# About the convergence of an integral and a double series

Let $k>0$ a real number and $N>0$ a natural number. For a work, I need to prove the convergence of $$\sum_{l_{1}\geq1}\sum_{l_{2}\geq1}\int_{1/N}^{\infty}\log^{2}\left(2Ny\right)\frac{e^{-\left(l_{1}^{2}+l_{2}^{2}\right)/\left(Ny^{2}\right)}}{y^{k+3/2}}dy.$$ Question: I'm able to prove that this object converges if $k>3/2$, but I would prove the convergence for $k>1$. Is it possible?

I tried to write it in terms of gamma function or a Gaussian integral but I'm not able to broke the wall of $3/2$. Thank you.

Put $\displaystyle c=(l_1^2+l_2^2)/N$. Suppose $k\leq 3/2$. We have:

$$I=\int_{1/N}^{+\infty}\log(2Ny)^2\frac{\exp(-c/y^2)}{y^{k+3/2}}dy \geq \int_{1}^{+\infty}\log(2Ny)^2\frac{\exp(-c/y^2)}{y^{k+3/2}}dy$$

Now, as $2Ny\geq 2N$ if $y\geq 1$ and $k\leq 3/2$, we get: $$I\geq (\log(2N))^2\int_{1}^{+\infty}\frac{\exp(-c/y^2)}{y^{k+3/2}}dy \geq (\log(2N))^2\int_{1}^{+\infty}\frac{\exp(-c/y^2)}{y^{3}}dy$$

As the derivative of $\displaystyle \exp(-c/y^2)$ is $\displaystyle \frac{2c}{y^3}\exp(-c/y^2)$, we get, as $c\geq 2/N$: $$I\geq \frac{(\log(2N))^2}{2c}(1-\exp(-c))\geq \frac{(\log(2N))^2}{2c}(1-\exp(-2/N))=\frac{A}{l_1^2+l_2^2}$$

for a positive constant $A$ independant of $l_1,l_2$.

Now put for $n\geq 1$, $\displaystyle u_n=\frac{1}{n}\sum_{1}^n \frac{1}{1+(k/n)^2}$. Then $u_n>0$ for all, and $\displaystyle u_n\to \int_0^1\frac{dt}{1+t^2}$ as $n\to +\infty$. Hence there exists a positive constant $B$ such that $u_n\geq B>0$ for all $n$.

Now:

$$\sum_{l_1=1}^{+\infty}\frac{1}{l_1^2+l_2^2}\geq \sum_{l_1=1}^{l_2}\frac{1}{l_1^2+l_2^2}=\frac{u_{l_2}}{l_2}\geq \frac{B}{l_2}$$ Hence we get $$\sum_{l_1=1}^{+\infty}\sum_{l_2=1}^{+\infty}\frac{1}{l_1^2+l_2^2} =+\infty$$

and your expression is divergent for all $k\leq 3/2$.

• This is very nice!!! Thank you so much! – User Oct 28 '15 at 6:08