2. Proof of Theorem 1·1

Theorem 1·1 follows easily from the next lemma.

Lemma 2·1. Assume the Riemann hypothesis and that Hypothesis $\mathscr{H}_{\delta}$ holds for some ${\delta}\in(0, 1]$ . Let $m_{\gamma }$ denote the multiplicity of the zero ${\rho }=1/2+i{\gamma }$ . Then for all sufficiently large T,

(2·1) \begin{equation}\begin{split} \# \bigg\{ 0<{\gamma }\leq T \, :\, & \frac{ \log(\big| \zeta^{(m_{\gamma })} (\frac12+ i{\gamma }) \big| / (\!\log{T} )^{m_{\gamma }})}{\sqrt{\frac12\log\log T}} \in [a, b] \bigg\} \\=& \frac{ N(T)}{\sqrt{2\pi} } \int_a^b e^{-{x^2}/{2}} \, dx +O\bigg( N(T)\frac{(\!\log\log\log T)^2}{\sqrt{\log\log T}} \bigg) .\end{split}\end{equation}

This follows from the proof of Theorem 1·4 combined with corollary 2·3 of Çiçek [ Reference Çiçek2 ].

Note that

\begin{equation}\notag N_{[a, b]}(T)= \sum_{0 < {\gamma } \leq T} \unicode{x1D7D9}_{[a,b]}\Bigg(\frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log {\gamma })^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log {\gamma }}}\Bigg).\end{equation}

Thus, to prove (1·2), we need to show that we may replace $\log {\gamma }$ and $\log\log {\gamma }$ here by $\log T$ and $\log\log T$ , respectively, at the cost of a reasonable error term. To see this, note that by (1·1), the terms in the sum with $0 <{\gamma }\leq T/\log T$ contribute at most O(T), hence

(2·2) \begin{equation}N_{[a, b]}(T) = \sum_{T/\log T < {\gamma } \leq T} \unicode{x1D7D9}_{[a,b]}\Bigg(\frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log {\gamma })^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log {\gamma }}}\Bigg)+O(T).\end{equation}

For $T/\log T < {\gamma } \leq T$ we easily find that

\begin{equation}\notag \begin{split} \frac{\log\! \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log {\gamma })^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log {\gamma }} }= &\frac{ \log\! \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log T)^{m_{\gamma }} \big)\! +O(m_{\gamma } \log\log T/ \log T) }{\sqrt{\frac12\log\log T} }\\ &\hskip1in \times\bigg(1 + O\Big(\frac{1}{\log T }\Big) \bigg).\end{split}\end{equation}

Using (1·1) again, we see that $m_{\gamma }\ll \log T/\log\log T$ . Thus, if we impose the condition that $\max(|a|, |b|)\ll (\!\log \log \log T)^{\frac12-{\epsilon}}$ , then when the expression on the left lies in the interval [a, b], the right-hand side equals

\begin{equation*} \frac{ \log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log T)^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log T} }+ O\Big(\frac{1}{ \sqrt{\log\log T} } \Big). \end{equation*}

Using this with (2·1), we see that replacing $\log {\gamma }$ and $\log\log {\gamma }$ in (2·2) by $\log T$ and $\log\log T$ , respectively, changes (2·2) by no more than $\displaystyle O( N(T) {(\!\log\log\log T)^2}/{\sqrt{\log \log T} }).$ Hence

\begin{align*}\notag N_{[a, b]}(T) = & \sum_{T/\log T < {\gamma } \leq T} \unicode{x1D7D9}_{[a,b]}\Bigg(\frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log T)^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log T}}\Bigg)\\&+O\Bigg(N(T)\frac{(\!\log\log\log T)^2}{\sqrt{\log \log T} }\Bigg).\end{align*}

Extending the sum back to the full range $0<{\gamma }\leq T$ and using (2·1) again, we see that for sufficiently large T,

\begin{equation}\notag N_{[a, b]}(T) =\frac{N(T)}{\sqrt{2\pi} } \int_a^b e^{-{x^2}/{2}}\, dx +O\Bigg( N(T)\frac{(\!\log\log\log T)^2}{\sqrt{\log \log T} }\Bigg),\end{equation}

provided $\max(|a|, |b|)\ll (\!\log \log \log T)^{\frac12-{\epsilon}}$ . This proves (1·2). It has already been noted that (1·3) follows from (1·2) and Hypothesis $\mathscr{H}_{\delta}$ , so the proof of Theorem 1·1 is complete.

3. Proof of Theorem 1·2

We assume the Riemann hypothesis, Hypothesis $\mathscr{H}_{\delta}$ , and that

\begin{equation}\notag \max(|a|, |b|)\ll (\!\log \log \log T)^{\frac12-{\epsilon}} \ \text{with}\ {\epsilon}>0.\end{equation}

Our assumption that a and b are either fixed, or functions of T for which $\int_a^b e^{-x^2/2} dx \gg 1$ means, by Theorem 1·1, that $N_{[a, b]}(T) \gg N(T)$ . Hence, by Weyl’s criterion [ Reference Weyl10 ], the sequence $ {\Gamma }_{[a, b]}$ is uniformly distributed modulo one if, for each fixed positive integer $\ell$ ,

(3·1) \begin{equation} \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) \,\, \unicode{x1D7D9}_{[a,b]}\Bigg(\frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log {\gamma })^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log {\gamma }}}\Bigg)=o(N(T))\end{equation}

as $T\to\infty$ . By the same argument we used in the last section, replacing $\log {\gamma }$ and $\log\log {\gamma }$ here by $\log T$ and $\log\log T$ , respectively, changes the sum by at most o(N(T). Thus, it suffices to show that

\begin{equation}\notag \begin{split} \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) \,\, \unicode{x1D7D9}_{[a,b]}\Bigg( \frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log T)^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log T}}\Bigg) = o( N(T)).\end{split}\end{equation}

Now let $ \displaystyle P({\gamma }) =\sum_{p \leq X^2} \frac{1}{ p^{1/2+i{\gamma }} },$ where p runs over the primes. The gist of corollary 2·3 in [ Reference Çiçek2 ] and some of the analysis following it, is that $P({\gamma })$ is on average a good approximation to

\begin{equation*} \frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log T)^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log T}}, \end{equation*}

provided X is sufficiently large. Indeed, from the discussion in sections 5·5 and 6 of [ Reference Çiçek2 ] it follows that

\begin{align*}\sum_{0<{\gamma } \leq T}\unicode{x1D7D9}_{[a,b]}\Bigg( \frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log T)^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log T}} \Bigg)= \sum_{0<{\gamma } \leq T} & \unicode{x1D7D9}_{[a,b]} \Bigg(\frac{\Re {P({\gamma })}}{\sqrt{\frac12\log\log T}}\Bigg) \\ &+O \bigg(N(T) \frac{(\!\log\log\log T)^2}{\sqrt{\log\log T}} \bigg). \end{align*}

This is the key result that allows us to prove our theorem. An immediate consequence is that

\begin{equation*}\unicode{x1D7D9}_{[a,b]} \Bigg(\frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log T)^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log T}}\Bigg)= \unicode{x1D7D9}_{[a,b]} \Bigg(\frac{\Re{P({\gamma })}}{\sqrt{\frac12\log\log T}}\Bigg)\end{equation*}

for all but $\displaystyle O\big(N(T) (\!\log\log\log T)^2/\sqrt{\log\log T} \big)=o(N(T))$ values of ${\gamma }$ in (0, T]. Therefore,

(3·2) \begin{equation}\begin{split} \sum_{0<{\gamma } \leq T} & e(\ell{\gamma }) \unicode{x1D7D9}_{[a,b]} \Bigg(\frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log T)^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log T}}\Bigg) \\=& \sum_{0<{\gamma } \leq T} e(\ell{\gamma }) \unicode{x1D7D9}_{[a,b]} \Bigg(\frac{\Re P({\gamma })}{\sqrt{\frac12\log\log T}}\Bigg)+ O \bigg(N(T) \frac{(\!\log\log\log T)^2}{\sqrt{\log\log T}} \bigg).\end{split}\end{equation}

Writing

\begin{equation}\notag A=A(T)=a\sqrt{\frac12\log\log T} \quad \text{and} \quad B=B(T)=b\sqrt{\frac12\log\log T},\end{equation}

we see that to prove our theorem we must show that

(3·3) \begin{equation}\sum_{0<{\gamma } \leq T}e(\ell{\gamma })\unicode{x1D7D9}_{[A, B]} ( \Re P({\gamma })) = o(N(T))\end{equation}

for each positive integer $\ell$ . To do this, we replace the characteristic function $\unicode{x1D7D9}_{[A, B]}$ by an approximation. Let $\Omega > 0$ and set

(3·4) \begin{equation} F_{\Omega}(x) =\Im \int_0^\Omega G\Big(\frac \omega \Omega\Big)\exp{(2\pi ix\omega)}\frac{\mathop{d\omega}}{\omega}, \end{equation}

where

\begin{equation}\notag G(u)= \frac{2u}{\pi}+2u(1-u)\cot{(\pi u)} \quad \text{for} \quad u\in[0, 1].\end{equation}

Then

(3·5) \begin{equation} \operatorname{sgn}{(x)} =F_{\Omega}(x) +O\bigg(\frac{\sin^2(\pi\Omega x)}{(\pi\Omega x)^2}\bigg), \end{equation}

(see [ Reference Tsang9 , pp. 26–29]). It follows that

(3·6) \begin{equation} \unicode{x1D7D9}_{[A, B]}(x) =\frac12F_\Omega(x-A)-\frac12F_\Omega(x-B) +O\bigg(\frac{\sin^2(\pi\Omega(x-A))}{(\pi\Omega(x-A))^2}\bigg) +O\bigg(\frac{\sin^2(\pi\Omega(x-B))}{(\pi\Omega(x-B))^2}\bigg). \end{equation}

This is the desired approximation of $ \unicode{x1D7D9}_{[A, B]}$ . Here we take $\displaystyle x=\Re{ {P}({\gamma })} = \Re \sum_{p \leq X^2} \frac{1}{ p^{1/2+i{\gamma }} }$ and

(3·7) \begin{equation} X = T^{\frac{1}{(\!\log\log T)^{20}}}, \qquad \Omega=(\!\log\log T)^2. \end{equation}

Now, it was shown in the course of the proof of proposition 5·5 in [ Reference Çiçek2 ] (with slightly different notation and parameters) that

\begin{equation}\notag \sum_{0 < {\gamma } \leq T}\frac{\sin^2\big(\pi \Omega(\Re {P}({\gamma })-A)\big)}{\big(\pi \Omega(\Re {P}({\gamma })-A)\big)^2} \ll \frac{N(T)}{\Omega},\end{equation}

and similarly for the sum with A replaced by B. Thus, by (3·6),

(3·8) \begin{equation}\begin{split}& \sum_{0<{\gamma } \leq T}e(\ell{\gamma }) \unicode{x1D7D9}_{[A, B]} ( \Re {P}({\gamma })) \\ = &\, \frac{1}{2} \sum_{0<{\gamma } \leq T} e(\ell{\gamma }) F_{\Omega} (\Re{ P({\gamma })}-A)-\frac{1}{2} \sum_{0<{\gamma } \leq T} e(\ell{\gamma }) F_{\Omega}(\Re {P}({\gamma })-B)+O\Big( \frac{N(T)}{\Omega}\Big) .\end{split}\end{equation}

From this and (3·3) we see that it suffices to prove that

(3·9) \begin{equation}\sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) F_{\Omega}(\Re {P}({\gamma })-A) =o(N(T))\end{equation}

for each positive integer $\ell$ , and similarly for the sum with A replaced by B.

To this end we use (3·4) to write

(3·10) \begin{align} \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) F_{\Omega}(\Re {P}({\gamma })-A) =\sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) \Im \int_0^\Omega G\Big(\frac \omega \Omega\Big) e^{-2\pi i A \omega} \exp{\big(2\pi i\omega\Re {P}({\gamma }) \big)}\frac{\mathop{d\omega}}{\omega}. \end{align}

By Taylor’s theorem, for any positive integer K,

\begin{equation}\notag \exp\big(2\pi i\omega\Re{P}({\gamma })\big) =1+\sum_{1\leq k < K} \frac{(2\pi i\omega\Re {P}({\gamma }))^k}{k!} + O\Big(\frac{(2\pi \omega|\Re {P}({\gamma })|)^K}{K!}\Big). \end{equation}

Inserting this in (3·10) and taking

(3·11) \begin{equation}K=2\big[(\!\log\log T)^6\big],\end{equation}

where [x] denotes the greatest integer less than or equal to x, we obtain

(3·12) \begin{align} \notag \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) F_{\Omega}(\Re {P}({\gamma })-A) =&\, F_\Omega(A) \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) \\ +\sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) \Im \int_0^\Omega & G\Big(\frac \omega \Omega\Big) e^{-2\pi i A \omega} \sum_{1\leq k < K}\frac{(2\pi i\omega)^k}{k!} (\Re {P}({\gamma }))^k \frac{\mathop{d\omega}}{\omega}\\ \notag &+ O\bigg( \sum_{0 < {\gamma } \leq T} |\Re {P}({\gamma })|^K \int_0^\Omega G\Big(\frac \omega \Omega\Big) \frac{(2\pi\omega)^K}{K!} \frac{\mathop{d\omega}}{\omega} \bigg). \end{align}

We estimate the sums over ${\gamma }$ on the right-hand side of the equation by means of the following result, which is an immediate consequence of an unconditional theorem and its corollary in [ Reference Gonek5 ] (see [ Reference Gonek4 ] also).

Lemma 3·1. Assume the Riemann hypothesis and let $x, T>1$ . Then

(3·13) \begin{equation} \begin{split} \sum_{0<{\gamma }\leq T} x^{i{\gamma }}= - &\frac{T}{2\pi}\frac{\Lambda(x)}{\sqrt x} +O(\sqrt{x} \log 2xT \log\log 3x) +O\Big(\!\log x\min\Big(\frac{T}{\sqrt x}, \frac{\sqrt x}{\langle x \rangle}\Big)\Big) \\ &+O\Big(\!\log 2T \min\Big(\frac{T}{\sqrt x}, \frac{1}{\sqrt x \log x} \Big)\Big). \end{split} \end{equation}

Here $\Lambda(x)=\log p$ if x is a positive integral power of a prime p and $\Lambda(x)=0$ for all other real numbers x, and $\langle x\rangle$ denotes the distance from x to the nearest prime power other than x itself. If $0<x<1$ , (3·13) also holds provided we replace x on the right-hand side by $1/x$ .

When $x>1$ , we will write (3·13) as

(3·14) \begin{equation} \sum_{0<{\gamma }\leq T} x^{i{\gamma }} =M(x) + E_1(x) + E_2(x)+ E_3(x), \end{equation}

where M(x) and $E_i(x), \ i=1, 2, 3,$ also depend on T. When $0<x<1$ , all the x’s on the right-hand side of (3·14) are to be replaced by $1/x$ . Note that when $x=1$ , $ \sum_{0<{\gamma }\leq T} x^{i{\gamma }}=N(T)$ .

Returning to (3·12), observe that by (3·5), $F_{\Omega}(A) \ll 1$ . Furthermore, taking $x=e^{2\pi \ell }>1$ in (3·13), we find that $\sum_{0<{\gamma }\leq T}e(\ell {\gamma }) \ll T$ . Thus, the first term on the right-hand side of (3·12) is

(3·15) \begin{equation}F_\Omega(A) \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) \ll T.\end{equation}

For the final term in (3·12) we use lemma 5·2 of [ Reference Çiçek2 ], which says that

\begin{equation}\notag \sum_{0<{\gamma } \leq T} | \Re P({\gamma }) |^K \ll (cK\Psi)^{K/2} N(T),\end{equation}

where $ \Psi =\log\log T.$ From this and Stirling’s approximation, we find that the O-term in (3·12) is

\begin{align*} &\ll N(T) \int_0^\Omega G\Big(\frac \omega \Omega\Big) \frac{(2\pi \omega)^K}{K!}(cK\Psi)^{K/2} \frac{\mathop{d\omega}}{\omega} \\ &\ll N(T) \int_0^\Omega G\Big(\frac \omega \Omega\Big) \frac{\omega(2\pi e)^K\omega^{K-1}}{K^K} (cK\Psi)^{K/2} \frac{\mathop{d\omega}}{\omega}. \end{align*}

By (3·7) and (3·11), and since G is bounded, this is

\begin{equation*} \ll N(T)\int_0^\Omega G\Big(\frac \omega \Omega\Big) \bigg(\frac{c\, \Omega\sqrt{\Psi}}{\sqrt{K}}\bigg)^K \mathop{d\omega} \ll \frac{N(T) }{2^{K}}\int_0^\Omega G\Big(\frac \omega \Omega\Big) \mathop{d\omega} \ll T. \end{equation*}

Combining this and (3·15), we may rewrite (3·12) as

(3·16) \begin{equation} \begin{split} & \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) F_{\Omega}(\Re P({\gamma })-A) \\ =& \int_0^\Omega G\Big(\frac \omega \Omega\Big) \sum_{1\leq k < K} \Im{\big(e^{-2\pi i A \omega}\, i^k\big)} \frac{(2\pi\omega)^k}{k!} \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) (\Re P({\gamma }))^k \frac{\mathop{d\omega}}{\omega} +O(T). \end{split} \end{equation}

To estimate the right-hand side we next bound the sums

(3·17) \begin{equation} S(k) = \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) (\Re P({\gamma }))^k.\end{equation}

By the binomial theorem

\begin{equation*} S(k) = \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) \Bigg(\sum_{p\leq X^2}\frac{1}{p^{1/2+i{\gamma }}}\Bigg)^j \Bigg(\sum_{p\leq X^2}\frac{1}{p^{1/2-i{\gamma }}}\Bigg)^{k-j}. \end{equation*}

Let $a_{r}(p_1\dots p_{r})$ denote the number of permutations of the primes $p_1,\dots, p_{r}$ , which might or might not be distinct. Also, for the rest of the paper, n will always denote a product of j primes, each of which is at most $X^2$ , while m denotes a product of $k-j$ primes, again each of size at most $X^2$ . We may thus write

\begin{align*} S(k) =&\, \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) \sum_n \frac{a_j(n)}{n^{1/2+i{\gamma }}} \sum_m \frac{a_{k-j}(m)}{m^{1/2-i{\gamma }}} \\ =&\,\frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{\sqrt n} \sum_m \frac{a_{k-j}(m)}{\sqrt m} \sum_{0 < {\gamma } \leq T} \Big(\frac{m e^{2\pi\ell}}{n}\Big)^{i{\gamma }}. \end{align*}

Since $e^{2\pi \ell} = ({-}1)^{-2i\ell}$ is of the form ${\alpha} _0^{{\beta}_0}$ with ${\alpha} _0, {\beta}_0$ algebraic, ${\alpha} _0\neq 0,1$ and $-2i\ell$ not rational, the Gelfond–Schneider theorem implies that $e^{2\pi \ell}$ is transcendental. Thus, $m e^{2\pi\ell}/{n}$ can neither be a positive integer nor the reciprocal of a positive integer. The M term in integer. The M term in (3·14) is therefore always zero. Hence, we may write

(3·18) \begin{equation} \begin{split}S(k) &= \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{\sqrt n} \sum_{\substack{m\\{me^{2\pi\ell}}/{n} >1}} \frac{a_{k-j}(m)}{\sqrt m} \bigg( \sum_{i=1}^3 E_i \Big(\frac{m e^{2\pi\ell}}{n}\Big)\bigg)\\ &\quad + \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{\sqrt n} \sum_{\substack{m\\{me^{2\pi\ell}}/{n} <1}} \frac{a_{k-j}(m)}{\sqrt m} \bigg( \sum_{i=1}^3 E_i \Big(\frac{n}{m e^{2\pi\ell}}\Big)\bigg) \\& \,=\!:\, S_1(k) + S_2(k). \end{split} \end{equation}

To estimate $S_1(k)$ , we insert the bounds for $E_1, E_2,$ and $E_3$ from (3·13) in to obtain

(3·19) \begin{equation}S_1(k) =\mathcal E_1(k)+\mathcal E_2(k)+\mathcal E_3(k),\end{equation}

where

\begin{equation}\notag \begin{split} \mathcal E_1(k)&= \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{\sqrt n} \sum_{\substack{m\\{me^{2\pi\ell}}/{n} >1}} \frac{a_{k-j}(m)}{\sqrt m} \sqrt{\frac{me^{2\pi \ell}}{n}} \log\Big(\frac{me^{2\pi \ell}T}{n}\Big) \log\log\Big(\frac{3me^{2\pi \ell}}{n}\Big), \\ \mathcal E_2(k)&= \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{\sqrt n} \sum_{\substack{m\\{me^{2\pi\ell}}/{n} >1}} \frac{a_{k-j}(m)}{\sqrt m} \frac{ \log(me^{2\pi \ell}/n)}{\sqrt{me^{2\pi \ell}/n}} \min\bigg(T, \frac{me^{2\pi \ell}/n }{\langle me^{2\pi \ell}/n \rangle} \bigg), \\ \mathcal E_3(k)&= \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{\sqrt n} \sum_{\substack{m\\{me^{2\pi\ell}}/{n} >1}} \frac{a_{k-j}(m)}{\sqrt m} \frac{\log T}{\sqrt{me^{2\pi \ell}/n} } \min \bigg(T, \frac{ 1}{\log(me^{2\pi \ell}/n)} \bigg). \end{split} \end{equation}

First consider $\mathcal E_1(k)$ . Since $e^{2\pi \ell}$ is fixed, we see that

\begin{align*} \mathcal E_1(k) &= \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{n} \sum_{\substack{m\\{me^{2\pi\ell}}/{n} >1}} a_{k-j}(m) \log\Big(\frac{mT}{n}\Big) \log\log \Big(\frac{3m}{n}\Big) \\ &\ll \frac{\log T\log\log T }{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n {a_j(n)} \sum_{\substack{m\\{me^{2\pi\ell}}/{n} >1}} a_{k-j}(m). \end{align*}

Here we have dropped the n in the denominator in the sum over n and used (3·7) and (3·11) to deduce that $m \leq X^{2k} \leq X^{2K} \leq T$ . Next, from the definitions of $a_j(n)$ and $a_{k-j}(m)$ and by the binomial theorem, we see that

(3·20) \begin{equation} \begin{split} & \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n a_j(n) \sum_{\substack{m\\{me^{2\pi\ell}}/{n} >1}} a_{k-j}(m) \\ &\ll \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \Bigg( \sum_{p\leq X^2}1\Bigg)^j \Bigg( \sum_{p\leq X^2} 1\Bigg)^{k-j} = \pi(X^2)^k, \end{split} \end{equation}

where $\pi(X^2)$ denotes the number of primes up to $X^2$ . By the prime number theorem $\pi(X^2)\ll {X^{2}}/{\log X}, $ so

\begin{equation}\notag \mathcal E_1(k) \ll \log T \log\log T \frac{X^{2k}}{(\!\log X)^k}. \end{equation}

To estimate

(3·21) \begin{equation} \mathcal E_2(k)= \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{\sqrt n} \sum_{\substack{m\\{me^{2\pi\ell}}/{n} >1}} \frac{a_{k-j}(m)}{\sqrt m} \frac{ \log(me^{2\pi \ell}/n)}{\sqrt{me^{2\pi \ell}/n}} \min\bigg(T, \frac{me^{2\pi \ell}/n }{\langle me^{2\pi \ell}/n \rangle} \bigg),\end{equation}

we require a lower bound for $\displaystyle \langle {me^{2\pi\ell}}/{n}\rangle$ . Recall that $\langle x\rangle $ denotes the distance from x to the nearest prime power other than x itself. As $e^{2\pi\ell}$ is transcendental, $me^{2\pi \ell}/n$ is not an integer. Now, for any positive non integral real number x, we have

\begin{equation*} \langle x \rangle \geq \min_{r\in \mathbb Z} |x-r| = \min\{ x-[x], 1-x+[x] \}.\end{equation*}

Thus

\begin{equation*}\Big\langle \frac{me^{2\pi\ell}}{n}\Big\rangle\geq \min\bigg\{ \frac{me^{2\pi\ell}}{n}- \Big[\frac{me^{2\pi\ell}}{n} \Big], \;1-\frac{me^{2\pi\ell}}{n} + \Big[\frac{me^{2\pi\ell}}{n} \Big] \bigg\}.\end{equation*}

We now recall a special case of Baker’s theorem [ Reference Baker1 , p. 24]. Since $e^{2\pi \ell} = ({-}1)^{-2i \ell}$ , for a given positive integer $\ell$ ,

\begin{equation}\notag \Big|e^{2\pi \ell}-\frac pq \Big| >q^{-C \log\log q} \end{equation}

for all rationals $p/q$ ( $p\geq 0, q\geq 4$ ), where C is a constant that depends on $\ell$ . Thus

(3·22) \begin{equation} \Big| \frac{me^{2\pi \ell}}{n} - \frac mn\frac pq \Big| > \frac mn q^{-C \log\log q}. \end{equation}

If we let $q=m$ and $\displaystyle p=n\Big[\frac{me^{2\pi\ell}}{n} \Big] $ , we obtain

\begin{equation*} \frac{me^{2\pi\ell}}{n}- \Big[\frac{me^{2\pi\ell}}{n} \Big] \geq \frac mn \, m^{-C \log\log m}.\end{equation*}

Similarly, taking $q=m$ and $\displaystyle p=n\Big[\frac{me^{2\pi\ell}}{n} \Big] +n$ , we see that $\displaystyle\Big( 1- \frac{me^{2\pi\ell}}{n} + \Big[\frac{me^{2\pi\ell}}{n} \Big] \Big)$ has the same lower bound. Hence

(3·23) \begin{equation}\Big\langle \frac{me^{2\pi\ell}}{n}\Big\rangle \geq \frac mn \, m^{-C \log\log m},\end{equation}

provided $m\geq 4$ . Now the terms in (3·21) with $1\leq m \leq 3$ contribute

\begin{equation}\notag \begin{split} & \ll_{\ell} \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_{n<6 e^{2\pi\ell}} \frac{a_j(n)}{n} \sum_{m\leq 3} a_{k-j}(m) \\ &\ll \frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \Bigg(\sum_{p\leq 6 e^{2\pi\ell} }1 \Bigg)^j \Bigg(\sum_{p\leq 3}1 \Bigg)^{k-j} \ll e^{ C k},\end{split}\end{equation}

where C is a constant depending on $\ell$ . Using (3·23) to estimate the terms in (3·21) with $m\geq 4$ , we find that they contribute

\begin{equation}\notag \begin{split}& \ll \frac{\log T}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n a_j(n) \sum_m {a_{k-j}(m)} m^{C \log\log m} \\ & \ll\frac{\log T}{2^k} X^{3Ck \log\log X}\sum_{j=0}^k \binom{k}{j} \Bigg(\sum_{p\leq X^2 }1\Bigg)^j \Bigg(\sum_{p\leq X^2 }1\Bigg)^{k-j} \\ & \ll (\!\log T ) \pi(X^2 )^{k} X^{3Ck \log\log X} \ll (\!\log T ) X^{4Ck \log\log X}.\end{split}\end{equation}

Thus,

\begin{equation}\notag \mathcal E_2(k) \ll_{\ell} (\!\log T ) X^{4Ck \log\log X} +e^{C k} \ll_{\ell} (\!\log T ) X^{4Ck \log\log X}.\end{equation}

Next consider

(3·24) \begin{equation} \mathcal E_3(k) =\frac{\log T}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n a_j(n) \sum_{\substack{m\\{me^{2\pi\ell}}/{n} >1}} \frac{a_{k-j}(m)}{m} \min \bigg(T, \frac{ 1}{\log(me^{2\pi\ell }/n)} \bigg).\end{equation}

Since $m, n \leq X^{2k}$ , by (3·22) with $p=n, q=m$ and $m\geq 4$ , we have

(3·25) \begin{equation} \frac{me^{2\pi \ell}}{n} - 1 > \frac mn m^{-C \log\log m} \geq X^{-2k-2kC\log\log (X^{2k})} \gg X^{-3kC\log\log X }. \end{equation}

Thus,

\begin{equation*}\frac{1}{\log(me^{2\pi\ell }/n)}\ll X^{3kC\log\log X}\end{equation*}

for $m\geq 4$ . The terms in (3·24) with $m\leq 3$ contribute

\begin{equation}\notag \begin{split}&\ll \frac{\log T}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_{n<6 e^{2\pi\ell} } a_j(n) \sum_{m\leq 3} {a_{k-j}(m)} \\&=\frac{\log T}{2^k}\sum_{j=0}^k \binom{k}{j} \Bigg(\sum_{p\leq 6 e^{2\pi\ell} }1 \Bigg)^j \Bigg(\sum_{p\leq 3}1 \Bigg)^{k-j}\ll e^{Ck} \log T.\end{split}\end{equation}

The contribution of the terms with $m\geq 4$ is

\begin{align*} \begin{split} \mathcal E_3(k) & \ll X^{3kC\log\log X} \ \frac{\log T }{2^k} \sum_{j=0}^k \binom{k}{j} \sum_n a_j(n) \sum_{m} a_{k-j}(m) \\ &=X^{3kC\log\log X} \ \pi(X^2)^k \log T \ll \frac{X^{2k} \log T}{(\!\log X)^{k}} X^{3kC\log\log X} \ll X^{4kC\log\log X} \log T. \end{split} \end{align*}

Thus

\begin{equation}\notag \mathcal E_3(k) \ll X^{4kC\log\log X} \log T. \end{equation}

Combining our estimates for $\mathcal E_1(k), \mathcal E_2(k) $ , and $\mathcal E_3(k)$ in (3·19), we see that

\begin{equation*} S_1(k) \ll_\ell (\!\log T ) X^{4Ck \log\log X}.\end{equation*}

The estimation of $S_2(k)$ is very similar and leads to the same bound. Hence, by (3·18)

\begin{equation*}S(k) \ll_\ell (\!\log T ) X^{4Ck \log\log X}.\end{equation*}

Inserting this into (3·16), we find that

\begin{equation}\notag \begin{split} \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) F_{\Omega}(\Re P({\gamma })-A) \ll_\ell X^{4CK\log\log X} \log T\int_0^\Omega G\Big(\frac \omega \Omega\Big) \sum_{1\leq k < K} \frac{(2\pi\omega )^k}{k!} \frac{\mathop{d\omega}}{\omega} +T. \end{split} \end{equation}

As G is bounded over [0, 1], this is

\begin{equation*}\ll_\ell \Omega \,e^{2\pi \Omega} X^{4CK\log\log X} \log T +T.\end{equation*}

By the choice of parameters $X, \Omega,$ and K in (3·7) and (3·11), it follows that for a fixed $\ell$ ,

\begin{equation*} \sum_{0 < {\gamma } \leq T} e(\ell {\gamma }) F_{\Omega}(\Re P({\gamma })-A) \ll e^{7(\!\log\log T)^2} T^{8C /(\!\log\log T)^{13}} +T \ll T =o(N(T)). \end{equation*}

This establishes (3·9), and the same estimate clearly holds when A is replaced by B. This completes the proof that ${\Gamma }_{[a, b]}$ is uniformly distributed modulo one.

Since the number of ${\gamma }$ ’s in ${\Gamma }_{[a, b]}$ with $0<{\gamma }\leq T$ that are not elements of ${\Gamma }_{[a, b]}^*$ (in other words, that are not simple) is at most o(N(T)), we see that ${\Gamma }_{[a, b]}^*$ is also uniformly distributed modulo one.

4. Proof of Theorem 1·3

By the Erdös–Turán inequality (see [ Reference Montgomery6 , Chapter 1, Corollary 1·1]), if L is a positive integer and $[\alpha,\beta]$ is a subinterval of [0,1], then

(4·1) \begin{equation}\left| \sum_{\substack{0<{\gamma } \leq T, \\ \{ {\gamma } (\!\log T)/2\pi \} \in [\alpha,\beta]}} 1 \ \ - \ (\beta-\alpha)N(T) \right| \ \leq \ \frac{N(T) }{L+1} \ + \ 3\sum_{\ell\leq L} \frac{1}{\ell}\Bigg| \sum_{0<{\gamma }\leq T } e\Big(\ell {\gamma } \frac{\log T}{2\pi}\Big)\Bigg|.\end{equation}

By Conjecture 1, for each integer $\ell >0$ ,

\begin{equation}\notag \sum_{0<{\gamma }\leq T } e\Big(\ell {\gamma } \frac{\log T}{2\pi}\Big)= \sum_{0 < {\gamma } \leq T} T^{i\ell \gamma} \ll T^{\frac12+\epsilon \ell}.\end{equation}

Hence, the right-hand side of (4·1) is

\begin{equation*}\ll \frac{N(T)}{L} + (\!\log L )\ T^{\frac12+\epsilon L}.\end{equation*}

Taking $ \displaystyle L=[{1}/{2\epsilon}]$ and assuming that $0<\epsilon <1/2$ , we see that this is $\ll \epsilon N(T)$ . Since $\epsilon$ can be arbitrarily small, this establishes (1·5).

5. Proof of Theorem 1·4

By the Erdös–Turán inequality again, if L is a positive integer and $[\alpha,\beta]$ is a subinterval of [0,1], then

(5·1) \begin{equation}\begin{split} & \left| \sum_{\substack{0<{\gamma } \leq T, {\gamma }\in {\Gamma }_{[a, b]}\\ \{ {\gamma } (\!\log T)/2\pi \} \in [\alpha,\beta]}} 1 \ \ - \ (\beta-\alpha)N_{[a, b]}(T) \right| \ \\& \qquad\qquad\qquad \leq \frac{N_{[a, b]}(T) }{L+1} \ + \ 3\sum_{\ell\leq L} \frac{1}{\ell}\Bigg| \sum_{0<{\gamma }\leq T, {\gamma }\in {\Gamma }_{[a, b]}} e\Big(\ell {\gamma } \frac{\log T}{2\pi}\Big)\Bigg|.\end{split}\end{equation}

Thus, to prove (1·6), we need to estimate

(5·2) \begin{equation} \begin{split} \sum_{0<{\gamma }\leq T, {\gamma }\in {\Gamma }_{[a, b]}} e\Big(\ell {\gamma } \frac{\log T}{2\pi}\Big)=& \sum_{0 < {\gamma } \leq T} T^{i\ell {\gamma } } \,\, \unicode{x1D7D9}_{[a,b]}\Bigg(\frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log {\gamma })^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log {\gamma }}}\Bigg)\end{split} \end{equation}

for positive integers $\ell$ . We do this, for the most part, by following the procedure of estimating the corresponding sum in (3·1) in the previous section. To start with, the same analysis that led to (3·2) leads to

(5·3) \begin{equation} \begin{split} & \sum_{0 < {\gamma } \leq T} T^{i\ell {\gamma } } \,\, \unicode{x1D7D9}_{[a,b]}\Bigg(\frac{\log \big(\big|\zeta^{(m_{\gamma })}(\frac12+i{\gamma }) \big|/(\!\log {\gamma })^{m_{\gamma }} \big) }{\sqrt{\frac12\log\log {\gamma }}}\Bigg)\\=&\sum_{0<{\gamma } \leq T} T^{i\ell {\gamma } } \,\, \unicode{x1D7D9}_{[a,b]} \Bigg(\frac{\Re P({\gamma })}{\sqrt{\frac12\log\log T}}\Bigg)+ O \bigg(N(T) \frac{(\!\log\log\log T)^2}{\sqrt{\log\log T}} \bigg).\end{split} \end{equation}

Similarly, the analysis that led to (3·8), with the same choices of the parameters A, B, X, and $\Omega$ , and Dirichlet polynomial P, shows that

(5·4) \begin{equation}\begin{split}\sum_{0<{\gamma } \leq T}&T^{i\ell {\gamma } } \unicode{x1D7D9}_{[A, B]} ( \Re {P}({\gamma })) \\ = &\frac{1}{2} \sum_{0<{\gamma } \leq T} T^{i\ell {\gamma } } F_{\Omega} (\Re{ P({\gamma })}-A)-\frac{1}{2} \sum_{0<{\gamma } \leq T} T^{i\ell {\gamma } } F_{\Omega}(\Re {P}({\gamma })-B)+O\Big( \frac{N(T)}{\Omega}\Big) .\end{split}\end{equation}

And, similarly to (3·12), we find that

(5·5) \begin{align} \notag \sum_{0 < {\gamma } \leq T} T^{i\ell {\gamma } } F_{\Omega}(\Re {P}({\gamma })-A) = &\, F_\Omega(A) \sum_{0 < {\gamma } \leq T} T^{i\ell {\gamma } } \\ +\sum_{0 < {\gamma } \leq T} T^{i\ell {\gamma } } \Im \int_0^\Omega & G\Big(\frac \omega \Omega\Big) e^{-2\pi i A \omega} \sum_{1\leq k < K}\frac{(2\pi i\omega)^k}{k!} (\Re {P}({\gamma }))^k \frac{\mathop{d\omega}}{\omega}\\ \notag &+ O\bigg( \sum_{0 < {\gamma } \leq T} |\Re {P}({\gamma })|^K \int_0^\Omega G\Big(\frac \omega \Omega\Big) \frac{(2\pi\omega)^K}{K!} \frac{\mathop{d\omega}}{\omega} \bigg), \end{align}

where $K=2\big[(\!\log\log T)^6\big]$ .

By (3·5), $F_{\Omega}(A) \ll 1$ and, by Conjecture 1,

\begin{equation}\notag \sum_{0 < {\gamma } \leq T} T^{i\ell \gamma} \ll T^{\frac12+\epsilon \ell}\end{equation}

for any $\epsilon>0$ . Thus, the first term on the right-hand side of (5·5) is $O(T^{\frac12+\epsilon \ell})$ . The third term is estimated in the same way as the third term in (3·12) and is likewise O(T). Hence

(5·6) \begin{equation} \begin{split} & \sum_{0 < {\gamma } \leq T} T^{i\ell\gamma} F_{\Omega}(\Re{P}({\gamma })-A) \\ = &\int_0^\Omega G\Big(\frac \omega \Omega\Big) \sum_{1\leq k < K} \Im{\big(e^{-2\pi i A \omega}\, i^k\big)} \frac{(2\pi\omega)^k}{k!} \sum_{0 < {\gamma } \leq T} T^{i\ell\gamma} (\Re{P}({\gamma }))^k \frac{\mathop{d\omega}}{\omega} +O(T)+O(T^{\frac12+\epsilon \ell}). \end{split} \end{equation}

The remaining term here is handled in the same way as the corresponding term in (3·16), except that we use Conjecture 1 rather than Lemma 3·1 to estimate the sums over ${\gamma }$ . We carry this out now.

Similarly to the analysis of the sum S(k) in (3·17) that gave (3·18), we find that

\begin{equation}\notag \begin{split} \sum_{0 < {\gamma } \leq T} T^{i\ell\gamma} (\Re{P}({\gamma }))^k =&\frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_{0 < {\gamma } \leq T} T^{i\ell\gamma} \sum_n \frac{a_j(n)}{n^{1/2+i{\gamma }}} \sum_m \frac{a_{k-j}(m)}{m^{1/2-i{\gamma }}} \\ =&\frac{1}{2^k}\sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{\sqrt n} \sum_m \frac{a_{k-j}(m)}{\sqrt m} \sum_{0 < {\gamma } \leq T} \Big(\frac{m T^\ell}{n}\Big)^{i{\gamma }}, \end{split} \end{equation}

where, as before, $a_{r}(p_1\dots p_{r})$ denotes the number of permutations of the primes $p_1,\dots, p_{r}$ , which might or might not be distinct. By Conjecture 1, for each m and n

\begin{equation*}\sum_{0 < {\gamma } \leq T} \Big(\frac{m T^\ell}{n}\Big)^{i{\gamma }}\ll T^{1+\ell \epsilon-\ell/2} \Big(\frac{m}{n}\Big)^{-\frac12+\epsilon}+T^{\frac12+\ell \epsilon} \Big(\frac{m}{n}\Big)^{\epsilon}.\end{equation*}

Thus

(5·7) \begin{equation} \begin{split} & \sum_{0 < {\gamma } \leq T} T^{i\ell\gamma} (\Re{P}({\gamma }))^k \\ \ll \, & \frac{T^{1+\ell \epsilon-\frac{\ell}{2}}}{2^k} \sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{n^{\epsilon}} \sum_m \frac{a_{k-j}(m)}{m^{1-\epsilon}} + \frac{T^{\frac12+\ell \epsilon} }{2^k} \sum_{j=0}^k \binom{k}{j} \sum_n \frac{a_j(n)}{n^{1/2+\epsilon}} \sum_m \frac{a_{k-j}(m)}{m^{1/2-\epsilon}}\\ &= \mathcal T_1(k)+ \mathcal T_2(k),\end{split}\end{equation}

say. Now

\begin{equation} \notag \begin{split}\mathcal T_1(k)&\ll \frac{T^{1+\ell \epsilon-\frac{\ell}{2}}}{2^k} \sum_{j=0}^k \binom{k}{j} \sum_n {a_j(n)} \sum_m {a_{k-j}(m)} \\& \ll \frac{T^{1+\ell \epsilon-\frac{\ell}{2}}}{2^k}\sum_{j=0}^k \binom{k}{j} \Bigg( \sum_{p\leq X^2}1\Bigg)^j \Bigg( \sum_{p\leq X^2} 1\Bigg)^{k-j} \ll T^{\frac12+\ell \epsilon }\pi(X^2)^k, \end{split} \end{equation}

since $\ell \geq 1$ . Similarly,

\begin{equation} \notag \begin{split}\mathcal T_2(k)&\ll \frac{T^{\frac12+\ell \epsilon} }{2^k} \sum_{j=0}^k \binom{k}{j} \sum_n {a_j(n)} \sum_m {a_{k-j}(m)} = T^{\frac12+\ell \epsilon} \pi(X^2)^k. \end{split} \end{equation}

Combining our estimates for $\mathcal T_1(k)$ and $\mathcal T_2(k)$ in (5·7), we see that

\begin{equation*} \sum_{0 < {\gamma } \leq T} T^{i\ell\gamma} (\Re{P}({\gamma }))^k \ll T^{\frac12+\ell \epsilon}\ \pi(X^2)^k\ll T^{\frac12+\ell \epsilon} X^{2k} .\end{equation*}

Using this in (5·6), we find that

\begin{equation} \notag \begin{split} \sum_{0 < {\gamma } \leq T} T^{i\ell\gamma} F_{\Omega}(\Re{P}({\gamma })-A) &\ll T^{\frac12+\ell \epsilon} \ X^{2K} \! \int_0^\Omega\! G\Big(\frac \omega \Omega\Big) \sum_{1\leq k < K} \frac{(2\pi\omega)^k}{k!} \frac{\mathop{d\omega}}{\omega} + O(T) +O(T^{\frac12+\epsilon \ell})\\ &\ll \Omega e^{2\pi \Omega} \ T^{\frac12+\ell \epsilon} \ X^{2K} + O(T) .\end{split}\end{equation}

By (3·7) and (3·11) this is

\begin{equation*}\ll e^{7(\!\log\log T)^2} \ T^{\frac12+\epsilon \ell} \ T^{\frac{5}{(\!\log\log T)^{14}}}+T \ll T, \end{equation*}

provided $\ell\leq L$ and $\epsilon$ is small enough relative to L. Inserting this (and the same estimate when A is replaced by B) into (5·4), we find that

\begin{equation}\notag \begin{split}\sum_{0<{\gamma } \leq T}T^{i\ell {\gamma } } \unicode{x1D7D9}_{[A, B]} &( \Re{P}({\gamma }))\ll T+ \frac{N(T)}{\Omega} \ll \frac{N(T)}{(\!\log\log T)^2} .\end{split}\end{equation}

By (5·2) and (5·3) we then obtain

\begin{equation}\notag \sum_{0<{\gamma }\leq T, {\gamma }\in {\Gamma }_{[a, b]}} e\Big(\ell {\gamma } \frac{\log T}{2\pi}\Big) \ll N(T) \frac{(\!\log\log\log T)^2}{\sqrt{\log\log T}}\end{equation}

for $\ell\leq L$ . Using this bound in (5·1), we see that

\begin{equation}\notag \left| \sum_{\substack{0<{\gamma } \leq T, {\gamma }\in {\Gamma }_{[a, b]}\\ \{ {\gamma } (\!\log T)/2\pi \} \in [\alpha,\beta]}} 1 \ \ - \ (\beta-\alpha)N_{[a, b]}(T) \right|\ \ll \ \frac{N_{[a, b]}(T) }{L+1} \ + \ N(T) (\!\log L) \frac{(\!\log\log\log T)^2}{\sqrt{\log\log T}}.\end{equation}

By our hypotheses on a and b, $N_{[a,b]}(T)\gg N(T)$ . Hence, since L may be arbitrarily large, we find that this equals $o(N_{[a, b]}(T))$ . This proves (1·6). The analogous inequality when ${\Gamma }_{[a, b]}$ is replaced by ${\Gamma }_{[a, b]}^*$ follows from this on noting that the number of ${\gamma }$ in ${\Gamma }_{[a, b]}$ with $0<{\gamma }\leq T$ that are not elements of ${\Gamma }_{[a, b]}^*$ is at most $o(N_{[a,b]}(T))$ .