Find the value $\sum_{m=1}^{\infty}{\frac{e^{-a m^2}}{m^2}}$ How to find the following series' value?
$$\sum_{m=1}^{\infty}{\frac{e^{-a m^2}}{m^2}}$$
I know that this series converges. I check it by ratio test or comparison test for 
$$  a \in \left [ 10^{-15} , 10^{15}\right ]$$ 
(I need "a" just in this interval.)
thanks.
 A: There is no 'closed-form expression' for the series 
$$f(a):=\sum_{m=1}^{\infty}{\frac{e^{-a\,m^2}}{m^2}}$$
since this would imply that the derivative, essentially a Jacobi theta function, admits a closed-form too!
Concerning lower and upper bounds you may consider for example :
$$e^{-a}\ <\ f(a)\ \le\ \frac{\pi^2}6+\frac a2-\sqrt{\pi\,a}\quad\quad \text{for}\ a\ge 0$$
The lower bound is very precise for $a\gg 1$ (for more precision use $\,e^{-a}+\frac 14 e^{-4a}\left[+\frac 19e^{-9a}+\cdots\right]$)
The upper bound is very precise for $a\ll 1$ (the error is of order $\,\left(\frac a{\pi}\right)^{\frac 32}e^{-\frac{\pi^2}a}$ as we will see) and may be obtained from the previous lower bound using the functional equation from the derivative $f'(a)=-\psi\bigl(\frac a{\pi}\bigr)$ where $\psi$ is the Jacobi theta function defined by (with Riemann's notation) :
$$\psi(x)=\sum_{n=1}^\infty e^{-\pi\, x\, n^2}$$
and the functional equation (previous reference or tag $(43)$ from the link) is :
$$1+2\,\psi(x)=\frac 1{\sqrt{x}}\left(1+2\,\psi\left(\frac 1x\right)\right)$$
Since the lower bound is straightforward let's obtain the upper bound :
\begin{align}
f(a)&=C-\int \psi\left(\frac a{\pi}\right) da=C+\frac 12\int 1-\sqrt{\frac{\pi}a}\left(1+2\,\psi\left(\frac {\pi}a\right)\right) da\\
&=C+\frac 12\int 1-\sqrt{\frac{\pi}a}\left(1+2 \sum_{m=1}^\infty\frac{e^{-\frac {\pi^2\,m^2}a}}{m^2}\right) da\\
&=\frac{\pi^2}6+\frac a2-\sqrt{\pi\,a}+2\sum_{m=1}^\infty \left(\frac{\pi^2}m\operatorname{erfc}\left(\frac{\pi\,m}{\sqrt{a}}\right)-\sqrt{\pi\,a}\frac{e^{-\frac {\pi^2\,m^2}a}}{m^2}\right)\\
&\sim \frac{\pi^2}6+\frac a2-\sqrt{\pi\,a}+2\sqrt{\pi\,a}\sum_{m=1}^\infty \frac{e^{-\frac {\pi^2\,m^2}a}}{m^2}\sum_{n>0}(-1)^n\,(2n-1)!!\left(\frac a{2\pi^2}\right)^n\\
\end{align}
using the asymptotic expansion of the complementary error function '$\operatorname{erfc}$' (where the $n=0$ term disappears and only a few $n>0$ terms may be used).
The double sum at the right is negative so that the upper bound is indeed :
$$\frac{\pi^2}6+\frac a2-\sqrt{\pi\,a}$$
The most important contribution from the double sum (for $m=n=1$) would subtract $\,\displaystyle\left(\frac a{\pi}\right)^{3/2}e^{-\frac {\pi^2}a}$ to this expression.
Hoping all this this helped more,
