How to find the integral $\int_{-\infty}^{\infty}\frac{dx}{1+ae^{bx^2}}$ Could somebody tell me how to find the integral
$$\int_{-\infty}^{\infty}\frac{dx}{1+ae^{bx^2}}$$
for constants $a$ and $b$?
Thanks!
 A: I will assume that $a, b$ are positive constants, in order to circumvent singularity issues. By the substitution $z = \sqrt{b} \, x$, the integral in question is equal to
$$ \frac{1}{\sqrt{b}} \int_{-\infty}^{\infty} \frac{e^{-z^2}}{a + e^{-z^2}} \; dz.$$
From the identity
$$ 1 + x^{2n+1} = (1 + x)(1 - x + \cdots - x^{2n-1} + x^{2n}), $$
we obtain
$$ \frac{1}{1 + x} = 1 - x + \cdots - x^{2n-1} + x^{2n} - \frac{x^{2n+1}}{1 + x}.$$
Now we temporary assume further that $a > 1$, so that $\alpha = a^{-1} \in (0, 1)$. Then
$$\begin{align*}
\int_{-\infty}^{\infty} \frac{e^{-z^2}}{1 + \alpha e^{-z^2}} \; dz
&= \int_{-\infty}^{\infty} \left( \sum_{k=1}^{2n+1} \alpha^{k-1} e^{-kz^2} - \frac{\alpha^{2n+1}e^{-(2n+2)z^2}}{1 + e^{-z^2}} \right) \; dz \\
&= \sum_{k=1}^{2n+1} (-1)^{k-1} \alpha^{k-1}  \int_{-\infty}^{\infty} e^{-kz^2} \, dz - \alpha^{2n+1} \int_{-\infty}^{\infty} \frac{e^{-(2n+2)z^2}}{1 + e^{-z^2}} \; dz \\
&= \sum_{k=1}^{2n+1} (-1)^{k-1} \alpha^{k-1} \sqrt{\frac{\pi}{k}} - \alpha^{2n+1} \int_{-\infty}^{\infty} \frac{e^{-(2n+2)z^2}}{1 + e^{-z^2}} \; dz
\end{align*}$$
Now taking $n\to\infty$, the remainder term vanishes. Hence we have
$$ \int_{-\infty}^{\infty} \frac{e^{-z^2}}{1 + \alpha e^{-z^2}} \; dz
= \sum_{k=1}^{\infty} (-1)^{k-1} \alpha^{k-1} \sqrt{\frac{\pi}{k}}
= -a \sqrt{\pi} \, \mathrm{Li}_{1/2} \left( -\tfrac{1}{a}\right),$$
where
$$ \mathrm{Li}_{s}(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^s}$$
is the polylogarithm of order $s$, primarily defined on $|z| < 1$. Although we have proved this identity only for $a > 1$, the equality above can be used to define an analytic continuation of the right hand side, thus (by tautology) it holds for all $a > 0$.
It has special value at $\alpha = 1$, given by
$$ \int_{-\infty}^{\infty} \frac{1}{1 + e^{z^2}} \; dz
= -\sqrt{\pi} \, \mathrm{Li}_{1/2}(-1)
= \sqrt{\pi} (1 - \sqrt{2}) \zeta \left( \tfrac{1}{2} \right)$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{-\infty}^{\infty}{\dd x \over 1 + a\expo{bx^{2}}}:\ {\large ?}\,.
     \qquad a>0\,,\quad b> 0.}$

\begin{align}&\color{#c00000}{%
\int_{-\infty}^{\infty}{\dd x \over 1 + a\expo{b x^{2}}}}
={2 \over \root{b}}\
\overbrace{\int_{0}^{\infty}{\dd x \over 1 + a\expo{x^{2}}}}
^{\ds{x \equiv t^{1/2}}}\ =\
{2 \over \root{b}}\int_{0}^{\infty}{1 \over  a\expo{t} + 1}
\,\half\,t^{1/2 - 1}\dd t
\\[3mm]&={1 \over \root{b}}\int_{0}^{\infty}
{t^{\color{#f00}{\large 1/2} - 1}\over
\expo{t}/\color{#f00}{\large \pars{1/a}} + 1}\,\dd t
={1 \over \root{b}}\,\bracks{-\Gamma\pars{\half}{\rm Li}_{1/2}\pars{-\,{1 \over a}}}
\end{align}
  The last integral is a
  well known PolyLogarithm $\ds{{\rm Li_{s}}\pars{z}}$ integral representation.

Since $\ds{\Gamma\pars{\half} = \root{\pi}}$:
$$
\color{#77f}{\large%
\int_{-\infty}^{\infty}{\dd x \over 1 + a\expo{b x^{2}}}
=-\ \root{\pi \over b}\ {\rm Li}_{1/2}\pars{-\,{1 \over a}}}
$$
