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 $n$.


$$\sum_{l_1=1}^{+\infty}\frac{1}{l_1^2+l_2^2}\geq \sum_{l_1=1}^{l_2}\frac{1}{l_1^2+l_2^2}=\frac{u_{l_2}}{l_2}\geq \frac{B}{l_2}$$ 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 $k\leq 3/2$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.