]$ then $T^{(k)}(\xi+it) \ll_k e^{\xi(u)}/m$ .

Proof. We have $T'(s) = I'(s) g(s/m)$ for $g(z)\;:\!=\;z/(1-e^{-z}) - 1$ . Repeated differentiation shows

(2·8) \begin{equation} {}T^{(k)}(s) = I^{(k)}(s) g(s/m)+O_k\bigg( \sum_{i=1}^{k-1}m^{-i} |I^{(k-i)}(s)g^{(i)}(s/m)|\bigg).\end{equation}

We suppose that $s=\xi(u) + it$ and that the inequalities $\xi(u)\le \pi m$ and $|t|\le \pi m$ are satisfied. From (2·8),

(2·9) \begin{equation} T^{(k)}(s) =I^{(k)}(s) g(s/m)+O_k\bigg( \frac{1}{m}\sum_{i=1}^{k-1} |I^{(k-i)}(s)|\bigg) {} {}\end{equation}

as our assumptions on s guarantee that $g^{(i)}(s/m)\ll_i 1$ . We first suppose $t=0$ and establish (2·7). Using (2·9) and the definition of g, the first equality in (2·7) will follow if we show that $I^{(k-i)}(\xi(u)) \ll_k I^{(k)}(\xi(u))$ , which is a consequence of Corollary 2·5. The second equality in (2·7) follows from another application of Corollary 2·5. Finally, the claim for $T^{(k)}(s)$ (when $|t|\le \pi m$ ) follows from (2·9) since $|I^{(k)}(s)| \ll_k e^{\xi(u)}/|s|$ and $|I^{(k-i)}(s)| \ll_k u$ ( $1\le i \le k-1$ ) by Corollary 2·5 and $|g(s/m)| \ll |s|/m$ .

2·4. $H_n$

Let $H_n\;:\!=\; \sum_{i=1}^{n} ({1}/{i})$ be the nth harmonic sum. By the Euler–Maclaurin formula,

(2·10) \begin{equation}H_n = \log n + \gamma + \frac{1}{2n} +O(n^{-2}).\end{equation}

3·1. Proof of the first part of Proposition 3·1

Here we prove (3·1). If $m=O(\log (u+1))$ then (3·1) follows from (1·4). Hence we suppose that $m \ge C \log (u+1)$ from now on. Similarly, if $u =O(1)$ then (3·1) is already in (1·4), so we suppose $u \ge C$ . We define a function

\[ D(z) \;:\!=\;z^{-n} \exp\!\left(\sum_{i=1}^{m} \frac{z^i}{i}\right), \]

so that x defined in (1·11) solves $D'(z)=0$ , and (2·1) can be written as

(3·5) \begin{equation} {}p_{n,m}= \frac{D(x)}{\sqrt{2 \pi \lambda}} (1+O( u^{-1})).\end{equation}

We can write $\log D(x)$ as

\[ \log D(x)=-n \log x + \sum_{j=1}^{m} \frac{1}{j}+I(m\log x)+T(m\log x),\]

see [ Reference Manstavičius and Petuchovas13 , equation (40)] for the details. We set

\[ \widetilde{\xi} \;:\!=\; m \log x\]

and define $\widetilde{u} \ge 1$ as

\[ \widetilde{u} \;:\!=\; \frac{e^{\widetilde{\xi}}-1}{\widetilde{\xi}}.\]

In this notation, (3·5) implies (using (2·4) with $\widetilde{u}$ in place of u, and (2·10)) that

\[ {}p_{n,m} = \rho(\widetilde{u}) \exp\!\left( (\widetilde{u}-u)\widetilde{\xi}\right) \exp \left(T(\widetilde{\xi})\right)\sqrt{ \frac{m^2 I''(\xi(\widetilde{u}))}{\lambda}}\exp\!\left( O\left(u^{-1}+ \widetilde{u}^{-1}+m^{-1}\right)\right)\]

for $u \ge C$ . A short computation shows that $u=I'(\xi)=(I+T)'(\widetilde{\xi})$ holds by the definition of $\xi$ , $\widetilde{\xi}$ and x. Since $T'(t) \ge 0$ for $t \ge 0$ we have $\widetilde{\xi}\le \xi$ and so, by the monotonicity of $I'(t)=(e^t-1)/t$ , it follows that

\[\widetilde{u}=I'(\widetilde{\xi})\le I'(\xi)=u.\]

By [ Reference Manstavičius and Petuchovas13 , equation (22)],

(3·6) \begin{equation}\widetilde{\xi} = \xi + O\left(\frac{\log (u+1)}{m}\right)\end{equation}

uniformly in $m \ge \log u \gt 0$ . From (3·6) we see that, uniformly in $m \ge \log u \gt 0$

(3·7) \begin{equation}\widetilde{u} = u + O\left( \frac{u\log(u+1)}{m}\right).\end{equation}

We will be using (3·6) and (3·7) frequently. For $m \ge \log u \gt 0$ we have

\[ \lambda = \frac{m^2}{\xi'(u)}\left(1+O\left( \frac{\log (u+1)}{m}\right)\right)\]

by [ Reference Manstavičius and Petuchovas13 , equation (23)]. Hence, since $I''(\xi(u))=1/\xi'(u)$ ,

\[ \sqrt{ \frac{m^2 I''(\xi(\widetilde{u}))}{\lambda }}=\sqrt{ \frac{m^2}{\lambda \xi'(\widetilde{u})}}=\sqrt{\frac{\xi'(u)}{\xi'(\widetilde{u})}} \exp\!\left(O\left(\frac{\log(u+1)}{m}\right)\right)\]

for $m \ge C\log (u+1) \gt 0$ . Since $\xi'(t)\asymp 1/t$ and $\xi''(t) =-\xi'(t)I^{(3)}(\xi(t))/I''(\xi(t))^2 \ll 1/t^2$ for $t \ge C$ ,

\[ \frac{\xi'(u)}{\xi'(\widetilde{u})} =1 +O\left( \frac{u-\widetilde{u}}{\widetilde{u}}\right) = \exp\!\left(O\left( \frac{\log(u+1)}{m}\right)\right)\]

holds for $m \ge C\log (u+1) \gt 0$ and $u \ge C$ and so

\[ \sqrt{ \frac{m^2I''(\xi(\widetilde{u}))}{\lambda}}= \exp\!\left(O\left(\frac{\log(u+1)}{m}\right)\right),\]

when $m \ge C\log (u+1) \gt 0$ and $u \ge C$ . At this point, we have established that

\[p_{n,m} = \rho(\widetilde{u}) \exp\big((\widetilde{u}-u) \widetilde{\xi}\big) \exp \left(T(\widetilde{\xi})\right)\exp\bigg( O\bigg( \frac{1}{u}+ \frac{\log(u+1)}{m}\bigg)\bigg)\]

holds in our range. We have $\widetilde{\xi} \le \xi \le 2\log u \le \pi m$ by Lemma 2·3 and our assumption $m \ge C\log(u+1)$ . By (2·6), (3·6) and (3·7),

\[ {}T(\widetilde{\xi}) = \frac{\widetilde{u}\widetilde{\xi}}{2m} + O\bigg(\! \frac{u\log^2(u+1)}{m^2}+\frac{\log(u+1)}{m}\!\bigg)= \frac{u\xi(u)}{2m} + O\bigg( \!\frac{u\log^2(u+1)}{m^2} + \frac{\log(u+1)}{m}\!\bigg)\]

in this range. We now have

\[ {}p_{n,m} = \exp\bigg(\frac{u\xi(u)}{2m}\bigg)\rho(\widetilde{u}) \exp\big( (\widetilde{u}-u)\widetilde{\xi}\big) \exp\bigg( O\bigg(\frac{1}{u}+ \frac{u\log^2(u+1)}{m^2}\bigg)\bigg) \]

for $m \ge C\log (u+1)$ , $u \ge C$ . Let $r(t)=-\rho'(t)/\rho(t)$ be the negative of the logarithmic derivative of $\log \rho$ . We have

\[\rho(\widetilde{u})=\rho(u)\exp\!\left(\int_{\widetilde{u}}^{u}r(t){\textrm{d}}t\right).\]

By integration by parts, we may write the integral in the exponent as

\[ {}\int_{\widetilde{u}}^{u}r(t){\textrm{d}}t {}= (u-\widetilde{u})r(u) - \int_{\widetilde{u}}^{u} (t-\widetilde{u})r'(t) {\textrm{d}}t. \]

By [ Reference de la Bretèche and Tenenbaum5 , lemma 3·7], the estimates

(3·8) \begin{equation}r(v) = \xi(v) + O\left(v^{-1}\right), \qquad r'(v) = \xi'(v)+O\left(v^{-2}\right)\end{equation}

hold uniformly for $v \ge 1$ . Hence, using (3·6) and (3·7),

\begin{align*}\rho(\widetilde{u}) \exp( (\widetilde{u}-u)\widetilde{\xi}) &= \rho(u) \exp((u-\widetilde{u}) (r(u)-\xi(u)+\xi(u)-\widetilde{\xi}) \\[5pt] & \qquad - \int_{\widetilde{u}}^{u}(t-\widetilde{u})(r'(t)-\xi'(t)+\xi'(t)){\textrm{d}}t )\\[5pt] &=\rho(u) \exp\bigg( O\bigg(\frac{1}{u} + \frac{u\log^2(u+1)}{m^2}\bigg)-\int_{\widetilde{u}}^{u} (t-\widetilde{u})\xi'(t){\textrm{d}}t\bigg).\end{align*}

To conclude the proof, it remains to bound $\int_{\widetilde{u}}^{u}(t-\widetilde{u})\xi'(t){\textrm{d}}t$ . Since $\xi'(t) \ll 1/t$ ,

\[ \int_{\widetilde{u}}^{u}(t-\widetilde{u})\xi'(t){\textrm{d}}t\ll \int_{\widetilde{u}}^{u} \frac{t-\widetilde{u}}{t} {\textrm{d}}t = \widetilde{u} \left( \frac{u}{\widetilde{u}}-1-\log \left(\frac{u}{\widetilde{u}}\right)\right) \ll \widetilde{u} \frac{(u-\widetilde{u})^2}{\widetilde{u}^2} \ll \frac{u\log^2(u+1)}{m^2}\]

if $m \ge C\log (u+1)$ , using (3·6) and (3·7). The proof of (3·1) is completed.

3·2. Preliminary lemmas for second part of Proposition 3·1

The following consequence of Cauchy’s integral formula is implicit in the proofs of [ Reference Manstavičius and Petuchovas13 , theorems 2,4].

Lemma 3·3. We have

\[ p_{n,m} = \frac{\exp\big(H_m -\gamma\big)}{m} {} \frac{1}{2\pi i} \int_{-\xi-im\pi}^{-\xi+im\pi} e^{us} \hat{\rho}(s) e^{T(-s)}{\textrm{d}}s.\]

The following lemma is implicit in the proof of [ Reference Manstavičius and Petuchovas13 , theorem 4].

Lemma 3·4. Suppose $\sqrt{n\log n}=O(m)$ . Then

\[\frac{1}{2\pi i} \int_{-\xi-im\pi}^{-\xi+im\pi} e^{us} \hat{\rho}(s){\textrm{d}}s = \rho(u) +O\left( \frac{e^{-(u-1)\xi}}{n}\right).\]

A variant of the next lemma is implicit in the proof of [ Reference Manstavičius and Petuchovas13 , theorem 2].

Lemma 3·5. Suppose $\sqrt{n\log n}=O(m)$ and $1+u\xi \le A \le m\pi$ . Then

\[ \frac{1}{2\pi i} \int_{-\xi+iA}^{-\xi+im\pi} e^{us} \hat{\rho}(s) \left(e^{T(-s)-T(\xi)}-1\right){\textrm{d}}s \ll \frac{e^{-u\xi}\log(u+1)}{m}. \]

The same bound holds if we integrate from $-\xi-im\pi$ to $-\xi-iA$ .

Proof. We have $\hat{\rho}'(s) = \rho(s) (e^{-s}-1)/s$ . Integration by parts shows that the integral is equal to

\begin{multline*} {}\frac{1}{2\pi i} \frac{e^{us}}{u}\hat{\rho}(s) \left(e^{T(-s)-T(\xi)}-1\right) \Big|^{s=-\xi+im\pi}_{s=-\xi+iA} \\[5pt] -\frac{1}{2\pi i} \int_{-\xi+iA}^{-\xi+im\pi} \frac{e^{us}}{u} \hat{\rho}(s)\left( \frac{e^{-s}-1}{s} \left(e^{T(-s)-T(\xi)}-1\right) -T'(-s) e^{T(-s)-T(\xi)}\right){\textrm{d}}s.\end{multline*}

By Corollary 2·8, $T(-s)-T(\xi)$ is bounded in our range of integration. In fact, $T(-s)-T(\xi) \ll (e^{\xi}+|\Im s|)/m$ . Hence, the third case of Lemma 2·7 shows that

\[ \frac{1}{2\pi i} \frac{e^{us}}{u}\hat{\rho}(s) \left(e^{T(-s)-T(\xi)}-1\right) \Big|^{s=-\xi+im\pi}_{s=-\xi+iA} \ll \frac{e^{-u\xi}}{n}\]

which is acceptable. As for the integral, we rearrange it using the definition of T:

\begin{align*} {} {}\frac{1}{2\pi i} &\int_{-\xi+iA}^{-\xi+im\pi} \frac{e^{us}}{u} \hat{\rho}(s)\left( \frac{e^{-s}-1}{s} \left(e^{T(-s)-T(\xi)}-1\right)-T'(-s) e^{T(-s)-T(\xi)}\right){\textrm{d}}s \\[5pt] {} {}&= {}\frac{1}{2\pi i} \int_{-\xi+iA}^{-\xi+im\pi} \frac{e^{us}}{u} \frac{e^{-s}-1}{s} \hat{\rho}(s)\left( e^{T(-s)-T(\xi)}\frac{\frac{-s}{m}}{1-e^{s/m}}-1\right){\textrm{d}}s. {}\end{align*}

By Corollary 2·8 and the Taylor expansion $z/(1-e^{-z})=1+z/2+O(z^2)$ ,

\[ e^{T(-s)-T(\xi)}\frac{\frac{-s}{m}}{1-e^{s/m}}-1 \ll \frac{|s|^2}{m^2} + \frac{e^{\xi}}{m}\]

in our range of integration. By the triangle inequality, the last integral is

\begin{multline*} \ll \frac{\log (u+1)}{e^{u\xi}} \int_{-\xi+iA}^{-\xi+im\pi} \frac{|\hat{\rho}(s)|}{|s|}\left( \frac{|s|^2}{m^2}+\frac{e^{\xi}}{m}\right) |{\textrm{d}}s|\\[5pt] \ll \frac{\log (u+1)}{e^{u\xi}} \int_{-\xi+iA}^{-\xi+im\pi} \frac{1}{|s|^2}\left( \frac{|s|^2}{m^2}+\frac{e^{\xi}}{m}\right) |{\textrm{d}}s| \ll \frac{\log (u+1),}{m e^{u\xi}}\end{multline*}

where in the second inequality we used the third case of Lemma 2·7.

Lemma 3·6. Suppose $0 \le B \le A$ . Then, for every $k \ge 0$ ,

\[\frac{1}{2\pi i} \int_{-\xi+iB}^{-\xi+iA} \left| (s+\xi)^k e^{us} \hat{\rho}(s)\right|| {\textrm{d}}s| \ll_k \rho(u) \big( u^{-\frac{k}{2}} e^{-cuB^2} + A^{k+1}\sqrt{u}\exp\big(-\frac{u}{\xi^2+\pi^2}\big)\big).\]

The same is true if we integrate from $-\xi-iA$ to $-\xi-iB$ . In particular, for $B=0$ and $A\le \exp(c_k u/(\log (u+1))^2)$ ,

\[\frac{1}{2\pi i} \int_{-\xi-iA}^{-\xi+iA} \left| (s+\xi)^k e^{us} \hat{\rho}(s)\right|\left| {\textrm{d}}s\right| \ll_k \rho(u)u^{-\frac{k}{2}}.\]

Proof. Using the first two parts of Lemma 2·7, the triangle inequality shows that the integral is

\begin{multline*} {} {}\ll e^{-u\xi +I(\xi)} \int_{B}^{A} |t|^k \left(\exp(\!-cut^2)+\exp\big(-\frac{u}{\xi^2+\pi^2}\big)\right) {\textrm{d}}t\\[5pt] \ll_k e^{-u\xi +I(\xi)} \bigg( u^{-\frac{k+1}{2}} \exp(\!-cuB^2)+ A^{k+1}\exp\big(-\frac{u}{\xi^2+\pi^2}\big)\bigg), {}\end{multline*}

and we now use (2·4).

Lemma 3·7. Fix $k\ge 0$ . Let $\mu_m\;:\!=\;m!/((m/2)!2^{m/2})$ for even m. Suppose $\sqrt{n\log n}=O(m)$ and $C_k\sqrt{\log(u+1)/u}\le A \le \exp(c_ku/(\log (u+1))^2)$ . Then, for even k,

\[\frac{1}{2\pi i} \int_{-\xi-iA}^{-\xi+iA} (s+\xi)^ke^{us} \hat{\rho}(s){\textrm{d}}s = (1+O_k(u^{-1}))\rho(u) \mu_k (\!-\!1)^{\frac{k}{2}} \xi'^{\frac{k}{2}}\]

and for odd k,

\[\frac{1}{2\pi i} \int_{-\xi-iA}^{-\xi+iA} (s+\xi)^ke^{us} \hat{\rho}(s){\textrm{d}}s = (1+O_k(u^{-1}))\rho(u) \mu_{k+3} (\!-\!1)^{\frac{k+1}{2}} \xi'^{\frac{k+3}{2}} \frac{I^{(3)}(\xi)}{6}.\]

Proof. The contribution of $A \ge |\Im s| \ge C_k\sqrt{\log (u+1)/u}$ is acceptable by Lemma 3·6. We may now assume that $A=C_k\sqrt{\log (u+1)/u}$ . Recall $\hat{\rho}(s)=\exp(\gamma+I(-s))$ . We may Taylor-expand $I(\xi+it)$ using Corollary 2·5, obtaining that

\begin{align*} {} {}\frac{1}{2\pi i} &\int_{-\xi-iA}^{-\xi+iA} (s+\xi)^k e^{us} \hat{\rho}(s){\textrm{d}}s \\[5pt] {} {}&= i^{k} \frac{e^{\gamma-u\xi+I(\xi)}}{2\pi}\int_{-A}^{A} t^k \exp\!\left( -\frac{t^2}{2} I''(\xi) +i\frac{t^3}{6} I^{(3)}(\xi)+\frac{t^4}{24}I^{(4)}(\xi)+O_k\left(u|t|^5\right) \right){\textrm{d}}t\\[5pt] {} {}&= i^{k}\frac{e^{\gamma-u\xi+I(\xi)}}{2\pi} \int_{-A}^{A} t^k \exp\big(-\frac{t^2}{2} I''(\xi)\big) \big(1+i \frac{t^3}{6} I^{(3)}(\xi) + \frac{t^4}{24}I^{(4)}(\xi) - \frac{t^6}{72}I^{(3)}(\xi)^2\big)\\[5pt] {} {}&\qquad \qquad \times \left(1+O_k\left( u|t|^5 + u^3 |t|^9\right)\right) {\textrm{d}}t. {}\end{align*}

Substituting $t^2 I''(\xi)=v^2$ , recalling $I''(\xi(u))=1/\xi'(u)$ and using (2·4), the last expression can be written as

\begin{multline*}= \rho(u)(1+O_k(u^{-1})) i^k \xi'^{k/2} \frac{1}{\sqrt{2\pi}}\int_{-A/\sqrt{\xi'}}^{A/\sqrt{\xi'}} v^k \exp(\!-v^2/2)\\[5pt] \times \bigg(1+i \frac{v^3 \xi'^{3/2}}{6} I^{(3)}(\xi) + \frac{v^4 \xi'^2 }{24}I^{(4)}(\xi) - \frac{v^6 \xi'^3}{72}I^{(3)}(\xi)^2 \bigg) \bigg( 1+ O_k\bigg( \frac{|v|^5 +|v|^9}{u^{3/2}} \bigg)\bigg) {\textrm{d}}v.\end{multline*}

To conclude we apply estimates of gaussian integrals: $\int_{-R}^{R} v^{2k+1}\exp(\!-v^2/2){\textrm{d}}v$ is 0, $\int_{\mathbb{R}}|v|^k \exp(\!-v^2/2){\textrm{d}}v \ll_k 1$ and $\int_{-R}^{R}v^{2k}\exp(\!-v^2/2){\textrm{d}}v =\sqrt{2\pi}\mu_{2k} + O_k(\exp(\!-R^2/4))$ .

3·3. Proof of the second part of Proposition 3·1

Here we prove (3·2) using the material in Section 3·2. If $1+u\xi \gt m \pi$ the result is already included in the first part of Proposition 3·1, so we assume from now on that $1+u\xi \le m \pi$ . From Lemma 3·3 we have

\[ p_{n,m} = \frac{\exp\big(H_m -\gamma\big)}{m} e^{T(\xi)} {} \frac{1}{2\pi i} \int_{-\xi-im\pi}^{-\xi+im\pi} e^{us} \hat{\rho}(s) e^{T(-s)-T(\xi)}{\textrm{d}}s.\]

We have $\exp(H_m-\gamma)/m =1+1/(2m)+O(m^{-2})$ by (2·10), and

\[ e^{T(\xi)} = e^{\frac{u\xi}{2m}} \left( 1- \frac{\xi}{2m}+ O\left(\frac{u\log^2(u+1)}{m^2}\right)\right)\]

by (2·6). Hence,

\[ p_{n,m} = e^{\frac{u\xi}{2m}} \left(1 + \frac{1-\xi}{2m} + O\left(\frac{u\log^2(u+1)}{m^2}\right)\right) \frac{1}{2\pi i} \int_{-\xi-im\pi}^{-\xi+im\pi} e^{us} \hat{\rho}(s) e^{T(-s)-T(\xi)}{\textrm{d}}s.\]

We separate the integral into 3 parts, $S_1+S_2+S_3$ , where

\begin{align*} {}S_1 &= \frac{1}{2\pi i} \int_{-\xi-im\pi}^{-\xi+im\pi} e^{us} \hat{\rho}(s) {\textrm{d}}s,\\[5pt] {}S_2 &= \frac{1}{2\pi i} \int_{-\xi-i(1+u\xi)}^{-\xi+i(1+u\xi)} \left(e^{T(-s)-T(\xi)}-1\right)e^{us} \hat{\rho}(s) {\textrm{d}}s,\\[5pt] {}S_3 &= \frac{1}{2\pi i} \left(\int_{-\xi-im\pi}^{-\xi-i(1+u\xi)}+\int_{-\xi+i(1+u\xi)}^{-\xi+im\pi}\right) \left(e^{T(-s)-T(\xi)} - 1 \right)e^{us} \hat{\rho}(s) {\textrm{d}}s.\end{align*}

The integral $S_1$ was estimated in Lemma 3·4 and it gives the main term $\rho(u)$ as well an absolute error of size $\ll e^{-(u-1)\xi}/n \ll e^{-u\xi} \log(u+1)/m$ . The integral $S_3$ was estimated in Lemma 3·5 and it contributes $\ll e^{-u\xi}\log(u+1)/m$ . We now study $S_2$ . We use Lemma 2·9 to Taylor-expand $e^{T(-s)-T(\xi)}-1=e^{T(\xi-it)-T(\xi)}-1$ at 0:

(3·9) \begin{multline}e^{T(-s)-T(\xi)}-1 = -T'(\xi)(s+\xi) + \frac{(s+\xi)^2}{2}\left( T''(\xi)+T'(\xi)^2\right)\\[5pt] -(T^{(3)}(\xi)+3T'(\xi)T''(\xi)+(T'(\xi))^3) \frac{(s+\xi)^3}{6}+O\left( \frac{u\log(u+1)}{m} |s+\xi|^4\right)\end{multline}

for $|\Im s| \le 1+u\xi$ . Applying Lemma 3·6 with $k=4$ , $B=0$ and Lemma 3·7 with $k=1,2,3$ and collecting the terms gives

(3·10) \begin{multline} {}S_2 = \rho(u)\bigg(\frac{T'(\xi)(1+O(\log (u+1)^{-1}))-T''(\xi)(1+O(\log(u+1)^{-1}))}{2u} \\[5pt] + O\big( \frac{\log (u+1)}{n}+\frac{u\log^2(u+1)}{m^2}\big)\bigg). \end{multline}

All in all, since $T'(\xi)$ and $T''(\xi)$ are both $({u\log(u+1)}/{2m}) (1+o_{u\to\infty}(1))$ in our range by Lemma 2·9,

\[ p_{n,m} =\exp\big(\frac{u\xi}{2m}\big) \rho(u)\bigg( 1-\frac{\log(u+1)}{2m}(1+o_{u\to\infty}(1)) + O\bigg(\frac{u\log^2 (u+1)}{m^2}\bigg)\bigg). \]

We are done, as we established, in stronger form, the second part of Proposition 3·1.

3·4. Proof of Lemma 3·2

We first show (3·3). Let $r(t)=-\rho'(t)/\rho(t)\ge 0$ . We have

\[ {}\rho(u)= \rho\bigg( u-\frac{u}{2m}\bigg) \exp\bigg( -\int_{u-\frac{u}{2m}}^{u} r(t){\textrm{d}}t\bigg).\]

If u is bounded then the bound $r(t)\ll 1$ for $t \ll 1$ finishes the proof. If $u \ge C$ we may differentiate r (it is differentiable for $t \ge 2$ ) and obtain

\[ \rho(u) = \rho\bigg( u-\frac{u}{2m}\bigg) \exp\bigg( -\frac{ur(u)}{2m} + \int_{u-\frac{u}{2m}}^{u} \bigg(u-\frac{u}{2m}-t\bigg)r'(t) {\textrm{d}}t\bigg)\]

by integration by parts. To conclude, we use (3·8) to find

\[ {}-\frac{ur(u)}{2m} + \int_{u-\frac{u}{2m}}^{u} \bigg(u-\frac{u}{2m}-t\bigg)r'(t) {\textrm{d}}t = -\frac{u\xi(u)}{2m} + O\bigg( \frac{u}{m^2} + \frac{1}{m}\bigg).\]

It remains to show (3·4). Set $t_1 = n/(m+1/2)$ and $t_2=u-u/(2m)$ . It suffices to bound $\log (\rho(t_2)/\rho(t_1))$ . Observe $t_1\ge t_2$ . We have

\[ \log \rho(t_2)-\log \rho(t_1) = \int_{t_2}^{t_1} r(t)\textrm{d}t \le (t_1-t_2) \max_{t\in [t_2,t_1]} r(t) \ll \frac{n}{m^3} \log(u+1) = \frac{u\log(u+1)}{m^2} \]

since $r(t)\ll \log(t+1)$ by (3·8) and Lemma 2·2. This completes the proof.

4·1. Proof of Theorem 1·2

Let

\[ S\;:\!=\;\sum_{i=1}^{m-1} d_{m,i} n^{1-\frac{i}{m}}.\]

In view of Lemma 4·1 we have

(4·1) \begin{equation} {}p_{n,m} = \left(\frac{e}{n}\right)^u \exp\!\left( S + O(\log n)\right),\end{equation}

when $m=O(\log n)$ , and it remains to estimate S. We have

\[\frac{1}{m-i} = \frac{1}{m} \left(1+O\left( \frac{i}{m-i}\right)\right), \qquad \frac{1}{\Gamma\left(1+\frac{i}{m}\right)} = 1+O\left( \frac{i}{m}\right)\]

uniformly for $1 \le i \le m-1$ . The estimate $\Gamma(x+y)/\Gamma(x) \in [x(x+y)^{y-1},x^y]$ for $y \in (0,1)$ [ Reference Wendel14 , equation (7)] implies

\[\frac{\Gamma\left(i+\frac{i}{m}\right)}{i!} =\frac{1}{i} \frac{\Gamma\left( i +\frac{i}{m}\right)}{\Gamma(i)}= \frac{i^{\frac{i}{m}}}{i} \left(1+O\left( \frac{1}{m}\right)\right).\]

Hence,

\[d_{m,i}=\frac{i^{\frac{i}{m}}}{mi}\left(1+O\left( \frac{i}{m-i}\right)\right).\]

Since $i^{i/m}=1+O({i\log i}/{m})$ if $i \lt m/\log (m+1)$ , we may write S as $S= u(S_1+O(S_2+S_3))$ , where

\[S_1 =\sum_{i=1}^{m-1} \frac{n^{-\frac{i}{m}}}{i},\,\quad S_2=\sum_{1 \le i \lt \frac{m}{\log (m+1)}} \frac{n^{-\frac{i}{m}}}{i} \frac{i\log(i+1)}{m},\, \quad S_3=\sum_{\frac{m}{\log(m+1)} \le i \le m-1} (i/n)^{\frac{i}{m}}\frac{m}{i}.\]

We have

\[ S_1 = -\log(1-n^{-\frac{1}{m}}) + O\left( \frac{1}{nm}\frac{1}{1-n^{-\frac{1}{m}}}\right)\]

by bounding the tail of the Taylor series of $-\log(1-n^{-1/m})$ by a geometric series with ratio $n^{-1/m}$ . Similarly,

\[ S_2 \ll \frac{1}{m} \frac{1}{n^{\frac{1}{m}}-1}\ll \frac{n^{-\frac{1}{m}}}{m}\]

because the contribution of $i \in [2^k,2^{k+1})$ to $S_2$ is $\ll (k+1)n^{-2^k/m}/(m(1-n^{-1/m}))$ and $\sum_{k \ge 0}(k+1)n^{-2^k/m}\ll n^{-1/m}$ when $m\ll \log n$ . As for $S_3$ ,

\[S_3\ll m \sum_{\frac{m}{\log(m+1)}\le i \le m-1}\frac{1}{i} \max_{\frac{m}{\log(m+1)}\le i \le m-1} (i/n)^{i/m} \ll m\log \log (m+2) n^{-\frac{1}{\log(m+1)}} \]

since $i\mapsto (i/n)^{i/m}$ decreases for $i \le n/e$ . Both $S_3$ and the error term in $S_1$ are dominated by our bound for $S_2$ . It follows that

\[ S = u \big( -\log\big(1-n^{-\frac{1}{m}}\big) + O\big( \frac{n^{-\frac{1}{m}}}{m}\big)\big).\]

The error $O(\log n)$ in (4·1) is absorbed in the error term in S when $m \gt 1$ , and we are done.

4·2. Proof of Corollary 1·3

We first consider $n \ge m \gt n/2$ . For $m=n$ , $p_{n,n}=\rho(1)=1$ , so we may assume $m\le n-1$ . We have exact formulas: $p_{n,m}=1-\sum_{i=m+1}^{n}1/i$ and $\rho(u)=1-\log u$ . Hence

\[p_{n,m}-\rho(u)= \sum_{i=m+1}^{n}( \log i - \log(i-1) - \frac{1}{i}) \asymp \sum_{i=m+1}^{n} \frac{1}{i^2} \asymp \int_{m}^{n} \frac{{\textrm{d}}t}{t^2}.\]

We are done since, in our range,

\[\int_{m}^{n} \frac{{\textrm{d}}t}{t^2} =\frac{1}{m}-\frac{1}{n}\asymp\frac{n-m}{m^2} =\frac{u-1}{m}\asymp \frac{\log u}{m}.\]

We now suppose $n/2 \ge m \ge C \log n$ . We shall need the lower bound [ Reference Gorodetsky8 , theorem A·1]

(4·2) \begin{equation}p_{n,m} \ge \rho(u)\bigg(1+\frac{cu\log u}{m}\bigg)\end{equation}

which holds uniformly for $n/2 \ge m \ge 1$ . We may assume $n\ge C$ , since for bounded n we just want to show $B_2\rho(u) \ge p_{n,m}\ge B_1\rho(u)$ for constants $B_2\ge B_1 \gt 1$ which follows from (4·2) and (1·4). If u is sufficiently large then it follows from Proposition 3·1 that, using the same definition for A as in the statement of the corollary,

\[ {}A \asymp \frac{u\xi(u)}{m}\asymp \frac{u\log u}{m} \asymp u \log \left( 1+ \frac{\log u}{m}\right) \]

as needed. If $u=O(1)$ then the same argument establishes $A \le Cu\log ( 1+ (\log u)/m)$ and a matching lower bound in this range follows from (4·2).

Finally we suppose $m=O(\log n)$ . From (1·1) we have

(4·3) \begin{equation} {}\left(\frac{e}{n}\right)^u \frac{1}{\rho(u)} = \exp\!\left( u \left( \log\left( \frac{\log u}{m}\right) + O\left( \frac{\log \log u}{\log u}\right) \right)\right). {}\end{equation}

From Theorem 1·2 and (4·3),

\[ u^{-1}A = \log\left( \frac{\log n}{m}\right) - \log\left(1-n^{-\frac{1}{m}}\right) +O\left(\frac{\log \log n}{\log n}\right).\]

If n is sufficiently large and $1\le m\le C\log n$ ,

\[\log\bigg( \frac{\log n}{m}\bigg) - \log(1-n^{-\frac{1}{m}}) = \log \bigg( \frac{(\log n)/m}{1-e^{-(\log n)/m}}\bigg) \asymp \log\bigg(1+\frac{\log n}{m}\bigg)\]

since $t/(1-e^{-t}) \ge 1+c$ when $t=(\log n)/m\ge 1/C$ . This finishes the proof.

Remark 3. When $m=O(\log n)$ the proof above shows more: setting $t\;:\!=\;(\log n)/m$ ,

\[ A= u \left( \log \frac{t}{1-e^{-t}} +O\left ( \frac{\log\log (n+2)}{\log (n+2)}\right)\right)\]

holds, where A is as in the statement of Corollary 1·3.

5·1. Proof of Theorem 1.5

We suppose that $n \ge m \ge \sqrt{10n\log n}$ . This guarantees $2^{m/3} \gt 1+ m(\log 2)/3 \ge 1 + 2u\log u \ge 1+u\xi = e^{\xi}$ , where we used Lemma 2·3 in the last inequality. Hence

(5·3) \begin{equation}e^{\xi/m} \lt \sqrt[3]{2}.\end{equation}

An application of Cauchy’s integral formula allows us to express $p_{n,m}$ as

(5·4) \begin{equation}p_{n,m} = \frac{\exp(H_m-\gamma)}{m} \frac{1}{2\pi i}\int_{-\xi-im\pi}^{-\xi+im\pi} \hat{\rho}(s)e^{us} e^{T(-s)}{\textrm{d}}s,\end{equation}

see [ Reference Manstavičius and Petuchovas13 , section 4]. In the same way,

(5·5) \begin{equation}p_{n,m,q} = \frac{\exp(H_m-\gamma)}{m} \frac{1}{2\pi i}\int_{-\xi-im\pi}^{-\xi+im\pi} \hat{\rho}(s)e^{us} e^{T(-s)} G_q(e^{-s/m}) {\textrm{d}}s,\end{equation}

see e.g. [ Reference Gorodetsky8 , section 4]. Since $G_q$ has radius of convergence equal to q, we must ensure that $e^{\xi/m} \lt q$ in order for the last integral to be valid, and this holds by (5·3). By taking a linear combination of (5·4) and (5·5), and using (2·10) we have

\[ p_{n,m,q}-G_q(e^{\xi/m})p_{n,m} = G_q(e^{\xi/m}) \frac{1+O(m^{-1})}{2\pi i}X \]

for

\[ X\;:\!=\; \int_{-\xi-im\pi}^{-\xi+im\pi} \hat{\rho}(s)e^{us} e^{T(-s)} \bigg(\frac{G_q(e^{-s/m})}{G_q(e^{\xi/m})}-1\bigg) {\textrm{d}}s.\]

Since $p_{n,m} \gg \rho(u)$ for $m \ge \sqrt{10n\log n}$ by (1·4), it follows that

\[ \frac{p_{n,m,q}}{p_{n,m}}=G_q(e^{\xi/m})\bigg(1+O\bigg(\frac{|X|}{\rho(u)}\bigg)\bigg).\]

We must show $X \ll \rho(u) \log(u+1)/(mq^{\lceil(m+1)/2\rceil})$ . To bound X, we consider separately the contribution of $|\Re s| \le 1+u\xi$ and $m\pi \ge |\Re s| \ge 1+u\xi$ . We start with $|\Re s | \le 1+u\xi$ . Let

\[ H(s)\;:\!=\; \frac{G_q(e^{-s/m})}{G_q(e^{\xi/m})}-1, \qquad H_T(s)\;:\!=\; H(s) e^{T(-s)}\]

so that the integrand in X is $\hat{\rho}(s) e^{us} H_T(s)$ . We Taylor-expand $H_T(s)$ at $s=-\xi$ , where it attains 0: $H_T(s) = (s+\xi)b_1 + O(|s+\xi|^2 b_2)$ for

\[b_1= H_T'(-\xi) \qquad \text{and} \qquad b_2=\max_{|t| \le 1+u\xi} |H_T''(-\xi+it)|.\]

Applying Lemma 3·6 with $k=2$ , $B=0$ and Lemma 3·7 with $k=1$ , it follows that

\[ \int_{-\xi-i(1+u\xi)}^{-\xi+i(1+u\xi)} \hat{\rho}(s)e^{us} H_T(s) {\textrm{d}}s \ll \left(|b_1|+b_2\right)\frac{\rho(u)}{u}.\]

We now consider the contribution of $\pi m \ge |\Re s| \ge 1+u\xi$ . We focus on $m\pi \ge \Re s \ge 1+u\xi$ , and negative $\Re s$ is handled the same say. We have $\hat{\rho}(s) = 1/s + O(u\log (u+1) /|s|^2)$ in this range by the third part of Lemma 2·7. The contribution of $O(u\log (u+1)/|s|^2)$ is

\[ \ll \int_{-\xi+i(1+u\xi)}^{-\xi+i\pi m}|e^{us} \frac{u\log(u+1)}{s^2} H_T(s)| |{\textrm{d}}s|\ll b_3e^{-u\xi}, \]

where

\[ b_3 = \max_{|t| \le \pi m} |H_T(-\xi+it)|.\]

We study the last piece of the integral,

\[ \int_{-\xi+i(1+u\xi)}^{-\xi+i\pi m} \frac{e^{us}}{s} e^{T(-s)} H(s) {\textrm{d}}s .\]

We write $e^{T(-s)}$ as $1+(e^{T(-s)}-1)$ . When $s=-\xi+it$ for $1+u\xi \le t \le \pi m$ we have $e^{T(-s)}-1=O(|s|/m + |s|^2/m^2)$ by Corollary 2·8. Applying the triangle inequality we get

\[ \int_{-\xi+i(1+u\xi)}^{-\xi+i\pi m} \frac{e^{us}}{s} (e^{T(-s)}-1) H(s){\textrm{d}}s\ll b_4 \int_{-\xi+i(1+u\xi)}^{-\xi+i\pi m}\ \bigg| \frac{e^{us}}{s}\bigg| \bigg(\frac{|s|}{m}+\frac{|s|^2}{m^2}\bigg) |{\textrm{d}}s| \ll b_4 e^{-u\xi}, \]

where

\[ b_4 = \max_{|t| \le \pi m} |H(-\xi+it)|.\]

We estimate

(5·6) \begin{equation} \int_{-\xi+i(1+u\xi)}^{-\xi+i\pi m} \frac{e^{us}}{s} H(s) {\textrm{d}}s .\end{equation}

Using the notation in (5·1) and the bound in (5·2), it follows that

(5·7) \begin{equation} {}\begin{split}H(s) &= \log G_q(e^{-s/m}) - \log G_q(e^{\xi/m}) +O\left( \frac{u^2 \log^2(u+1)}{m^2 q^{2\lceil \frac{m+1}{2}\rceil}}\right)\\[5pt] &=\sum_{j=m+1}^{2m} \frac{a_j}{j}(e^{-js/m}-e^{j\xi/m}) + O\left(\frac{u^2\log^2(u+1)}{mq^{m+1}}\right).\end{split}\end{equation}

Hence the integral in (5·6) equals

\[ =\sum_{j=m+1}^{2m} \frac{a_j}{j} \int_{-\xi+i(1+u\xi)}^{-\xi+i\pi m} \frac{e^{us}}{s} (e^{-js/m}-e^{j\xi/m}) {\textrm{d}}s+O\left( \frac{u^2 \log^2(u+1) e^{-u\xi}\log(m+1)}{mq^{m+1}} \right).\]

If $u \lt 3$ then $G_q(e^{\xi/m})-1\ll 1/(mq^{\lceil (m+1)/2\rceil})$ , implying Theorem 1·5 is already in (1·6). From now on we assume $u \ge 3$ . We have the estimate

\[ \int_{-\xi+i(1+u\xi)}^{-\xi+i\pi m} \frac{e^{(u-j/m)s}}{s}{\textrm{d}}s \ll \frac{e^{-(u-j/m)\xi}}{(1+u\xi)|u-j/m|},\]

which is valid for all $2m\ge j \ge 0$ and $u \ge 3$ , and is established by integration by parts. It follows that the integral (5·6) contributes

\[\ll \frac{u^2 \log^2(u+1) e^{-u\xi}\log (m+1)}{mq^{m+1}} + \sum_{j=m+1}^{2m} \frac{a_j}{j} \frac{e^{-u\xi}}{1+u\xi} \frac{e^{j\xi/m}}{|u-j/m|} \ll \frac{e^{-u\xi}}{n q^{\lceil \frac{m+1}{2}\rceil}}.\]

Collecting the estimates,

(5·8) \begin{equation}X \ll \frac{\rho(u)}{u}(|b_1|+b_2)+(b_3+b_4) e^{-u\xi}+ \frac{e^{-u\xi}}{n q^{\lceil \frac{m+1}{2}\rceil}}.\end{equation}

Recall we want to show $X \ll \rho(u) \log(u+1)/(mq^{\lceil(m+1)/2\rceil})$ . We have $e^{-u\xi}\ll_k \rho(u) u^{-k}$ for any k by (2·4), showing that the last term in (5·8) is acceptable and that it suffices to show $b_i \ll u\log(u+1)/(mq^{\lceil(m+1)/2\rceil})$ for $i=1,2,3,4$ .

Since $T(-\xi+it)$ is bounded when $|t|\le \pi m$ by Corollary 2·8, it follows that $b_3 \ll b_4$ . By (5·7) and the triangle inequality, $b_4 \ll u\log(u+1)/(mq^{\lceil(m+1)/2\rceil})$ . To bound $b_1$ and $b_2$ we use the fact that T and its derivatives are bounded by Lemma 2·9 in order to reduce the problem to showing

\[ H^{(i)}(\!-\xi+it) \ll u \log(u+1)/(mq^{\lceil(m+1)/2\rceil})\]

holds for $i=0,1,2$ . For $i=0$ this is in (5·7), for $i=1$ we have

\[ H'(s) = -\frac{e^{-s/m}}{m} \frac{G_q'(e^{-s/m})}{G_q(e^{\xi/m})} \ll \frac{|G_q'(e^{-s/m})|}{m} \ll \frac{\sum_{i=m+1}^{\infty} a_i e^{(i-1)\xi/m}}{m} \ll \frac{u\log(u+1)}{mq^{\lceil(m+1)/2\rceil}}\]

and a similar computation holds for $i=2$ . This completes the proof of Theorem 1·5.

5·2. Proof of Corollary 1·6

Fix $\varepsilon \gt 0$ . We assume $n \ge m \ge (2+\varepsilon)\log_q n$ and we want to establish (1·10) in this range. We already know (1·10) holds in $n/(\log n \log^3\log(n+1)) \ge m \ge (2+\varepsilon)\log_q n$ , so we may suppose $n \ge m \gt n/(\log n \log^3\log(n+1))$ .

If $n=O(1)$ then (1·10) becomes $p_{n,m,q}=p_{n,m}G_q(x) + O_{n,\varepsilon}(G_q(x)/q^{\lceil (m+1)/2\rceil})$ . To establish this, recall $p_{n,m,q}=p_{n,m} + O(1/q^{\lceil (m+1)/2\rceil})$ by [ Reference Gorodetsky8 , proposition 1·5] and observe that $G_q(x) = 1+O_{n,\varepsilon}(1/q^{\lceil (m+1)/2\rceil})$ by (5·1) and (5·2).

From now on we may suppose n is sufficiently large. In particular, $n \gt \sqrt{10 n \log n}$ holds, and $n \ge m \gt n/(\log n \log^3\log(n+1))$ implies $n \ge m \ge \sqrt{10 n \log n}$ . We apply Theorem 1·5, and see that it suffices to show that

(5·9) \begin{equation} {}\frac{G_q(x)}{G_q(e^{\xi(u)/m})} = 1 +O\left( \frac{u\log^2 (u+1)}{m^2 q^{\lceil \frac{m+1}{2}\rceil}}\right)\end{equation}

holds for $n\ge m \ge \sqrt{10 n \log n}$ . First we verify that $x \lt \sqrt[3]{2}$ . Since $x \le n^{1/m}$ , it suffices to show that $2^m \gt n^3$ , which follows from $n \ge m \ge \sqrt{10n\log n}$ . Now that we know $G_q$ converges absolutely at x we proceed. The relation $x^m = e^{\xi}(1+O(\log(u+1)/m))$ from (3·6) implies

\[ \log G_q(x) - \log G_q(e^{\xi/m}) = \sum_{i=m+1}^{\infty} \frac{a_i (x^i - e^{\xi i /m})}{i} \ll \frac{u\log^2 (u+1)}{m^2 q^{\lceil \frac{m+1}{2}\rceil}}\]

using (5·2), and (5·9) follows by exponentiating.

5·3. Auxiliary computation

Recall $F_q$ defined in (1·7). We define $x_q\;:\!=\;x_{n,m,q} \lt q$ as the unique positive solution to $z F_q'(z)/F_q(z) = n$ . The next lemma gives precise results on $x-x_q$ in certain ranges.

Lemma 5·1. If $n \ge m \ge 1$ then $x_q \le x$ . Fix q and $\varepsilon \gt 0$ . For $n \ge C_{q,\varepsilon}$ and $m \in [(3/4)\log_q n, (2-\varepsilon)\log_q n]$ ,

\[ C_{q,\varepsilon}\frac{\min\{n,q^m\}}{q^m m} \big(1-\frac{x_q}{q}\big)^{-1} \ge x-x_q \ge c_q \frac{\min\{n,q^m\}^{2}}{q^m nm}.\]

Proof. We shall use the form of $G_q$ given in (5·1). Let $f(z)\;:\!=\;z(\log F(z))' = \sum_{i=1}^{m} z^i$ and $g_q(z) \;:\!=\; z(\log G_q(z))'= \sum_{i=m+1}^{\infty}a_i z^i$ . By definition,

(5·10) \begin{equation} {}n = f(x)= f(x_q)+g_q(x_q).\end{equation}

Since f and $g_q$ have nonnegative coefficients, the first part of the lemma follows at once. We now assume $(2-\varepsilon)\log_q n \ge m \ge (3/4)\log_q n $ . We use (5·10) to determine the size of $x-x_q$ . By the mean value theorem,

\[ 0 \le x-x_q = f^{-1}(n) - f^{-1}(n-g_q(x_q)) = \frac{g_q(x_q)}{f'(f^{-1}(n-t))} \]

for some $t \in [0,g_q(x_q)]$ , where $f^{-1}$ is the inverse of f. Since $f^{-1}$ and f’ are monotone increasing we have $f'(f^{-1}(n-t)) \in [f'(x_q), f'(x)]$ and so

\[ {}x-x_q \in \left[ \frac{g_q(x_q)}{f'(x)}, \frac{g_q(x_q)}{f'(x_q)} \right]. \]

We have the trivial bounds $1 \le x\le n^{1/m} \ll_q 1$ and so $f'(x) = \lambda/x \asymp_q nm$ by Lemma 2·1. We also have

\[ g_q(x_q) \ge a_{2m}x_q^{2m} \gg q^{-m} x_q^{2m}\]

since $a_i \ge 0$ and $a_{2m} \gg q^{-m}$ by [ Reference Gorodetsky8 , lemma 2·1]. It follows that

\[ x-x_q \gg_q \frac{x_q^{2m}}{q^m nm}.\]

[ Reference Manstavičius12 , lemma 2 and theorem 2] estimate $x_q$ and yield that, when $2\log_q n \ge m \ge (3/4)\log_q n$ ,

(5·11) \begin{equation} {}x_q^m \asymp_q \min\{n,q^m\},\end{equation}

implying the desired lower bound on $x-x_q$ . To prove the upper bound, first observe that $m \le (2-\varepsilon)\log_q n$ implies, via (5·11), that $x_q \ge q^{1/2}(1+c_{q,\varepsilon})$ if n is sufficiently large, and so

\[ g_q(x_q) \ll \sum_{i=m+1}^{2m} \frac{x_q^i}{q^{i/2}}+q^m \sum_{i \ge 2m}\frac{x_q^i}{q^i} \ll_{q,\varepsilon} \frac{ x_q^{2m}}{q^m}\big(1-\frac{x_q}{q}\big)^{-1}\]

using the bounds in (5·2). To conclude we need to lower bound $f'(x_q)$ . We have $f'(x_q) = x_q^{-1} \sum_{i=1}^{m} i x_q^i \gg_{q} m x_q^m$ .

5·4. Proof of Theorem 1·7

Recall the functions F, $F_q$ and $G_q$ are defined in (1·8), (1·7) and (1·9). If $m \to \infty$ and $u\ge (\log \log m)^3 \log m$ , Manstavičius proved in [ Reference Manstavichyus11,Reference Manstavičius12 ] that

(5·12) \begin{equation} {}p_{n,m,q} = \frac{F_q(x_{q})}{x_q^n\sqrt{2\pi \lambda_q}}\left( 1+O_q\left( \frac{m}{n}+\frac{m}{q^m}\right)\right),\end{equation}

where

\[ \lambda_q\;:\!=\;\lambda_{n,m,q}= z \left(\frac{zF'_q(z)}{F_q(z)}\right)'|_{z=x_q}.\]

We divide (5·12) by (2·1), and use the fact that $\lambda$ is asymptotic to nm by Lemma 2·1 as long as $u \to \infty$ , and

(5·13) \begin{equation}\lambda_q \sim nm \left(1+ \frac{n}{q^m}\left(1-\frac{1}{q}\right)\right)\end{equation}

by [ Reference Manstavičius12 , theorem 2] as long as $u \to \infty$ and $m \to \infty$ , to obtain

(5·14) \begin{equation} \frac{p_{n,m,q}}{p_{n,m}G_q(x)} \sim \left(1+ \frac{n}{q^m}\left(1-\frac{1}{q}\right)\right)^{-1/2} \frac{x^nF(x_{q})G_q(x_{q})}{x_q^nF(x)G_q(x)}\end{equation}

as $n \to \infty$ if $(2-\varepsilon)\log_q n \ge m \gt \log_q n$ . Letting

\[ H_q(z) \;:\!=\;\log F_q(z)-n\log z = \log F(z) + \log G_q(z)-n \log z,\]

we can write (5·14) as

(5·15) \begin{equation} \frac{p_{n,m,q}}{p_{n,m}G_q(x)} \sim \exp\!\left(H_q(x_q)-H_q(x)\right) \left(1+ \frac{n}{q^m}\left(1-\frac{1}{q}\right)\right)^{-1/2}.\end{equation}

By definition, $H'_q(x_q)=0$ . By Taylor-expanding $H_q$ at $x_q$ ,

\[ H_q(x)-H_q(x_{q}) = \frac{H''_q(t)}{2}(x-x_q)^2 \]

for some $t \in [x_q,x]$ (recall $x_q \le x$ by Lemma 5·1). In the range $(2-\varepsilon)\log_q n \ge m \gt \log_q n$ we have $x\asymp_q x_q \asymp_q 1$ by Lemma 2·1 and (5·11). In the notation of (5·1),

\[ t^2 H_q''(t) = \sum_{i=2}^{m}(i-1)t^i + \sum_{i=m+1}^{\infty}(i-1)a_i t^i+ n \gt 0\]

is increasing for $t \gt 0$ since $a_i \ge 0$ . It follows that

\[ x_q^2 H_q''(x_q) \ll_q H_q''(t)\ll_q x^2 H_q''(x).\]

A short computation shows that $x_q^2 H_q''(x_q)=\lambda_q$ holds by the definition of $x_q$ and $\lambda_q$ . By (5·13) it then follows that $x_q^2 H_q''(x_q) \asymp_q nm$ when $(2-\varepsilon)\log_q n \ge m \gt \log_q n$ . In the same range we find that $x^2 H_q''(x) = \lambda +\sum_{i=m+1}^{\infty}(i-1)a_i x^i \ll_{q,\varepsilon} nm$ using Lemma 2·1 and the bounds in (5·2). Hence $H_q''(t) \asymp_{q,\varepsilon} nm$ and

\[ H_q(x)-H_q(x_q) \asymp_{q,\varepsilon} nm (x-x_q)^2 \asymp_{q,\varepsilon} \frac{n^3}{mq^{2m}},\]

where we used Lemma 5·1 in the second estimate. Plugging this estimate in (5·15), and observing that $(1+ ({n}/{q^m})(1-{1}/{q}))^{-1/2} \sim 1$ holds in the range $m\ge (1+\varepsilon)\log_q n$ considered in Theorem 1·7, concludes the proof.

5·5. A variant

We prove a variant of Theorem 1·7 with the main term $G_q(x)$ replaced by $G_q(x_q)$ .

Theorem 5·2. Fix q and $\varepsilon \gt 0$ . For $m \in [(3/4)\log_q n, (2-\varepsilon)\log_q n]$ we have

(5·16) \begin{equation} {} {}\begin{split} {} {} {}\frac{p_{n,m,q}}{p_{n,m}} &\le (1+o(1))G_q(x_q)\exp\!\left( \frac{C_{q,\varepsilon} \min\{n,q^m\}^2 n}{mq^{2m} } \left(1-\frac{x_q}{q}\right)^{-2}\right),\\[5pt] {} {} {}\frac{p_{n,m,q}}{p_{n,m}} &\ge (1+o(1)) G_q(x_q)\exp\!\left( \frac{c_{q,\varepsilon} \min\{n,q^m\}^5}{n^2 mq^{2m}}\right) {} {}\end{split} {}\end{equation}

as $n \to \infty$ . If furthermore $n \ge C_{q,\varepsilon}$ then

\[C_{q,\varepsilon}\frac{\min\{n,q^m\}^2}{mq^m}\big(1-\frac{x_q}{q}\big)^{-1} \ge \log G_q(x_q) \ge c_q \frac{\min\{n,q^m\}^2}{mq^m}.\]

Proof. In very much the same way (5·14) is proved, we also have

(5·17) \begin{equation} \frac{p_{n,m,q}}{p_{n,m}G_q(x_q)} \sim \frac{x^nF(x_{q})G_q(x_{q})}{x_q^nF(x)G_q(x_q)}\left(1+ \frac{n}{q^m}\left(1-\frac{1}{q}\right)\right)^{-1/2}\end{equation}

if $u \ge (\log \log m)^3 \log m$ and $m\to\infty$ . Letting

\[ H(z) \;:\!=\; \log F(z) -n \log z,\]

we can write (5·17) as

\[ \frac{p_{n,m,q}}{p_{n,m}G_q(x_q)} \sim \exp\!\left(H(x_q)-H(x)\right)\left(1+ \frac{n}{q^m}\left(1-\frac{1}{q}\right)\right)^{-1/2}.\]

By definition, $H'(x) = 0$ . By Taylor-expanding H at x,

\[ H(x_q)-H(x) = \frac{H''(t)}{2}(x-x_q)^2\]

for some $t \in [x_q,x]$ . Since $t^2 H''(t) = \sum_{i=2}^{m} (i-1)t^i +n$ is increasing in t, and $x \asymp_q x_q \asymp_q 1$ by Lemma 2·1 and (5·11), it follows that

\[ x_q^2 H''(x_q) \ll_q H''(t) \ll_q x^2 H''(x).\]

We have $x^2 H''(x)=\lambda \asymp nm$ by Lemma 2·1 when $(2-\varepsilon)\log_q n \ge m \ge (3/4)\log_q n$ . Lemma 5·1 bounds $x-x_q$ . This gives the first part of (5·16) (we bound the factor $(1+ ({n}/{q^m})(1-{1}/{q}))^{-1/2}$ by 1). We have $x_q^2 H''(x_q) \ge (m-1)x_q^m \gg_q m \min\{n,q^m\}$ by (5·11), which leads to the second part of (5·16) (we absorb $(1+ ({n}/{q^m})(1-{1}/{q}))^{-1/2}$ in the exponential factor in right-hand side of (5·16)).

It remains to estimate $G_q(x_q)$ . For a lower bound we use $\log G_q(x_q) \ge a_{2m} x_q^{2m}/(2m) \gg x_q^{2m}/(mq^m)$ . Here we used $a_i \ge 0$ and $a_{2m} \asymp q^{-m}$ [ Reference Gorodetsky8 , lemma 2·1]. This is simplified using (5·11). For the upper bound we use $a_{i} \ll \min\{q^{-\lceil i/2\rceil},q^{m-i}\}$ and $x_q \ge q^{1/2}(1+c_{q,\varepsilon})$ to obtain $\log G_q(x_q) \ll_{q,\varepsilon} x_q^{2m}/( mq^m(1-x_q/q))$ . We again simplify $x_q^m$ using (5·11).