The integral $\int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^x)^2} dx$. Let
$$T(n) = \int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^{x})^n} dx.$$
We have that
$$ T(0) = \sqrt{\pi} \text{ and } T(1) = \frac{\sqrt{\pi}}{2}$$
and also that
$$ T(3) = \tfrac{3}{2} T(2) - \frac{\sqrt{\pi}}{4}.$$
Can we find a closed form for $T(2)$? It would also give us $T(3)$. Perhaps something in terms of special functions?
 A: An expression for $T(2)$ as a series of Bernoulli numbers:
\begin{align}
T(2) &= \int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^{x})^2} d\,dx\\
 &=\frac{1}{4}\int_{-\infty}^\infty \frac{e^{-x^2-x}}{\cosh^2 \frac{x}{2}} \,dx\\
&=\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2-2x}}{\cosh^2 x} \,dx\\
 &=\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2+2x}}{\cosh^2 x} \,dx
\end{align}
Summing the last two expressions and using the formula $\cosh 2x=2\cosh^2x-1$, it comes
\begin{align}
T(2)&=\frac{1}{2}\int_{-\infty}^\infty \frac{\cosh 2x}{\cosh^2 x}e^{-4x^2}\,dx\\
&=\int_{-\infty}^\infty e^{-4x^2}\,dx-\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2}}{\cosh^2 x}\,dx\\
&=\frac{\sqrt{\pi}}{2}-\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2}}{\cosh^2 x}\,dx
\end{align}
We integrate by parts,
\begin{equation}
T(2)=\frac{\sqrt{\pi}}{2}-4\int_{-\infty}^\infty xe^{-4x^2}\tanh x\,dx
\end{equation} 
Now, using the series expansion for $\tanh$ DLMF:
\begin{equation}
\tanh z=\sum_{n=1}^\infty\frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}z^{2n-1}
\end{equation} 
one can express, 
\begin{align}
T(2)&=\frac{\sqrt{\pi}}{2}-4\sum_{n=1}^\infty\frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}\int_{-\infty}^\infty x^{2n}e^{-4x^2}\,dx\\
&=\frac{\sqrt{\pi}}{2}-2\sqrt{\pi}\sum_{n=1}^\infty\frac{2^{2n}-1}{2^{2n}}\frac{B_{2n}}{n!}
\end{align} 
where we have used the tabulated formula
\begin{equation}
\int_0^\infty x^{2n}e^{-px^2}\,dx=\frac{(2n-1)!!}{2(2p)^n}\sqrt{\frac{\pi}{p}}
\end{equation} 
