The integral $\int\limits_{-\infty}^{\infty}e^{-x^2-e^x}dx$ converges. This can be seen e.g. since $0<e^{-e^x}<1$ for all $x$ and therefore $$ \int\limits_0^ae^{-x^2}e^{-e^x}dx <\int\limits_{0}^{a}e^{-x^2}dx $$ and since the latter converges, also the former, and the same for the $-\infty$ bound.
I am trying to evaluate this integral. I thought of doing a series expansion: $$ e^{-e^x} = \sum\limits_{n=0}^{\infty}\frac{(-1)^n}{n!}e^{nx} $$ The exponential function is analytic, so this series converges for all $x$. When plugging in, I get $$ \int\limits_{-\infty}^{\infty}\sum\limits_{n=0}^{\infty}\frac{(-1)^n}{n!}e^{-x^2+nx}dx = \sum\limits_{n=0}^{\infty}\frac{(-1)^n}{n!}\int\limits_{-\infty}^{\infty}e^{-x^2+nx}dx = \sum\limits_{n=0}^{\infty}\frac{(-1)^n}{n!}e^{\frac{n^2}{4}} $$ but this series diverges, e.g. from Stirling's approximation, $n!\sim n^{n+1/2}/e^n = \exp\left({(n+0.5)\ln{n}-n}\right)$ and therefore this fails convergence test.
What's going on? Is there something wrong with exchanging the integral and the summation? Shouldn't this series be uniformly convergent?