Nondiagonal terms in the second moment of automorphic $L$-functions

Let $H_{2k}$ be the normalized Hecke basis of the space of holomorphic cusp forms of level one and weight $2k \geq 2$. For every function $f(z)\in H_{2k}$ we can use the Fourier expansion

$$\begin{equation} f(z)=\sum_{n\geq 1}\lambda_{f}(n)n^{k-1/2}\exp(2\pi inz), \qquad \lambda_{f}(1)=1, \end{equation} \tag{ 1 }$$

in order to define the associated $L$-function

$$\begin{equation} L(f,s):=\sum_{n=1}^{\infty}\frac{\lambda_f(n)}{n^s}, \qquad \operatorname{Re}{s}>1. \end{equation} \tag{ 2 }$$

It is possible to construct an analytic continuation of $L(f,s)$ to the whole complex plane. The so-called completed $L$-function

$$\begin{equation} \Lambda_f(s)=\biggl(\frac{1}{2 \pi}\biggr)^s\Gamma\biggl(s+\frac{2k-1}{2}\biggr)L(f,s) \end{equation} \tag{ 3 }$$

satisfies the functional equation

$$\begin{equation} \Lambda_f(s)=\varepsilon_f\Lambda_f(1-s), \qquad \varepsilon_f=i^{2k}. \end{equation} \tag{ 4 }$$

One of the central problems in the theory of $L$-functions is to determine the proportion of nonvanishing central $L$-values. Note that the functional equation (4) implies that $L(f,1/2)=0$ for all odd $k$. Therefore, it is interesting to consider only even $k$.

Nonvanishing results for central values of $L$-functions are not only of theoretical interest but also have important applications (see [1] for references). One of the most interesting examples of such applications (discovered by Iwaniec and Sarnak [5]) is the nonexistence of Landau-Siegel zeros for Dirichlet $L$-functions of real primitive characters.

The most common approach to nonvanishing problems is the method of moments. In particular, the given technique was used in [1] to prove that $20\%$ of the $L$-functions associated to holomorphic cusp forms of weight $2k \geq 2$ and level one have nonzero central values. The proof is based on the following exact formula for the twisted second moment, which was first obtained by Kuznetsov (preprint 1994, see also [1], Theorem 4.2 for the proof) and independently proved by Iwaniec and Sarnak (see [5], Theorem 17).

Theorem 1.1.  For $\operatorname{Re}{v}=0$, $\operatorname{Im}{v}\neq 0$, $|\operatorname{Re}{u}|<k-1$ we have

$$\begin{equation} \begin{aligned} \, \notag M_2(l;u,v) &:=\sum_{f \in H_{2k}(1)}^{h}\lambda_f(l)L_f\biggl(\frac12+u+v\biggr)L_f\biggl(\frac12+u-v\biggr) \\ \notag &=\tau_v(l)\biggl(\frac{\zeta(1+2u)}{l^{1/2+u}} +\frac{(2\pi)^{4u}}{l^{1/2-u}}\zeta(1-2u) \frac{\Gamma(k-u+v)\Gamma(k-u-v)}{\Gamma(k+u+v)\Gamma(k+u-v)}\biggr) \\ \notag &\qquad+(-1)^k\tau_u(l) \frac{\zeta(1+2v)}{(2\pi)^{-2u+2v}l^{1/2+v}}\,\frac{\Gamma(k-u+v)}{\Gamma(k+u-v)} \\ &\qquad+(-1)^k\tau_u(l)\frac{\zeta(1-2v)}{(2\pi)^{-2u-2v}l^{1/2-v}}\, \frac{\Gamma(k-u-v)}{\Gamma(k+v+u)}+\sum_{i=1}^3E_i(l;u,v). \end{aligned} \end{equation} \tag{ 5 }$$

Each summand $E_i(l;u,v)$ can be expressed in terms of hypergeometric functions:

$$\begin{equation} E_1(l;u,v)=\frac{(-1)^k}{\sqrt{l}}\sum_{1 \leq n \leq l-1}\tau_v(n)\tau_u(l-n)\varphi_k\biggl(\frac{n}{l};u,v\biggr), \end{equation} \tag{ 6 }$$

$$\begin{equation} E_2(l;u,v)=\frac{1}{\sqrt{l}}\sum_{ n \geq l+1}\tau_v(n)\tau_u(n-l)\Phi_k\biggl(\frac{l}{n};u,v\biggr) \end{equation} \tag{ 7 }$$

and

$$\begin{equation} E_3(l;u,v)=\frac{(-1)^k}{\sqrt{l}}\sum_{ n \geq 1}\tau_v(n)\tau_u(n+l)\psi_k\biggl(\frac{l}{n};u,v\biggr), \end{equation} \tag{ 8 }$$

where

$$\begin{equation} \varphi_k(x;u,v)=\widetilde{\varphi}_k(x;u,v)+\widetilde{\varphi}_k(x;u,-v), \end{equation} \tag{ 9 }$$

$$\begin{equation} \begin{split} &\widetilde{\varphi}_k(x;u,v) =\frac{(2\pi)^{2u+1}}{2\cos(\pi(1/2+v))}\, \frac{\Gamma(k-u+v)}{\Gamma(2v+1)\Gamma(k+u-v)} \\ &\qquad\qquad\times x^{v}(1-x)^{-u} F(k-u+v,1-k-u+v,1+2v;x), \end{split} \end{equation} \tag{ 10 }$$

$$\begin{equation} \begin{split} &\Phi_k(x;u,v) =2(2\pi)^{2u}\frac{\Gamma(k-u+v)\Gamma(k-u-v)}{\Gamma(2k)} \\ &\qquad\qquad\times \sin\biggl(\pi\biggl(\frac12+u\biggr)\biggr)x^k(1-x)^{-u}F(k-u+v,k-u-v,2k;x) \end{split} \end{equation} \tag{ 11 }$$

and

$$\begin{equation} \begin{split} &\psi_k(x;u,v) =2(2\pi)^{2u}\frac{\Gamma(k-u+v)\Gamma(k-u-v)}{\Gamma(2k)} \\ &\qquad\qquad\times \sin\biggl(\pi\biggl(\frac12+v\biggr)\biggr)x^k(1+x)^{-u}F(k-u+v,k-u-v,2k;-x), \end{split} \end{equation} \tag{ 12 }$$

$\Gamma(z)$ is the Gamma function, $F(a,b,c;x)$ is the Gauss hypergeometric function ( definitions and properties of these functions can be found in [7], Chs. 5 and 15) and

$$\begin{equation} \tau_v(n)=\sum_{n_1n_2=n}\biggl(\frac{n_1}{n_2}\biggr)^v=n^{-v}\sigma_{2v}(n), \end{equation} \tag{ 13 }$$

where $\sigma_v(n)=\sum_{d\mid n}d^v.$

For simplicity, let $\widetilde{\varphi}_k(x)=\widetilde{\varphi}_k(x;0,0)$ and $\Phi_k(x)=\Phi_k(x;0,0)$. Using the asymptotic expansions for $\widetilde{\varphi}_k(x)$ and $\Phi_k(x)$ obtained in [1], §5, and then estimating the absolute values of everything in the error terms $E_i(l;0,0)$, the following formula was proved in [1], Theorem 6.4:

$$\begin{equation} M_2(l;0,0)=\frac{2\tau(l)}{\sqrt{l}}(2\log{k}-\log{l}-2\log{2\pi}+2\gamma)+O\biggl( \frac{l^{1/2+\varepsilon}}{\sqrt{k}}\biggr). \end{equation} \tag{ 14 }$$

This is nontrivial for $l\ll\sqrt{k}$. We observe that taking an additional smooth average over weight $K<k<2K$, it is possible to obtain a nontrivial asymptotic formula for $l\ll K^2$ (see [1], Theorem 7.4, for details). Iwaniec and Sarnak [5] showed that a possible approach to demonstrate the nonexistence of Landau-Siegel zeros is to prove an asymptotic formula for the averaged twisted moment for $l\gg K^{2+\varepsilon}$. In this sense, $l=k^2$ is a natural barrier caused by the behaviour of the corresponding special functions. Indeed, if $l\ll k^2$, then according to [1], Lemma 6.2, we have

$$\begin{equation} \frac{1}{\sqrt{l}}\sum_{n=1}^{\infty}\tau(n)\tau(n+l)\Phi_{2k}\biggl( \frac{l}{n+l}\biggr)\ll \exp\biggl(-\frac{ck}{\sqrt{l}}\biggr) \biggl( \frac{l^{\varepsilon}}{l^{1/4}k^{1/2}}+\frac{l^{1/2+\varepsilon}}{k^{3/2}}\biggr), \end{equation} \tag{ 15 }$$

and thus the term $E_2(l;0,0)$ is negligibly small. For $l>k^2$ this term is no longer negligible. This is because for large $x$ the function $\Phi_{2k}(x)$ decays exponentially, but for $x$ very close to 1 it does not (see [1], §5.3)! As a result, for $l>k^2$ the left-hand side of (15) will not be negligible. Since $\Phi_{2k}(x)$ does not have oscillatory behaviour, even averaging over $k$ will not make $E_2(l;0,0)$ smaller. We show below that

$$\begin{equation} \frac{1}{\sqrt{l}}\sum_{n=1}^{\infty}\tau(n)\tau(n+l)\Phi_{2k} \biggl( \frac{l}{n+l}\biggr)\approx\frac{\sqrt{l}}{k^2}. \end{equation} \tag{ 16 }$$

So it is reasonable to conjecture that there is a second main term in (14) for $l>k^2$. The problem is that this second main term is larger than the main term in (14). It might seem that this strange phenomenon makes it impossible to improve the $50\%$ nonvanishing result due to Iwaniec and Sarnak, since, as it is shown in [3], the first moment does not change its behaviour for $l>k$ (note that when proving results on nonvanishing the twist in the second moment should grow as the square of the twist in the first moment). However, it is widely believed that the percentage of nonvanishing should be $100\%$, and therefore it is reasonable to investigate the second main terms further.

According to (5), there are three convolution sums (6)–(8). Each of them has the following form of a binary additive divisor problem:

$$\begin{equation} A_f(\alpha,\beta;W):=\sum_{n=1}^{\infty}\sigma_{\alpha}(n)\sigma_{\beta}(n+f)W \biggl(\frac{n}{f}\biggr),\qquad \end{equation} \tag{ 17 }$$

$$\begin{equation} B_N(\alpha,\beta;W_0):=\sum_{n=1}^{N}\sigma_{\alpha}(n)\sigma_{\beta}(N-n)W_0 \biggl(\frac{n}{N}\biggr); \end{equation} \tag{ 18 }$$

see [6], pp. 531 and 533. This problem has a long history and is very well studied (for references see [2]). An exact formula with a precise form of both the main term and the error term was discovered by Kuznetsov. In this paper we are interested only in the shape of the main terms. In the notation of [6], (4.9) and (4.29), we have

$$\begin{equation} \begin{aligned} \, A_f(\alpha,\beta;W)&=U(\alpha,\beta;W)+E(\alpha,\beta;W), \\ B_N(\alpha,\beta;W_0)&=V(\alpha,\beta;W_0)+E(\alpha,\beta;W_0), \end{aligned} \end{equation} \tag{ 19 }$$

where $E(\alpha,\beta;W)$ and $E(\alpha,\beta;W_0)$ are error terms and

$$\begin{equation} U(\alpha,\beta;W)=\int_0^{\infty}W(x)\mu_f(x;\alpha,\beta)\,dx \end{equation} \tag{ 20 }$$

and

$$\begin{equation} V(\alpha,\beta;W_0)=\int_0^{1}W_0(x)\rho_N(x;\alpha,\beta)\,dx, \end{equation} \tag{ 21 }$$

where

$$\begin{equation} \begin{split} &\mu_f(x;\alpha,\beta) =\sigma_{1+\alpha+\beta}(f) \frac{\zeta(1+\alpha)\zeta(1+\beta)}{\zeta(2+\alpha+\beta)}x^{\alpha}(1+x)^{\beta} \\ &\qquad\qquad +f^{\alpha}\sigma_{1-\alpha+\beta}(f) \frac{\zeta(1-\alpha)\zeta(1+\beta)}{\zeta(2-\alpha+\beta)}(1+x)^{\beta} \\ &\qquad\qquad +f^{\beta}\sigma_{1+\alpha-\beta}(f) \frac{\zeta(1+\alpha)\zeta(1-\beta)}{\zeta(2+\alpha-\beta)}x^{\alpha} +f^{\alpha+\beta}\sigma_{1-\alpha-\beta}(f) \frac{\zeta(1-\alpha)\zeta(1-\beta)}{\zeta(2-\alpha-\beta)} \end{split} \end{equation} \tag{ 22 }$$

and

$$\begin{equation} \begin{split} &\rho_N(x;\alpha,\beta) =\sigma_{1+\alpha+\beta}(N) \frac{\zeta(1+\alpha)\zeta(1+\beta)}{\zeta(2+\alpha+\beta)}x^{\alpha}(1-x)^{\beta} \\ &\qquad +N^{\alpha}\sigma_{1-\alpha+\beta}(N) \frac{\zeta(1-\alpha)\zeta(1+\beta)}{\zeta(2-\alpha+\beta)}(1-x)^{\beta} \\ &\qquad+ N^{\beta}\sigma_{1+\alpha-\beta}(N) \frac{\zeta(1+\alpha)\zeta(1-\beta)}{\zeta(2+\alpha-\beta)}x^{\alpha} +N^{\alpha+\beta}\sigma_{1-\alpha-\beta}(N) \frac{\zeta(1-\alpha)\zeta(1-\beta)}{\zeta(2-\alpha-\beta)}. \end{split} \end{equation} \tag{ 23 }$$

We conjecture that one can apply this result to each $E_i(l;u,v)$. For $i=1,2,3$ let

$$\begin{equation} E_i(l;u,v)=\mathrm{MT}_i(l;u,v)+\mathrm{ET}_i(l;u,v), \end{equation} \tag{ 24 }$$

where $\mathrm{MT}_i(l;u,v)$ is the main term.

Conjecture 1.2.  The main term $\mathrm{MT}_1(l;u,v)$ of the summand $E_1(l;u,v)$ can be evaluated using (21). For $i=2,3$ the main term $\mathrm{MT}_i(l;u,v)$ of $E_i(l;u,v)$ can be evaluated using (20).

Assuming that this conjecture holds, it is possible to evaluate the predicted main term $\mathrm{MT}_i(l;u,v)$ of each $E_i(l;u,v)$. Using the results of [6], it is also possible to evaluate the error terms, but we have not been successful in obtaining any reasonably good estimates for them. This is because the application of the results of [6] unfortunately leads to a dead loop by which we recover back the initial value of the second moment.

The main result of this paper is the following.

Theorem 1.3.  Assuming Conjecture 1.2 we have

$$\begin{equation} \mathrm{MT}_1(l;0,0)+\mathrm{MT}_2(l;0,0)+\mathrm{MT}_3(l;0,0)=0. \end{equation} \tag{ 25 }$$

This theorem asserts that for large $l$ there is no second main term in (14).

For the convenience of the reader we first recall the definitions and some properties of the Gamma function and the Gauss hypergeometric function. These facts, among others, can be found in [7], Chs. 5 and 15, for example.

For $\operatorname{Re}{z}>0$ the Gamma function $\Gamma(z)$ is defined by means of the Euler integral

$$\begin{equation*} \Gamma(z)=\int_0^{\infty}e^{-t}t^{z-1}\,dt \end{equation*}$$

([7], 5.2.1). The Gamma function is a meromorphic function with simple poles at negative integers $z=0,-1,-2,\dots$ . This function satisfies the following functional equations:

$$\begin{equation*} \Gamma(z+1)=z\Gamma(z) \end{equation*}$$

([7], 5.5.1) and

$$\begin{equation} \Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}, \qquad z\neq0,\pm1,\pm2,\dots \end{equation} \tag{ 26 }$$

([7], 5.5.3). For $z\neq0,-1,-2,\dots$ one can define the psi-function $\psi(z)$

$$\begin{equation*} \psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}. \end{equation*}$$

This is a meromorphic function with simple poles at the nonpositive integers $z=0,-1,-2,\dots$ . The psi-function satisfies the following functional equation:

$$\begin{equation} \psi(z+1)=\psi(z)+\frac{1}{z} \end{equation} \tag{ 27 }$$

(see [7], 5.5.2). For $|z|<1$ the Gauss hypergeometric function is defined by the absolutely convergent series

$$\begin{equation*} F(a,b,c;z)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \sum_{j=0}^{\infty}\frac{\Gamma(a+j)\Gamma(b+j)}{\Gamma(c+j)j!}z^j \end{equation*}$$

(see [7], 15.2.1). This function can be continued analytically to the whole complex plane. In the case of $z=1$ and $\operatorname{Re}{(c-a-b)}>0$ the Gauss hypergeometric function can be evaluated in terms of Gamma functions:

$$\begin{equation} F(a,b,c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} \end{equation} \tag{ 28 }$$

(see [7], 15.4.20).

Lemma 2.1.  For $d<0$ and $-1<c<-1-d$ we have

$$\begin{equation} \int_0^{\infty}x^c(1+x)^d\,dx=\frac{\Gamma(c+1)\Gamma(-1-c-d)}{\Gamma(-d)}. \end{equation} \tag{ 29 }$$

For $c,d>-1$ the following identity holds:

$$\begin{equation} \int_0^{1}x^c(1-x)^d\,dx=\frac{\Gamma(c+1)\Gamma(d+1)}{\Gamma(2+c+d)}. \end{equation} \tag{ 30 }$$

Equation (29) follows from [7], 5.12.3. To prove (30) we apply [7], 5.12.1.

Lemma 2.2.  For $0<x<1$ and $\operatorname{Re}{v}=0$, $0<\operatorname{Re}{u}<k-1$ we have

$$\begin{equation} \varphi_k(x;u,v)=(2\pi)^{2u}x^{-1/2+u}(1-x)^{-u}I_1(x), \end{equation} \tag{ 31 }$$

$$\begin{equation} \Phi_k(x;u,v)=(-1)^k(2\pi)^{2u}x^{1/2}(1-x)^{-u}I_1\biggl(\frac1x\biggr) \end{equation} \tag{ 32 }$$

and

$$\begin{equation} I_1(x)=\frac{1}{2\pi i}\int_{(\sigma)}\Gamma(k,u,v;s)\sin\biggl(\pi\biggl(u+\frac s2\biggr)\biggr)x^{s/2}\,ds, \end{equation} \tag{ 33 }$$

where $1-2k<\sigma<1-2\operatorname{Re}{u}$ and

$$\begin{equation} \Gamma(k,u,v;s)= \frac{\Gamma(k-1/2+s/2)}{\Gamma(k+1/2-s/2)}\Gamma\biggl(\frac12-u+v-\frac s2\biggr) \Gamma\biggl(\frac12-u-v-\frac s2\biggr). \end{equation} \tag{ 34 }$$

Proof.  It follows from [1], (4.15), that

$$\begin{equation} \varphi_k(x;u,v)=2\pi(4\pi)^{2u}(1-x)^{-u}\int_0^{\infty}k_0(y\sqrt{x},v)J_{2k-1}(y)y^{-2u}\,dy, \end{equation} \tag{ 35 }$$

where

$$\begin{equation*} k_0(x,v)=\frac{1}{2\cos(\pi (1/2+v))}\bigl(J_{2v}(x)-J_{-2v}(x) \bigr). \end{equation*}$$

In the same way, we obtain

$$\begin{equation} \Phi_k(x;u,v)=(-1)^k2\pi(4\pi)^{2u}\biggl(\frac 1x-1\biggr)^{-u} \int_0^{\infty}k_0\biggl(\frac{y}{\sqrt{x}},v\biggr)J_{2k-1}(y)y^{-2u}\,dy. \end{equation} \tag{ 36 }$$

Using [4], Lemma 5.7, to transform the right-hand sides of (35) and (36), we show that

$$\begin{equation*} \varphi_k(x;u,v)=x^{-1/2+u}(1-x)^{-u}(2\pi)^{2u}I_1(x) \end{equation*}$$

and

$$\begin{equation*} \Phi_k(x;u,v)=(-1)^kx^{1/2}(1-x)^{-u}(2\pi)^{2u}I_1\biggl(\frac 1x\biggr), \end{equation*}$$

where $I_1(x)$ is defined in [4], §5, and coincides with (33). The lemma is proved.

Replacing the function $\Phi_k(x;u,v)$ in (7) by (32) and using (13) we obtain

$$\begin{equation} \begin{aligned} \, \notag E_2(l;u,v) &=\frac{(-1)^k}{\sqrt{l}}(2\pi)^{2u}l^{u+v} \\ &\qquad\times \sum_{ n=1}^{\infty} \sigma_{-2u}(n)\sigma_{-2v}(n+l)\biggl(1+\frac{n}{l}\biggr)^{-1/2+u+v} I_1\biggl(1+\frac{n}{l}\biggr). \end{aligned} \end{equation} \tag{ 37 }$$

Assuming Conjecture 1.2, the main term $\mathrm{MT}_2(l;u,v)$ of (37) is given by

$$\begin{equation} \mathrm{MT}_2(l;u,v)= \frac{(-1)^k}{\sqrt{l}}(2\pi)^{2u}l^{u+v} \int_0^{\infty}(1+x)^{-1/2+u+v}I_1(1+x)\mu_l(x;-2u,-2v)\,dx \end{equation} \tag{ 38 }$$

(see (17), (19) and (20)). Let

$$\begin{equation} Q_1(u,v)= \frac{\Gamma(1-2u)\cos(\pi u)}{(k-1+u-v)(k-u+v)} \end{equation} \tag{ 39 }$$

and

$$\begin{equation} Q_2(u,v)=\Gamma(1+2u)\cos(\pi u) \frac{\Gamma(k-1-u-v)\Gamma(k-u+v)}{\Gamma(k+1+u+v)\Gamma(k+u-v)}. \end{equation} \tag{ 40 }$$

Lemma 3.1.  The following identity holds:

$$\begin{equation} \begin{aligned} \, \notag \mathrm{MT}_2(l;u,v) &=\frac{2}{\sqrt{l}}(2\pi)^{2u}l^{u+v}\biggl( \sigma_{1-2u-2v}(l)\frac{\zeta(1-2u)\zeta(1-2v)}{\zeta(2-2u-2v)}Q_1(u,-v) \\ \notag &\qquad +l^{-2u}\sigma_{1+2u-2v}(l)\frac{\zeta(1+2u)\zeta(1-2v)}{\zeta(2+2u-2v)}Q_2(u,-v) \\ \notag &\qquad +l^{-2v}\sigma_{1-2u+2v}(l)\frac{\zeta(1-2u)\zeta(1+2v)}{\zeta(2-2u+2v)}Q_1(u,v) \\ &\qquad +l^{-2u-2v}\sigma_{1+2u+2v}(l)\frac{\zeta(1+2u)\zeta(1+2v)}{\zeta(2+2u+2v)}Q_2(u,v) \biggr). \end{aligned} \end{equation} \tag{ 41 }$$

Proof.  To prove the lemma we substitute (33) and (22) into (38). In doing so we obtain four similar terms: each of them can be evaluated in the same way. We consider the first and second terms. The first term is given by

$$\begin{equation} \begin{aligned} \, \notag &\sigma_{1-2u-2v}(l)\frac{\zeta(1-2u)\zeta(1-2v)}{\zeta(2-2u-2v)}\, \frac{1}{2\pi i}\int_{(\sigma)}\Gamma(k,u,v;s)\sin\biggl(\pi\biggl(u+\frac s2\biggr)\biggr) \\ &\qquad\qquad\times\int_0^{\infty}x^{-2u}(1+x)^{-1/2+u-v+s/2}\,dx\,ds. \end{aligned} \end{equation} \tag{ 42 }$$

To evaluate the integral with respect to $x$ we apply (29), getting

$$\begin{equation*} \begin{aligned} \, & \frac{1}{2\pi i}\int_{(\sigma)}\Gamma(k,u,v;s)\sin\biggl(\pi\biggl(u\,{+}\,\frac s2\biggr)\biggr) \int_0^{\infty}x^{-2u}(1\,{+}\,x)^{-2v}(1\,{+}\,x)^{-1/2+u-v+s/2}\,dx\,ds \\ &\qquad= \frac{\Gamma(1-2u)}{2\pi i}\int_{(\sigma)}\sin\biggl(\pi\biggl(u+\frac s2\biggr)\biggr) \frac{\Gamma(k-1/2+s/2)}{\Gamma(k+1/2-s/2)} \\ &\qquad\qquad\times \Gamma\biggl(\frac12-u-v-\frac s2\biggr)\Gamma\biggl(-\frac12+u+v-\frac s2\biggr)\,ds. \end{aligned} \end{equation*}$$

Moving the line of integration to the left and evaluating the residues at $s={1-2k-2j}$ for $j=0,1,\dots$, we infer

$$\begin{equation} \begin{aligned} \, \notag &2(-1)^k\Gamma(1-2u)\cos(\pi u)\sum_{j=0}^{\infty} \frac{\Gamma(k-u-v+j)\Gamma(k-1+u+v+j)}{\Gamma(2k+j)j!} \\ \notag &\qquad =2(-1)^k\Gamma(1-2u)\cos(\pi u)\frac{\Gamma(k-u-v)\Gamma(k-1+u+v)}{\Gamma(2k)} \\ \notag &\qquad\qquad\times F(k-u-v,k-1+u+v,2k;1) \\ &\qquad =\frac{2(-1)^k\Gamma(1-2u)\cos(\pi u)}{(k-1+u+v)(k-u-v)}=2(-1)^kQ_1(u,-v), \end{aligned} \end{equation} \tag{ 43 }$$

where (28) was used to evaluate the hypergeometric function. Substituting (43) into (42), we obtain the first term in (41).

Let us briefly describe how the second term in (41) can be obtained. In this case, we need to compute

$$\begin{equation*} \frac{1}{2\pi i}\int_{(\sigma)}\Gamma(k,u,v;s)\sin\biggl(\pi\biggl(u+\frac s2\biggr)\biggr) \int_0^{\infty}(1+x)^{-1/2+u-v+s/2}\,dx\,ds. \end{equation*}$$

To evaluate the integral with respect to $x$ we apply (29). Consequently,

$$\begin{equation*} \frac{1}{2\pi i}\int_{(\sigma)}\sin\biggl(\pi\biggl(u+\frac s2\biggr)\biggr) \frac{\Gamma(k-1/2+s/2)}{\Gamma(k+1/2-s/2)} \Gamma\biggl(\frac12-u-v-\frac s2\biggr)\Gamma\biggl(-\frac12-u+v-\frac s2\biggr)\,ds. \end{equation*}$$

Moving the line of integration to the left and evaluating the residues at $s={1-2k-2j}$ for $j=0,1,\dots$, we show that

$$\begin{equation*} \begin{aligned} \, &2(-1)^k\cos(\pi u)\sum_{j=0}^{\infty}\frac{\Gamma(k-u-v+j)\Gamma(k-1-u+v+j)}{\Gamma(2k+j)j!} \\ &\quad= 2(-1)^k\cos(\pi u)\frac{\Gamma(k-u-v)\Gamma(k-1-u+v)}{\Gamma(2k)}F(k-u-v,k\,{-}\,1\,{-}\,u\,{+}\,v,2k;1) \\ &\quad=2(-1)^k\Gamma(1+2u)\cos(\pi u)\frac{\Gamma(k-u-v)\Gamma(k-1-u+v)}{\Gamma(k+u+v)\Gamma(k+1+u-v)} =2(-1)^kQ_2(u,-v), \end{aligned} \end{equation*}$$

where (28) was used to compute the hypergeometric function. The lemma is proved.

Our next goal is to evaluate $\mathrm{MT}_2(l;0,0)$ using (41). To this end, we need to study the properties of $Q_1(u,v)$ and $Q_2(u,v)$.

Lemma 3.2.  We have

$$\begin{equation} Q_1(0,0)=Q_2(0,0)=\frac{1}{k(k-1)}:=Q, \end{equation} \tag{ 44 }$$

$$\begin{equation} \frac{\partial}{\partial v}Q_1(0,v)\Big|_{v=0}= \frac{\partial}{\partial v}Q_2(0,v)\Big|_{v=0}=Q^2, \end{equation} \tag{ 45 }$$

$$\begin{equation} \frac{\partial}{\partial u}Q_1(u,0)\Big|_{u=0}= -2Q\psi(1)-Q^2:=Qq_1^u, \end{equation} \tag{ 46 }$$

$$\begin{equation} \frac{\partial}{\partial u}Q_2(u,0)\Big|_{u=0}= Q\bigl(2\psi(1)-4\psi(k)+Q\bigr):=Qq_2^u, \end{equation} \tag{ 47 }$$

$$\begin{equation} \frac{\partial^2}{\partial u\,\partial v}Q_1(u,v)\Big|_{u,v=0}= -2\psi(2)Q^2-2Q^3:=Qq_1^{u,v} \end{equation} \tag{ 48 }$$

and

$$\begin{equation} \frac{\partial^2}{\partial u\,\partial v}Q_2(u,v)\Big|_{u,v=0}= Q^2\bigl(2\psi(1)-4\psi(k)+Q\bigr)+Q^3\bigl(k^2+(k-1)^2\bigr):=Qq_2^{u,v}. \end{equation} \tag{ 49 }$$

Proof.  It follows from (39) that

$$\begin{equation} \frac{\partial}{\partial v}Q_1(u,v)=\Gamma(1-2u)\cos(\pi u) \frac{1-2u+2v}{(k-1+u-v)^2(k-u+v)^2} \end{equation} \tag{ 50 }$$

and

$$\begin{equation} \frac{\partial}{\partial u}Q_1(u,0)\Big|_{u=0}= \frac{\partial}{\partial u} \biggl(\frac{\Gamma(1-2u)\cos(\pi u)}{(k-u)(k-1+u)}\biggr)\Big|_{u=0}= -2Q\psi(1)-Q^2. \end{equation} \tag{ 51 }$$

Thus (46) and a part of (45) are proved. Computing the derivative of (50) with respect to $u$, we prove (48).

Formula

$$\begin{equation} \frac{\partial}{\partial u}Q_2(u,0)\Big|_{u=0}= Q\bigl(2\psi(1)-\psi(k-1)-\psi(k+1)-2\psi(k)\bigr) \end{equation} \tag{ 52 }$$

follows immediately from (40). Since $\psi(k+1)+\psi(k-1)=2\psi(k)-Q$ (see (27)), we prove (47).

Also as a consequence of (40) we have

$$\begin{equation} \begin{aligned} \, \notag &\frac{\partial}{\partial v}Q_2(u,v)=\Gamma(1+2u)\cos(\pi u) \frac{\Gamma(k-1-u-v)\Gamma(k-u+v)}{\Gamma(k+1+u+v)\Gamma(k+u-v)} \\ &\qquad \times\bigl(-\psi(k-1-u-v)+\psi(k-u+v)-\psi(k+1+u+v)+\psi(k+u-v)\bigr). \end{aligned} \end{equation} \tag{ 53 }$$

Letting $u=v=0$ on the right-hand side of (53), we show that

$$\begin{equation*} \frac{\partial}{\partial v}Q_2(0,v)\Big|_{v=0}=Q\bigl( \psi(k)-\psi(k-1)-(\psi(k+1)-\psi(k))\bigr). \end{equation*}$$

Since $\psi(z+1)-\psi(z)=1/z$ (see (27)), we complete the proof of (45). Letting $v=0$ in (53) and then computing the derivative of (53) with respect to $u$, we obtain

$$\begin{equation*} \begin{aligned} \, &\frac{\partial^2}{\partial u\,\partial v}Q_2(u,v)\Big|_{u,v=0}= Q\bigl(\psi(k)-\psi(k-1)-\psi(k+1)+\psi(k)\bigr) \\ &\qquad\qquad\times \bigl(2\psi(1)-\psi(k-1)-\psi(k+1)-2\psi(k)\bigr) \\ &\qquad+ Q\bigl(\psi'(k-1)-\psi'(k)-\psi'(k+1)+\psi'(k)\bigr). \end{aligned} \end{equation*}$$

Since $\psi'(z+1)-\psi'(z)=-1/z^2$, we finally prove (49). The lemma is proved.

Let

$$\begin{equation} D(u)=\frac{1}{\zeta(2+2u)}\biggl(\log\frac{d^2}{l}+2\gamma-2-2\frac{\zeta'(2+2u)}{\zeta(2+2u)} \biggr), \end{equation} \tag{ 54 }$$

$$\begin{equation} D_1=2\gamma-2\frac{\zeta'(2)}{\zeta(2)} \end{equation} \tag{ 55 }$$

and

$$\begin{equation} D_2=-2D_1\frac{\zeta'(2)}{\zeta(2)}-4\biggl(\frac{\zeta'(2)}{\zeta(2)}\biggr)'. \end{equation} \tag{ 56 }$$

Then

$$\begin{equation} D(0)=\frac{1}{\zeta(2)}\biggl(\log\frac{d^2}{l}+D_1\biggr) \quad\text{and}\quad D'(0)=-\frac{2}{\zeta(2)}\,\frac{\zeta'(2)}{\zeta(2)}\log\frac{d^2}{l}+\frac{D_2}{\zeta(2)}. \end{equation} \tag{ 57 }$$

Now we are ready to evaluate $\mathrm{MT}_2(l;0,0)$ using (41).

Theorem 3.3.  The following formula holds:

$$\begin{equation} \begin{aligned} \, \notag \mathrm{MT}_2(l;0,0) &=\frac{2Q}{\zeta(2)\sqrt{l}}\sum_{d\mid l}d\log^2\frac{d^2}{l} \\ \notag &\qquad+\frac{2Q}{\zeta(2)\sqrt{l}} \bigl(2D_1+2\psi(1)-2\psi(k)+2Q\bigr)\sum_{d\mid l}d\log\frac{d^2}{l} \\ \notag &\qquad +\frac{2Q}{\zeta(2)\sqrt{l}}\biggl( D_1(D_1+2\psi(1)-2\psi(k))-4\biggl(\frac{\zeta'(2)}{\zeta(2)}\biggr)' \\ &\qquad+Q\bigl(2D_1+\psi(1)+\psi(2)+1-2\psi(k)+2Q\bigr)\biggr)\sum_{d\mid l}d, \end{aligned} \end{equation} \tag{ 58 }$$

where $Q=1/(k^2-k)$ and $D_1$ is defined in (55).

Proof.  First, we continue (41) analytically to a region containing $v=0$. To do this, we rewrite (41) in the following form

$$\begin{equation} \begin{aligned} \, \notag &\mathrm{MT}_2(l;u,v) =\frac{2}{\sqrt{l}}(2\pi)^{2u}l^{u}\zeta(1-2u)\sum_{d\mid l}d^{1-2u} \biggl(d^{2v}l^{-v}\frac{\zeta(1+2v)}{\zeta(2-2u+2v)}Q_1(u,v) \\ \notag &\ +d^{-2v}l^{v}\frac{\zeta(1-2v)}{\zeta(2-2u-2v)}Q_1(u,-v) \biggr) +\frac{2}{\sqrt{l}}(2\pi)^{2u}l^{-u}\zeta(1+2u) \\ &\ \quad \times\sum_{d\mid l}d^{1+2u}\biggl( d^{2v}l^{-v}\frac{\zeta(1+2v)}{\zeta(2+2u+2v)}Q_2(u,v)\,{+}\, d^{-2v}l^{v}\frac{\zeta(1\,{-}\,2v)}{\zeta(2\,{+}\,2u\,{-}\,2v)}Q_2(u,-v) \biggr). \end{aligned} \end{equation} \tag{ 59 }$$

The next step is to evaluate the limits of two brackets in (59) as $v\to0.$ To this end, we use the Laurent expansion of the Riemann zeta-function

$$\begin{equation} \zeta(s)=\frac{1}{s-1}+\gamma+O(|s-1|) \end{equation} \tag{ 60 }$$

(see [7], 25.2.4). We compute the limit of the first bracket in (59). Using (60), we obtain that it is equal to

$$\begin{equation} \begin{aligned} \, \notag &2\gamma\frac{Q_1(u,0)}{\zeta(2-2u)}+ \frac{\partial}{\partial v}\biggl(\biggl(\frac{d^2}{l}\biggr)^v\frac{Q_1(u,v)}{\zeta(2-2u+2v)}\biggr)\Big|_{v=0} \\ &\qquad=Q_1(u,0)D(-u)+\frac{1}{\zeta(2-2u)}\,\frac{\partial}{\partial v}Q_1(u,v)\Big|_{v=0}. \end{aligned} \end{equation} \tag{ 61 }$$

Using (59) and (61), we show that

$$\begin{equation} \begin{aligned} \, \notag &\mathrm{MT}_2(l;u,0) \\ \notag &\qquad =\frac{2}{\sqrt{l}}\sum_{d\mid l}d\biggl[\zeta(1+2u)\biggl(\frac{2\pi d}{\sqrt{l}}\biggr)^{2u} \biggl(Q_2(u,0)D(u)+\frac{1}{\zeta(2+2u)}\,\frac{\partial}{\partial v}Q_2(u,v)\Big|_{v=0}\biggr) \\ &\qquad +\zeta(1-2u)\biggl(\frac{2\pi\sqrt{l}}{d}\biggr)^{2u} \biggl(Q_1(u,0)D(-u)+\frac{1}{\zeta(2-2u)}\,\frac{\partial}{\partial v}Q_1(u,v)\Big|_{v=0}\biggr) \biggr]. \end{aligned} \end{equation} \tag{ 62 }$$

We need to verify that the right-hand side of (62) is holomorphic at $u=0$. For this goal, it is enough to check that

$$\begin{equation} Q_2(0,0)D(0)+\frac{1}{\zeta(2)}\,\frac{\partial}{\partial v}Q_2(0,v)\Big|_{v=0}= Q_1(0,0)D(0)+\frac{1}{\zeta(2)}\,\frac{\partial}{\partial v}Q_1(0,v)\Big|_{v=0}, \end{equation} \tag{ 63 }$$

which follows from (44) and (45). Applying L'Hôpital's rule and (60), we compute the limit of the right-hand side of (62) at $u=0$:

$$\begin{equation} \begin{aligned} \, \notag &\mathrm{MT}_2(l;0,0) =\frac{4\gamma}{\sqrt{l}}\sum_{d\mid l}d \biggl(Q_2(0,0)D(0)+\frac{1}{\zeta(2)}\,\frac{\partial}{\partial v}Q_2(0,v)\Big|_{v=0}\biggr) \\ \notag &\qquad\qquad+\frac{2}{\sqrt{l}}\sum_{d\mid l}\,\frac{d}{2} \frac{\partial}{\partial u} \biggl[\biggl(\frac{2\pi d}{\sqrt{l}}\biggr)^{2u} \biggl(Q_2(u,0)D(u)+\frac{1}{\zeta(2+2u)}\,\frac{\partial}{\partial v}Q_2(u,v)\Big|_{v=0}\biggr) \\ &\qquad\qquad -\biggl(\frac{2\pi\sqrt{l}}{d}\biggr)^{2u} \biggl(Q_1(u,0)D(-u)+\frac{1}{\zeta(2-2u)}\,\frac{\partial}{\partial v}Q_1(u,v)\Big|_{v=0}\biggr) \biggr]\Big|_{u=0}. \end{aligned} \end{equation} \tag{ 64 }$$

Evaluating the derivative, using (63) and making some transformations, we obtain

$$\begin{equation*} \begin{aligned} \, &\mathrm{MT}_2(l;0,0) =\frac{2}{\sqrt{l}}\sum_{d\mid l}d\log\frac{d^2}{l} \biggl(Q_2(0,0)D(0)+\frac{1}{\zeta(2)}\,\frac{\partial}{\partial v}Q_2(0,v)\Big|_{v=0}\biggr) \\ &\qquad\qquad +\frac{2}{\sqrt{l}}\sum_{d\mid l}dD(0) \biggl[2\gamma Q_2(0,0)+\frac{1}{2}\,\frac{\partial}{\partial u}Q_2(u,0)\Big|_{u=0}- \frac{1}{2}\,\frac{\partial}{\partial u}Q_1(u,0)\Big|_{u=0}\biggr] \\ &\qquad\qquad +\frac{2}{\sqrt{l}}\sum_{d\mid l}dD'(0)\frac{Q_1(0,0)+Q_2(0,0)}{2}+ \frac{2}{\sqrt{l}}\sum_{d\mid l}d \biggl[\frac{2\gamma}{\zeta(2)}\,\frac{\partial}{\partial v}Q_2(0,v)\Big|_{v=0} \\ &\qquad\qquad -\frac{\zeta'(2)}{\zeta^2(2)}\,\frac{\partial}{\partial v}Q_2(0,v)\Big|_{v=0}+ \frac{1}{2\zeta(2)}\,\frac{\partial^2}{\partial u\,\partial v}Q_2(u,v)\Big|_{v=0} \\ &\qquad\qquad -\frac{\zeta'(2)}{\zeta^2(2)}\,\frac{\partial}{\partial v}Q_1(0,v)\Big|_{v=0}- \frac{1}{2\zeta(2)}\,\frac{\partial^2}{\partial u\,\partial v}Q_1(u,v)\Big|_{v=0}\biggr]. \end{aligned} \end{equation*}$$

Substituting (57) and using (44) and (45), we conclude that

$$\begin{equation} \begin{aligned} \, \notag &\mathrm{MT}_2(l;0,0) =\frac{2}{\sqrt{l}}\,\frac{Q_2(0,0)}{\zeta(2)}\sum_{d\mid l}d\log^2\frac{d^2}{l} \\ \notag &\qquad +\frac{2}{\sqrt{l}}\,\frac{1}{\zeta(2)} \sum_{d\mid l}d\log\frac{d^2}{l} \biggl[Q_2(0,0)\biggl(D_1+2\gamma-2\frac{\zeta'(2)}{\zeta(2)}\biggr) \\ \notag &\qquad+ \frac{\partial}{\partial v}Q_2(0,v)\Big|_{v=0}+ \frac{1}{2}\,\frac{\partial}{\partial u}Q_2(u,0)\Big|_{u=0} -\frac{1}{2}\,\frac{\partial}{\partial u}Q_1(u,0)\Big|_{u=0}\biggr] \\ \notag &\qquad +\frac{2}{\sqrt{l}}\,\frac{1}{\zeta(2)}\sum_{d\mid l}d \biggl[D_1\biggl(2\gamma Q_2(0,0)+\frac{1}{2}\,\frac{\partial}{\partial u}Q_2(u,0)\Big|_{u=0} -\frac{1}{2}\,\frac{\partial}{\partial u}Q_1(u,0)\Big|_{u=0}\biggr) \\ \notag &\qquad +D_2Q_2(0,0)+\biggl(2\gamma-2\frac{\zeta'(2)}{\zeta(2)}\biggr)\frac{\partial}{\partial v}Q_2(0,v)\Big|_{v=0} \\ &\qquad +\frac{1}{2}\,\frac{\partial^2}{\partial u\,\partial v}Q_2(u,v)\Big|_{u=v=0}- \frac{1}{2}\,\frac{\partial^2}{\partial u\,\partial v}Q_1(u,v)\Big|_{u=v=0} \biggr]. \end{aligned} \end{equation} \tag{ 65 }$$

Applying (45)–(49) we have

$$\begin{equation*} \begin{aligned} \, &\mathrm{MT}_2(l;0,0) =\frac{2}{\sqrt{l}}\,\frac{Q}{\zeta(2)}\sum_{d\mid l}d\log^2\frac{d^2}{l} +\frac{2}{\sqrt{l}}\,\frac{Q}{\zeta(2)}\sum_{d\mid l}d\log\frac{d^2}{l} \biggl[2D_1\,{+}\,Q\,{+}\,\frac{1}{2}q_2^u\,{-}\,\frac{1}{2}q_1^u\biggr] \\ &\qquad\qquad +\frac{2}{\sqrt{l}}\,\frac{Q}{\zeta(2)}\sum_{d\mid l}d \biggl[D_1\biggl(2\gamma+\frac{1}{2}q_2^u-\frac{1}{2}q_1^u\biggr)+D_2 +D_1Q+\frac{1}{2}q_2^{u,v}-\frac{1}{2}q_1^{u,v}\biggr]. \end{aligned} \end{equation*}$$

Substituting (55), (56) and (46)–(49), we finally prove (58). Theorem 3.3 is proved.

Substituting (31) into (6) and using (13), we have

$$\begin{equation} E_1(l;u,v)=\frac{(-1)^k}{\sqrt{l}}(2\pi)^{2u}l^{u+v}\sum_{ n=1}^{l-1} \sigma_{-2v}(n)\sigma_{-2u}(l-n)\biggl(\frac{n}{l}\biggr)^{-1/2+u+v} I_1\biggl(\frac{n}{l}\biggr). \end{equation} \tag{ 66 }$$

Assuming Conjecture 1.2, the main term $\mathrm{MT}_1(l;u,v)$ of (66) is given by

$$\begin{equation} \mathrm{MT}_1(l;u,v) =\frac{(-1)^k}{\sqrt{l}}(2\pi)^{2u}l^{u+v} \int_0^{1}x^{-1/2+u+v}I_1(x)\rho_l(x;-2v,-2u)\,dx \end{equation} \tag{ 67 }$$

(see (18), (19) and (21)). Let

$$\begin{equation} \begin{aligned} \, \notag S_1(u,v) &=-\frac{2\Gamma(1-2u)}{(k-1+u-v)(k-u+v)}\biggl( (-1)^k\cos(\pi u) \\ &\qquad +\cos(\pi v)\frac{\Gamma(1+2v)\Gamma(k-u-v)}{\Gamma(1-2u)\Gamma(k+u+v)}\biggr) \end{aligned} \end{equation} \tag{ 68 }$$

and

$$\begin{equation} \begin{aligned} \, \notag S_2(u,v) &=-2(-1)^k\Gamma(1+2u)\cos(\pi u) \frac{\Gamma(k-1-u-v)\Gamma(k-u+v)}{\Gamma(k+1+u+v)\Gamma(k+u-v)} \\ &\qquad -2\Gamma(1+2v)\cos(\pi v)\frac{\Gamma(k-1-u-v)}{\Gamma(k+1+u+v)}. \end{aligned} \end{equation} \tag{ 69 }$$

Lemma 4.1.  The following identity holds

$$\begin{equation} \begin{aligned} \, \notag \mathrm{MT}_1(l;u,v) &=\frac{(-1)^k}{\sqrt{l}}(2\pi)^{2u}l^{u+v}\biggl( \sigma_{1-2u-2v}(l)\frac{\zeta(1-2u)\zeta(1-2v)}{\zeta(2-2u-2v)}S_1(u,-v) \\ \notag &\qquad +l^{-2v}\sigma_{1-2u+2v}(l)\frac{\zeta(1-2u)\zeta(1+2v)}{\zeta(2-2u+2v)}S_1(u,v) \\ \notag &\qquad +l^{-2u}\sigma_{1+2u-2v}(l)\frac{\zeta(1+2u)\zeta(1-2v)}{\zeta(2+2u-2v)}S_2(u,-v) \\ &\qquad +l^{-2u-2v}\sigma_{1+2u+2v}(l)\frac{\zeta(1+2u)\zeta(1+2v)}{\zeta(2+2u+2v)}S_2(u,v) \biggr). \end{aligned} \end{equation} \tag{ 70 }$$

Proof.  The proof is similar to that of Lemma 3.1. We substitute (33) and (23) into (67). In this way we obtain four terms that can be evaluated in the same way. We consider the first and fourth terms. The first term is given by

$$\begin{equation} \begin{aligned} \, \notag &\sigma_{1-2u-2v}(l)\frac{\zeta(1-2u)\zeta(1-2v)}{\zeta(2-2u-2v)} \frac{1}{2\pi i}\int_{(\sigma)}\Gamma(k,u,v;s)\sin\biggl(\pi\biggl(u+\frac s2\biggr)\biggr) \\ &\qquad\qquad \times\int_0^{1}x^{-1/2+u-v+s/2}(1-x)^{-2u}\,dx\,ds. \end{aligned} \end{equation} \tag{ 71 }$$

To evaluate the integral with respect to $x$ we apply (30). In doing so, we obtain

$$\begin{equation*} \begin{aligned} \, &\frac{\Gamma(1-2u)}{2\pi i}\int_{(\sigma)}\sin\biggl(\pi\biggl(u+\frac s2\biggr)\biggr) \frac{\Gamma(k-1/2+s/2)}{\Gamma(k+1/2-s/2)}\Gamma\biggl(\frac 12-u-v-\frac s2\biggr) \\ &\qquad\qquad \times\Gamma\biggl(\frac 12-u+v-\frac s2\biggr) \frac{\Gamma(1/2+u-v+s/2)}{\Gamma(3/2-u-v+s/2)}\,ds. \end{aligned} \end{equation*}$$

Moving the line of integration to the left and evaluating the residues at $s={1-2k-2j}$, $s=-1-2u+2v-2j$ for $j=0,1,\dots$, we show that

$$\begin{equation} \begin{aligned} \, \notag &2(-1)^k\Gamma(1-2u)\cos(\pi u)\sum_{j=0}^{\infty} \frac{\Gamma(k-u+v+j)\Gamma(k-u-v+j)\Gamma(1-k+u-v-j)}{\Gamma(2k+j)\Gamma(2-k-u-v-j)j!} \\ &\qquad\qquad -2\Gamma(1-2u)\cos(\pi v)\sum_{j=0}^{\infty} \frac{\Gamma(k-1-u+v-j)\Gamma(1-2v+j)\Gamma(1+j)}{\Gamma(k+1+u-v+j)\Gamma(1-2u-j)j!}. \end{aligned} \end{equation} \tag{ 72 }$$

Consider the first sum in (72). Applying (26) twice, we prove that it is equal to

$$\begin{equation} \begin{aligned} \, \notag &2(-1)^k\Gamma(1-2u)\frac{\cos(\pi u)\sin\pi(u+v)}{\sin\pi(u-v)} \sum_{j=0}^{\infty} \frac{\Gamma(k-1+u+v+j)\Gamma(k-u-v+j)}{\Gamma(2k+j)j!} \\ \notag &\qquad =2(-1)^k\Gamma(1-2u)\frac{\cos(\pi u)\sin\pi(u+v)}{\sin\pi(u-v)} \frac{\Gamma(k-u-v)\Gamma(k-1+u+v)}{\Gamma(2k)} \\ \notag &\qquad\qquad \times F(k-u-v,k-1+u+v,2k;1) \\ &\qquad = \frac{\cos(\pi u)\sin\pi(u+v)}{\sin\pi(u-v)}\,\frac{2(-1)^k\Gamma(1-2u)}{(k-1+u+v)(k-u-v)}, \end{aligned} \end{equation} \tag{ 73 }$$

where we have used (28) to evaluate the hypergeometric function.

Consider the second sum in (72). Applying (26) twice, we show that it is equal to

$$\begin{equation} \begin{aligned} \, \notag &{-}2(-1)^k\Gamma(1\,{-}\,2u)\frac{\cos(\pi v)\sin(2\pi u)}{\sin\pi(u\,{-}\,v)} \sum_{j=0}^{\infty} \frac{\Gamma(1\,{-}\,2v\,{+}\,j)\Gamma(2u\,{+}\,j)\Gamma(1\,{+}\,j)}{\Gamma(k\,{+}\,1\,{+}\,u\,{-}\,v\,{+}\,j) \Gamma(2\,{-}\,k\,{+}\,u\,{-}\,v\,{+}\,j)j!} \\ \notag &\qquad =-2(-1)^k\Gamma(1-2u)\frac{\cos(\pi v)\sin(2\pi u)}{\sin\pi(u-v)}\, \frac{\Gamma(1-2v)\Gamma(2u)}{\Gamma(k+1+u-v)\Gamma(2-k+u-v)} \\ \notag &\qquad\qquad \times{}_3F_{2}(1,1-2v,2u;k+1+u-v,2-k+u-v;1) \\ \notag &\qquad =-2\Gamma(1-2u)\cos(\pi v)\frac{\Gamma(k-1-u+v)\Gamma(1-2v)}{\Gamma(k+1+u-v)\Gamma(1-2u)} \\ &\qquad\qquad \times {}_3F_{2}(1,1-2v,2u;k+1+u-v,2-k+u-v;1). \end{aligned} \end{equation} \tag{ 74 }$$

Applying [8], §7.4.4, formula 27 (for $a=1-2v,b=2u,c=2-k+u-v$), we obtain

$$\begin{equation} \begin{aligned} \, \notag & {}_3F_{2}(1,1-2v,2u;k+1+u-v,2-k+u-v;1)=\frac{1}{(k-u-v)(k-1+u+v)} \\ &\qquad\qquad \times\biggl((k+u-v)(k-1-u+v)+\frac{\Gamma(k+1+u-v)\Gamma(2-k+u-v)}{\Gamma(1-2v)\Gamma(2u)} \biggr). \end{aligned} \end{equation} \tag{ 75 }$$

Substituting (75) into (74) and using (26) twice, we prove that the second sum in (72) is equal to

$$\begin{equation} \begin{aligned} \, \frac{-2\Gamma(1-2u)\cos(\pi v)}{(k-u-v)(k-1+u+v)}\biggl( \frac{\Gamma(k-u+v)\Gamma(1-2v)}{\Gamma(k+u-v)\Gamma(1-2u)}+(-1)^k \frac{\sin(2\pi u)}{\sin\pi(u-v)} \biggr). \end{aligned} \end{equation} \tag{ 76 }$$

Summing (73) and (76), we conclude that the integral in (71) is equal to

$$\begin{equation} \begin{aligned} \, \notag &\frac{-2\Gamma(1-2u)}{(k-u-v)(k-1+u+v)}\biggl(\cos(\pi v) \frac{\Gamma(k-u+v)\Gamma(1-2v)}{\Gamma(k+u-v)\Gamma(1-2u)} \\ &\qquad\qquad +(-1)^k\frac{\sin(2\pi u)\cos(\pi v)-\cos(\pi u)\sin\pi(u+v)}{\sin\pi(u-v)}\biggr)=S_1(u,-v), \end{aligned} \end{equation} \tag{ 77 }$$

and therefore we recover the first summand in (70).

Let us briefly describe how we can obtain the fourth term in (70). In this case, we must compute

$$\begin{equation*} \begin{aligned} \, &\frac{1}{2\pi i}\int_{(\sigma)}\Gamma(k,u,v;s)\sin\biggl(\pi\biggl(u+\frac s2\biggr)\biggr) \int_0^{1}x^{-1/2+u+v+s/2}\,dx\,ds \\ &\qquad =\frac{1}{2\pi i}\int_{(\sigma)}\sin\biggl(\pi\biggl(u+\frac s2\biggr)\biggr) \frac{\Gamma(k-1/2+s/2)}{\Gamma(k+1/2-s/2)} \\ &\qquad\qquad \times\frac{\Gamma(1/2-u-v-s/2)\Gamma(1/2-u+v-s/2)}{1/2+u+v+s/2}\,ds. \end{aligned} \end{equation*}$$

Moving the line of integration to the left and evaluating the residues at $s={1-2k-2j}$ for $j=0,1,\dots$ and $s=-1-2u-2v$, we have

$$\begin{equation} \begin{aligned} \, \notag &{-}2\frac{\Gamma(k-1-u-v)}{\Gamma(k+1+u+v)}\Gamma(1+2v)\cos(\pi v) \\ \notag &\qquad\qquad -2(-1)^k\cos(\pi u)\sum_{j=0}^{\infty} \frac{\Gamma(k-u+v+j)\Gamma(k-1-u-v+j)}{\Gamma(2k+j)j!} \\ \notag &\qquad =-2\frac{\Gamma(k-1-u-v)}{\Gamma(k+1+u+v)}\Gamma(1+2v)\cos(\pi v) \\ \notag &\qquad\qquad -2(-1)^k\cos(\pi u)\frac{\Gamma(k-u+v)\Gamma(k-1-u-v)}{\Gamma(2k)} \\ &\qquad\qquad\times F(k-u+v,k-1-u-v,2k;1)=S_2(u,v), \end{aligned} \end{equation} \tag{ 78 }$$

where we have applied (28) to evaluate the hypergeometric function. Lemma 4.1 is proved.

Next, we compute $\mathrm{MT}_1(l;0,0)$ using (70). For this goal, we study the properties of $S_1(u,v)$ and $S_2(u,v)$.

Lemma 4.2.  The following identities hold:

$$\begin{equation} S_1(0,0)=S_2(0,0)=-2\frac{1+(-1)^k}{k(k-1)}=-2(1+(-1)^k)Q, \end{equation} \tag{ 79 }$$

$$\begin{equation} \frac{\partial}{\partial v}S_1(0,v)\Big|_{v=0}= \frac{\partial}{\partial v}S_2(0,v)\Big|_{v=0}= -4Q(\psi(1)-\psi(k))-2(1+(-1)^k)Q^2, \end{equation} \tag{ 80 }$$

$$\begin{equation} \frac{\partial}{\partial u}S_1(u,0)\Big|_{u=0}= 4Q(\psi(k)+(-1)^k\psi(1))+2(1+(-1)^k)Q^2, \end{equation} \tag{ 81 }$$

$$\begin{equation} \frac{\partial}{\partial u}S_2(u,0)\Big|_{u=0}= 4Q\psi(k)-2(-1)^k\bigl(2\psi(1)-4\psi(k)\bigr)Q-2(1+(-1)^k)Q^2, \end{equation} \tag{ 82 }$$

$$\begin{equation} \begin{split} &\frac{\partial^2}{\partial u\,\partial v}S_1(u,v)\Big|_{u,v=0}= 8\psi(k)\bigl(\psi(1)-\psi(k)\bigr)Q \\ &\qquad\qquad +4(1+(-1)^k)(1+\psi(1))Q^2+4(1+(-1)^k)Q^3 \end{split} \end{equation} \tag{ 83 }$$

and

$$\begin{equation} \begin{split} & \frac{\partial^2}{\partial u\,\partial v}S_2(u,v)\Big|_{u,v=0}= 8\psi(k)\bigl(\psi(1)-\psi(k)\bigr)Q \\ &\qquad\qquad +4(1+(-1)^k)\bigl(2\psi(k)-1-\psi(1)\bigr)Q^2-4(1+(-1)^k)Q^3. \end{split} \end{equation} \tag{ 84 }$$

Proof.  Equality (79) is a direct consequence of (68) and (69). Furthermore, it follows from (68) and (69) that

$$\begin{equation} \begin{aligned} \, \notag &\frac{\partial}{\partial v}S_1(u,v)\Big|_{v=0} =-2\biggl(\frac{(1-2u)}{(k-u)^2(k-1+u)^2}\,\frac{\Gamma(k-u)}{\Gamma(k+u)} \\ &\qquad\qquad +\frac{2\psi(1)-\psi(k-u)-\psi(k+u)}{(k-u)(k-1+u)}\,\frac{\Gamma(k-u)}{\Gamma(k+u)}+ \frac{(-1)^k\Gamma(2-2u)\cos(\pi u)}{(k-u)^2(k-1+u)^2} \biggr) \end{aligned} \end{equation} \tag{ 85 }$$

and

$$\begin{equation} \begin{aligned} \, \notag &\frac{\partial}{\partial v}S_2(u,v)\Big|_{v=0}= -2(-1)^k\Gamma(1+2u)\cos(\pi u) \frac{\Gamma(k-1-u)\Gamma(k-u)}{\Gamma(k+1+u)\Gamma(k+u)} \\ \notag &\qquad\qquad\qquad \times\bigl(\psi(k-u)-\psi(k-1-u)+\psi(k+u)-\psi(k+1+u)\bigr) \\ &\qquad\qquad -2\frac{\Gamma(k-1-u)}{\Gamma(k+1+u)}\bigl(2\psi(1)-\psi(k-1-u)-\psi(k+1+u)\bigr). \end{aligned} \end{equation} \tag{ 86 }$$

Substituting $v=0$ in (85) and (86) we obtain (80) by applying the relation

$$\begin{equation} \psi(k-1)+\psi(k+1)=2\psi(k)-Q, \end{equation} \tag{ 87 }$$

which follows from (27), in the case of (86).

Applying (87), the proof of (81) and (82) follows directly from (68) and (69), respectively. Moreover, we prove (83) by computing the derivative of (85). To show that (84) holds, we compute the derivative of (86) and obtain

$$\begin{equation*} \begin{aligned} \, &\frac{\partial^2}{\partial u\,\partial v}S_2(u,v)\Big|_{u,v=0}= -2Q\bigl(-\psi(k-1)-\psi(k+1)\bigr)\bigl(2\psi(1)-\psi(k-1)-\psi(k+1)\bigr) \\ &\ {-}\,2Q\bigl(\psi'(k-1)-\psi'(k+1)\bigr)(1+(-1)^k) \\ &\ {-}\,2(-1)^kQ\bigl[2\psi(1)-2\psi(k)-\psi(k-1)-\psi(k+1)\bigr] \bigl[2\psi(k)-\psi(k-1)-\psi(k+1)\bigr]. \end{aligned} \end{equation*}$$

Using (87) and the relation

$$\begin{equation} \psi'(k-1)-\psi'(k+1)=\psi'(k)+\frac{1}{(k-1)^2}-\biggl[\psi'(k)-\frac{1}{k^2}\biggr]=2Q+Q^2, \end{equation} \tag{ 88 }$$

we finally prove (84). The lemma is proved.

Theorem 4.3.  We have

$$\begin{equation} \begin{aligned} \, \notag \mathrm{MT}_1(l;0,0) &=-(1+(-1)^k)\frac{2Q}{\zeta(2)\sqrt{l}}\sum_{d\mid l}d\log^2\frac{d^2}{l} \\ \notag &\qquad -(1+(-1)^k)\frac{2Q}{\zeta(2)\sqrt{l}}\bigl(2D_1+2\psi(1)-2\psi(k)+2Q\bigr) \sum_{d\mid l}d\log\frac{d^2}{l} \\ \notag &\qquad -(1+(-1)^k)\frac{2Q}{\zeta(2)\sqrt{l}} \biggl(D_1(D_1+2\psi(1)-2\psi(k))-4\biggl(\frac{\zeta'(2)}{\zeta(2)}\biggr)' \\ &\qquad +Q\bigl(2D_1+\psi(1)+\psi(2)+1-2\psi(k)+2Q\bigr)\biggr)\sum_{d\mid l}d, \end{aligned} \end{equation} \tag{ 89 }$$

where $Q=1/(k^2-k)$ and $D_1$ is defined in (55).

Proof.  Identity (89) is a consequence of (70). More precisely, we proceed similarly to the proof of Theorem 3.3 by starting from (41) and repeating all the steps required to derive (58). Note that in order to prove analytic continuation to the point $v=0$ in Theorem 3.3 we require that (63) holds. Now we need the analogue of this equation:

$$\begin{equation*} S_2(0,0)D(0)+\frac{1}{\zeta(2)}\,\frac{\partial}{\partial v}S_2(0,v)\Big|_{v=0}= S_1(0,0)D(0)+\frac{1}{\zeta(2)}\,\frac{\partial}{\partial v}S_1(0,v)\Big|_{v=0}, \end{equation*}$$

which is a consequence of (79) and (80). Finally, applying (56), we derive the analogue of (65):

$$\begin{equation} \begin{aligned} \, \notag &\mathrm{MT}_1(l;0,0) =\frac{(-1)^k}{\sqrt{l}}\,\frac{S_2(0,0)}{\zeta(2)}\sum_{d\mid l}d\log^2\frac{d^2}{l} \\ \notag &\qquad +\frac{(-1)^k}{\sqrt{l}}\,\frac{1}{\zeta(2)} \sum_{d\mid l}d\log\frac{d^2}{l} \biggl[2D_1S_2(0,0)+\frac{\partial}{\partial v}S_2(0,v)\Big|_{v=0} +\frac{1}{2}\,\frac{\partial}{\partial u}S_2(u,0)\Big|_{u=0} \\ \notag &\qquad -\frac{1}{2}\,\frac{\partial}{\partial u}S_1(u,0)\Big|_{u=0}\biggr] +\frac{(-1)^k}{\sqrt{l}}\,\frac{1}{\zeta(2)} \sum_{d\mid l}d \biggl[D_1\biggl(D_1 S_2(0,0)+\frac{1}{2}\,\frac{\partial}{\partial u}S_2(u,0)\Big|_{u=0} \\ \notag &\qquad -\frac{1}{2}\,\frac{\partial}{\partial u}S_1(u,0)\Big|_{u=0} +\frac{\partial}{\partial v}S_2(0,v)\Big|_{v=0}\biggr) -4\biggl(\frac{\zeta'(2)}{\zeta(2)}\biggr)'S_2(0,0) \\ &\qquad +\frac{1}{2}\,\frac{\partial^2}{\partial u\,\partial v}S_2(u,v)\Big|_{u=v=0}- \frac{1}{2}\,\frac{\partial^2}{\partial u\,\partial v}S_1(u,v)\Big|_{u=v=0} \biggr]. \end{aligned} \end{equation} \tag{ 90 }$$

Using (80)–(84), we obtain (89). This completes the proof.

According to the proof of [1], Lemma 6.1, we have

$$\begin{equation*} E_3(l;0,0)+E_2(l;0,0)=(1+(-1)^k)E_2(l;0,0). \end{equation*}$$

Thus

$$\begin{equation*} \mathrm{MT}_1(l;0,0)+\mathrm{MT}_2(l;0,0)+\mathrm{MT}_3(l;0,0)=\mathrm{MT}_1(l;0,0)+(1+(-1)^k)\mathrm{MT}_2(l;0,0). \end{equation*}$$

Now Theorem 1.3 follows from Theorems 3.3 and 4.3.