What is the series $\sum_{n=1}^{\infty} \frac{e^{-n^2 x}}{n}$? Following Passare: How to compute $\sum 1/n^2$ by solving triangles
 I tried the following
$$
\int_0^{\infty}\frac{e^{-nx}}{n^2} dx = \frac{1}{n^3}
$$
So we can write (with some help of Wolfram Alpha)
$$
\sum_{n=1}^{\infty} \int_0^{\infty}\frac{e^{-nx}}{n^2} dx =  \int_0^{\infty} \sum_{n=1}^{\infty} \frac{e^{-nx}}{n^2} dx = \int_0^{\infty} Li_2(e^{-x}) dx = \zeta(3)
$$
where $Li_2$ is the https://en.wikipedia.org/wiki/Polylogarithm#Dilogarithm . 
But is also true that
$$
\int_0^{\infty}\frac{e^{-n^2x}}{n} dx = \frac{1}{n^3}
$$
so that one can write
$$
\sum_{n=1}^{\infty} \int_0^{\infty}\frac{e^{-n^2x}}{n} dx =  \int_0^{\infty} \sum_{n=1}^{\infty} \frac{e^{-n^2x}}{n} dx = \int_0^{\infty} ?? dx = \zeta(3)
$$
The problem here is the evaluation of the series
$$
\sum_{n=1}^{\infty} \frac{e^{-n^2x}}{n} = ??
$$
which I (and also Wolfram Alpha) don't know how to evaluate. Is this series known in the literature and is there any way to evaluate it or express it somehow in terms of some special functions?
 A: Let's replace:
$$x=y^2/4$$
Then we have:
$$e^{-n^2 y^2/4}= \frac{1}{\sqrt{\pi}} \int_{-\infty}^\infty e^{-t^2+i n y t} dt$$
Thus, provided the integral exists, we should have:
$$g(y)=\sum_{n=1}^\infty \frac{e^{-n^2 y^2/4}}{n}=-\frac{1}{\sqrt{\pi}} \int_{-\infty}^\infty e^{-t^2} \log \left(1-e^{i y t}  \right) dt$$
Extracting the real part and using symmetry, we obtain:
$$g(y)=-\frac{1}{\sqrt{\pi}} \int_0^\infty e^{-t^2} \log \left(1-\cos (y t)  \right) dt- \frac{\log 2}{2}$$
Now getting back to $x$:
$$f(x)=\sum_{n=1}^\infty \frac{e^{-n^2 x}}{n}$$
$$f(x)=- \frac{\log 2}{2}-\frac{1}{\sqrt{\pi}} \int_0^\infty e^{-t^2} \log \left(1-\cos (2 t \sqrt{x})  \right) dt$$
or:
$$f(x)=-\frac{1}{\sqrt{\pi}} \int_0^\infty e^{-t^2} \log \left(2-2\cos (2 t \sqrt{x})  \right) dt$$
This works numerically, even though the integrand has an infinite number of singularities.
Let's substitute:
$$t= \sqrt{x} u$$
$$f(x)=-\frac{\sqrt{x}}{\sqrt{\pi}} \int_0^\infty e^{-x u^2} \log \left(2-2\cos (2 u x)  \right) du$$
Now there might be a chance to integrate w.r.t. $x$ as well:
$$I=-\frac{1}{\sqrt{\pi}} \int_0^\infty \int_0^\infty  \sqrt{x} e^{-x u^2} \log \left(2-2\cos (2 u x)  \right) du dx$$
But I somehow doubt the integral converges, though I will check numerically later.
