An integral I just cannot do

Question

$$\int\sum_{n=-\infty}^\infty e^{-\pi{n^2}/ x} dx $$ I tried doing this $$\sum_{n=-\infty}^\infty \int e^{-\pi{n^2}/ x} dx $$ and then $$\sum_{n=-\infty}^\infty {-\pi{n^2} \ln(x) e^{(-\pi{n^2}/ x) +1}\over (-\pi{n^2}/ x) +1 } dx $$ and now I'm not sure what to do..how do I sum this up and if I take the definite integral do I sum it up first or evaluate at $0$ and $\infty$

$$\int_0^\infty \sum_{n=-\infty}^\infty e^{-\pi{n^2}/ x} dx $$ Does this integral diverge? If so,
What about this one $$\int_0^\infty \frac 1 {\sum_{n=-\infty}^\infty e^{-\pi{n^2}/ x}} dx $$

Answer 1

Note that when $x$ is small and negative the argument of the exponential is huge and positive. We can pick one of the terms in the sum, say $n=1$ and feed it to Alpha to be told it does not converge. The integral from $0$ to $\infty$ also does not converge because the integrand goes to $1$.

Answer 2

$$x>\epsilon\implies e^{-n^2\pi/x}>e^{-n^2\pi/\epsilon}$$

This is sufficient for every term of the summation to cause divergence.

Answer 3

The inner sum can be represented by a theta function: $$\sum_{n=-\infty}^{\infty} q^{n^2} = \mathcal{\theta}_3(0,q).$$

See http://mathworld.wolfram.com/JacobiThetaFunctions.html and https://en.wikipedia.org/wiki/Theta_function for documentation.

So $$\sum_{n=-\infty}^{\infty} e^{-\pi n^2/x} = \mathcal{\theta}_3(0,e^{-\pi/x}).$$

The first definite integral you write definitely diverges. This can be seen from the arguments of the other commenters (it's worth noting that they are switching the order of the sum and integral, which is allowed by Tonelli's theorem because all terms of the series are positive). It can also be seen by a quick look at the shape of the function on Wolfram Alpha. I'm not, however, so sure about the second integral.