sequences and series - Help with an infinite sum of exponential terms?

I've been trying to calculate the mean squared displacement of a particle confined to a one-dimensional box, and I managed to get an answer in terms of an infinite series of the basic form
$$
\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\exp(-an^2)\;\;.

$$
I can't figure out if this series has a simple solution; I can't find it in any table of series, nor can I seem to expand the exponential and collect like terms without running into $\sum(-1)^2$ terms. Does this have a solution or is this as far as I can go? Thanks in advance.

Answer

By using the Jacobian Theta function the resulting series can be placed into the form
\begin{align}
f(a) = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \, e^{- a \, n^{2}} = \frac{a}{2} - \frac{\pi^{2}}{12} - \frac{1}{2} \, \int_{0}^{a} \theta\left(\frac{1}{2}, \frac{i u}{\pi}\right) \, du.
\end{align}

This is developed by the following. Differentiate $f(a)$ to obtain
$$f^{'}(a) = \sum_{n=1}^{\infty} (-1)^{n+1} \, e^{-a \, n^{2}}.$$

Now,
$$ \theta(x, it) = 1 + 2 \, \sum_{n=1}^{\infty} e^{- \pi \, t \, n^{2}} \, \cos(2 n \pi x) $$
which yields
$$ f^{'}(a) = \frac{1}{2} \, \left( 1 - \theta\left(\frac{1}{2}, \frac{i \, a}{\pi}\right) \right).$$
Integrating with respect to $a$ yields
$$f(a) = \frac{a}{2} - \frac{1}{2} \, \int_{0}^{a} \theta\left(\frac{1}{2}, \frac{i \, u}{\pi}\right) \, du + c_{0}$$
Since $f(0) = - \frac{1}{2} \, \zeta(2)$ then the presented result is obtained.