Let $N>0$ an integer, $k>0$ a real parameter and let $\rho = \beta +i \gamma$ a non trivial zero of the Riemann zeta function. For a work I need to find the best possible $k$ such that $$I=\sum\_{l\_{1}\geq1}\sum\_{l\_{2}\geq1}\sum\_{\gamma>0}\gamma^{-k-3/2}\int\_{0}^{\gamma}e^{-N\left(l\_{1}^{2}+l\_{2}^{2}\right)v^{2}/\gamma^{2}}e^{-v}v^{k+\beta}dv \tag{1}$$ converges. I'm able to prove that $(1)$ converges if $k>3/2$ using some trivial bounds, but I would like to prove the convergence for $k>1$. And I'm stuck. Is it possible? Thank you.


**P.S.** I would prefer a non conditional answer, but I will accept also answer with the assumption of the Riemann hypothesis (namely $\beta=1/2$).



Convergence of a triple sum involving the imaginary part of the Riemann zeta function's non trivial zeros
Notation: write $ A \approx B$ for $(A\ll B) \wedge (B\ll A)$.


It's very tempting to apply Poisson sum, so we set up the following formalism:


We'll apply the identity (Poisson summation formula)
$$\sum\_{\ell\in\mathbb{Z}} f(\ell) = \sum\_{k\in\mathbb{Z}} \hat{f}(k)$$ to the Gaussian $f(x) = \exp(-zx^2)$ with Fourier transform $\hat{f}(k) = \sqrt{\frac{\pi}{z}}\exp (-\frac{4\pi^2}{z}k^2)$. This will allow us to control the blowup in the original equation of the sum over $\underline{\ell}$ when $v$ is close to zero.


Accordingly let $\Theta(z)=\sum\_{\ell\in\mathbb{Z}} f(\ell) = \sum\_{\ell\in\mathbb{Z}} \exp(-z\ell ^2)$. Our identity then reads
$$\Theta(z) = \sqrt\frac{\pi}{z}\Theta\left(\frac{4\pi^2}{z}\right)$$


Now for an integer $d$ let
\begin{align\*} S\_d(k) & = \sum\_{\underline{\ell}\in\mathbb{Z}^d}\sum\_{\gamma>0} \gamma^{-k-3/2}\int\_0^\gamma e^{-N\Vert\underline{\ell}\Vert^2v^2/\gamma^2}e^{-v}v^{k+\beta}dv \\
 & = \sum\_{\gamma>0} \gamma^{-k-3/2}\int\_0^\gamma \left(\Theta\left(Nv^2/\gamma^2\right)\right)^d e^{-v} v^{k+\beta}dv\,.
\end{align\*}


We will show that $S\_d(k)$ converges iff $k>d-1/2$.


To begin, apply the identity above and get
\begin{align\*} S\_d(k) & = \sum\_{\gamma>0} \gamma^{-k-3/2} \int\_0^\gamma \left(\frac{\pi\gamma^2}{Nv^2}\right)^{d/2} \left(\Theta\left(\frac{4\pi^2\gamma^2}{Nv^2}\right)\right)^d e^{-v} v^{k+\beta}dv\\
& \approx \sum\_{\gamma>0} \gamma^{d-k-3/2} \int\_0^\gamma \left(\Theta\left(\frac{4\pi^2\gamma^2}{Nv^2}\right)\right)^d e^{-v} v^{k+\beta}dv\\
\end{align\*}


Having extracted the main terms $\gamma^d v^{-d}$, we have $\left(\Theta\left(\frac{4\pi^2\gamma^2}{Nv^2}\right)\right)^d \approx 1$, since $\Theta(z)$ is continuous on $\left[\frac{4\pi^2}{N},\infty\right)$, is at least $1$ everywhere, and satisfies $\lim\_{z\to\infty}\Theta(z) = 1$.


It follows that
$$S\_d \approx \sum\_{\gamma>0} \gamma^{d-k-3/2} \int\_0^\gamma e^{-v} v^{k+\beta-d}dv \,.$$


We consider divergence first, using only the zeroes with $\beta = 1/2$. We must then have $k-d+1/2 > -1$ or the integral would diverge at $v=0$. Assume then that $k > d-3/2$. Then $\lim\_{\gamma\to\infty} \int\_0^\gamma e^{-v} v^{k+\beta-d}dv = \Gamma(k-d+3/2)$ and hence the integral is $\gg 1$. We conclude that
$$S\_d(k) \gg \sum\_{\beta=1/2} \gamma^{d-k-3/2} \gg \int\_1^\infty \gamma^{d-k-3/2}\log \gamma d\gamma$$
which diverges when $d-k-3/2\geq -1$, that is when $k\leq d-1/2$, since a positive proportion of the zeroes are on the line.


For convergence when $k> d-1/2$ we extend the integral to infinity getting
$$S\_k(d) \ll \sum\_{\gamma>0} \gamma^{d-k-3/2} \Gamma(k-d+\beta+1)\,.$$
Since $k-d+1>1/2$ and $0<beta<1$, the numbers $\Gamma(k-d+\beta+1)$ are uniformly bounded and since $d-k-3/2 < -1$ the sum over the zeroes converges.


Finally, the four times the indicated sum is $S\_2(k) - 2 S\_1(k) + 4 S\_0(k)$. For $k\in[1,3/2]$, the sums $S\_1(k),S\_0(k)$ converge while $S\_2(k)$ diverges so the indicated sum must diverge.


