TITLE: Can we find a closed form for $ \int\nolimits_{- \infty}^{\infty} \frac{\exp\left(-(a+bx)^2\right)}{1+\exp(x)}\mathrm dx$?
QUESTION [18 upvotes]: Can we find a closed form for this definite integral:
$$ 
\int\nolimits_{- \infty}^{\infty} \frac{\exp\left(-(a+bx)^2\right)}{1+\exp(x)}\mathrm dx
$$

REPLY [4 votes]: The result of Srivatsan
$$\frac{\sqrt{\pi}}{2|b|} = \int_{-\infty}^{\infty} \frac{e^{-b^2x^2}}{1+e^{x}} dx = \int_{-\infty}^{\infty} \frac{e^{-b^2x^2} \cdot e^x}{1+e^{x}} dx$$
gives the case $a=-\frac{1}{2b}$ because this is
$$
 =
 e^\frac{1}{4b^2}\int_{-\infty}^\infty\frac{e^{-(-\frac{1}{2b}+bx)^2}}{1+e^{x}} dx. 
$$