# 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!

-
Are you familiar with complex analysis ? – Selim Ghazouani Mar 20 '12 at 8:02

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{-\infty}^{\infty}{\dd x \over 1 + a\expo{bx^{2}}}:\ {\large ?}\,. \qquad a>0\,,\quad b> 0.}$

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

-

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

$$\int_{-\infty}^{\infty} \frac{1}{1 + e^{z^2}} \; dz = -\sqrt{\pi} \, \mathrm{Li}_{1/2}(-1) = \sqrt{\pi} (1 - \sqrt{2}) \zeta \left( \tfrac{1}{2} \right)$$

-