Can we find a closed form for $ \int\nolimits_{- \infty}^{\infty} \frac{\exp\left(-(a+bx)^2\right)}{1+\exp(x)}\mathrm dx$? 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$?
 A: Split integration region and expand $(1+\mathrm{e}^x)^{-1}$:
$$
  (1+\mathrm{e}^x)^{-1} = \left\{ 
    \begin{array}{ll} 
          \sum_{k \ge 0} (-1)^k \mathrm{e}^{-(k+1)x} & x > 0 \\ 
          \sum_{k \ge 0} (-1)^k \mathrm{e}^{k x} & x \le 0
     \end{array} \right.
$$
and then integrate term-wise, assuming $b>0$:
$$
   \int_0^\infty \mathrm{e}^{-(a+b x)^2} \mathrm{e}^{-(k+1) x} \, \mathrm{d} x =
      \frac{\sqrt{\pi}}{2 b} \exp \left( \frac{(k+1)(k+1+4 a b)}{4 b^2} \right) \mathrm{erfc}\left( a + \frac{k + 1}{2 b} \right)
$$
and
$$
   \int_{-\infty}^0 \mathrm{e}^{-(a+b x)^2} \mathrm{e}^{k x} \, \mathrm{d} x =
     \frac{\sqrt{\pi}}{2 b} \exp\left( \frac{k(k-4 a b)}{4 b^2} \right) \mathrm{erfc}\left( \frac{k}{2b} -a\right)
$$
A: Wolfram Alpha gives a result with $a=0$ of $\frac{\sqrt{\pi}}{2b}$ (note my $a$ is your $b^2$) but chokes when $a$ is nonzero.
A: 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
$$
 =
 e^\frac{1}{4b^2}\int_{-\infty}^\infty\frac{e^{-(-\frac{1}{2b}+bx)^2}}{1+e^{x}} dx. 
$$
A: The integral has a closed form when $a = 0$ and $b \neq 0$. (The integral diverges if $b=0$.) We have
$$
I = \int_{-\infty}^{\infty} \frac{e^{-b^2x^2}}{1+e^x} dx.
$$
Making the substitution $x \to -x$, we can rewrite it as
$$
I = \int_{-\infty}^{\infty} \frac{e^{-b^2x^2}}{1+e^{-x}} dx = \int_{-\infty}^{\infty} \frac{e^{-b^2x^2} \cdot e^x}{1+e^{x}} dx.
$$
Adding the two integrals and dividing by $2$,
$$
I = \frac{1}{2} \int_{-\infty}^{\infty} e^{-b^2x^2} dx = \frac{1}{2 |b|} \int_{-\infty}^{\infty} e^{-y^2} dy,
$$
through the substitution $y =|b|x$. Thus we end up with the familiar Gaussian integral, which evaluates to $\sqrt{\pi}$. Thus the given integral is equal to
$$
I = \frac{\sqrt{\pi}}{2|b|}.
$$
For general $a$, of course, this trick does not work. I think it should be difficult to compute the integral. 

Note. The same idea can be used to show that if $f(x)$ is a continuous even function and $c > 0$, then 
$$
\int_{-c}^c \frac{f(x)}{1+e^x} dx = \frac{1}{2} \int_{-c}^c f(x) dx = \int_0^c f(x) dx.  
$$
