Integration $\int_{-\infty}^{\infty} \frac{e^{-a^2x^2}}{1+e^{x+\eta}} dx$.

Question

I've encountered the following integral $\displaystyle \int_{-\infty}^{\infty} \frac{e^{-a^2x^2}}{1+e^{x+\eta}} dx$. When $\eta=0$, I have managed to solve it. However, when $\eta \neq 0$ I have not.

@Jack D'Aurizio has answered the question here:

$\displaystyle \int_{-\infty}^{+\infty}\frac{e^{-a^2 x^2}}{1+e^{x+\eta}}\,dx = \displaystyle \int_{-\infty}^{+\infty}\frac{e^{-a^2 (x-\eta)^2}}{1+e^x}\,dx = \displaystyle e^{-a^2\eta^2}\int_{0}^{\infty}e^{-a^2 x^2}\left(\frac{e^{2a^2\eta x}}{1+e^x}+\frac{e^{-2a^2\eta x}}{1+e^{-x}}\right)\,dx$

in combination with a geometric series expansion: $\frac{e^{\pm 2a^2\eta x}}{1+e^{\pm x}}=\sum_{m\geq 0}(-1)^m e^{\pm(2a^2\eta+m)x},\qquad \int_{0}^{+\infty}e^{-a^2 x^2}\cosh(\mu x)\,dx=\frac{\sqrt{\pi}}{2a}\,e^{\frac{\mu^2}{4a^2}}$.

However, I can't understand one step. Namely, I do not understand the step where $$ I=\frac{e^{\pm 2a^2\eta x}}{1+e^{\pm x}}=\sum_{m\geq 0}(-1)^m e^{\pm(2a^2\eta+m)x}.$$

Following the tip from @Kurt G. I ended up with the following:

1. $ \displaystyle \frac{e^{ 2a^2\eta x}}{1+e^{x}}=\sum_{m=0}^{\infty}(-1)^m e^{(2a^2\eta+m)x}, \text {for } e^{x} < 1$. Since the term (I) is part of an integral from 0 to $\infty$ why it is assumed that $e^{x} < 1$?

2. $\displaystyle \frac{e^{(-2a^2\eta \color{red}{ + 1}\color{black}{)x}}}{1+e^{x}}=\sum_{m=0}^{\infty}(-1)^m e^{-(2a^2\eta+m)x}, \text {for } e^{-x} < 1$ (because $x>0 \rightarrow -x<0 \rightarrow e^{-x}<1$). The red term is not given in (I). Is it skipped or I missed something?

Thank you very much.

Answer 1

As commented by the author, the approach with $\int_0^\infty e^{-a^2x^2}\cosh bx\,dx$ doesn't quite work.

For $x\geqslant 0$, we cannot expand $1/(1+e^x)$ as the geometric series (in $-e^x$), not even saying about termwise integration. An alternative (sketched also in my answer I'm linking to below) would be to use $1/(1+e^x)=e^{-x}/(1+e^{-x})$ and the geometric series (now in $-e^{-x}$), but this way we end up with a series involving the error function.

Observe that the given integral equals $-2\pi e^{-a^2(i\pi-\eta)^2}I(2\pi a^2(i\pi-\eta),4i\pi a^2)$ where $$I(z,\tau)=\int_{-\infty-i\epsilon}^{\infty-i\epsilon}\frac{e^{i\pi\tau w^2-2zw}}{e^{2\pi w}-1}\,dw,\qquad\left\{\begin{gathered}z,\tau\in\mathbb{C}\\\Im\tau>0\\0<\epsilon<1\end{gathered}\right.$$ is one of the integrals studied by L. J. Mordell, see this answer of mine for a reference.