I've encountered the following integral
$$\int_{-\infty}^{\infty} \frac{e^{-a^2x^2}}{1+e^{x+\eta}} dx$$
I tried mathematica, but it cannot be solved. I also tried by setting $a$ and $\eta$ to be different numbers, but mathematica still returns no results. Is this because the integral is not convergent for any vales of $a$ and $\eta$?
If this integral is convergent with certain restraint for $a$ and $\eta$, can anyone help derive this integral?
Thanks for any help.
If $\eta=0$ $$ \int_{-\infty}^{+\infty}\frac{e^{-a^2 x^2}}{1+e^{x}}\,dx = \int_{0}^{+\infty}e^{-a^2 x^2}\left(\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right)\,dx = \int_{0}^{+\infty} e^{-a^2 x^2}\,dx = \frac{\sqrt{\pi}}{2a}$$ and a similar approach can be applied when $\eta\neq 0$:
$$ \int_{-\infty}^{+\infty}\frac{e^{-a^2 x^2}}{1+e^{x+\eta}}\,dx = \int_{-\infty}^{+\infty}\frac{e^{-a^2 (x-\eta)^2}}{1+e^x}\,dx = e^{-a^2\eta^2}\int_{0}^{\infty}e^{-a^2 x^2}\left(\frac{e^{2a^2\eta x}}{1+e^x}+\frac{e^{-2a^2\eta x}}{1+e^{-x}}\right)\,dx $$ in combination with a geometric series expansion: $$ \frac{e^{\pm 2a^2\eta x}}{1+e^{\pm x}}=\sum_{m\geq 0}(-1)^m e^{\pm(2a^2\eta+m)x},\qquad \int_{0}^{+\infty}e^{-a^2 x^2}\cosh(\mu x)\,dx=\frac{\sqrt{\pi}}{2a}\,e^{\frac{\mu^2}{4a^2}}.$$