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I've encountered the following integral

$$\int_{-\infty}^{\infty} \frac{e^{ax^{2}}}{1+e^{x}} dx\; \; with\; a<0$$

Is this integrable? I have seen similar topics asked before was solved with contour method but do not know whether, and how can contour method works for this one.

Any help would be appreciated!

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  • $\begingroup$ Mathematica gives: $$\mathscr{I}=\frac{\sqrt{\pi}}{2\sqrt{-\text{a}}}$$ Where $\Re\left(\text{a}\right)<0$ $\endgroup$
    – Jan Eerland
    Commented May 23, 2017 at 9:44
  • $\begingroup$ Can detailed proof or resources be provided Jan? and what do you mean by $R(a)<0$ $\endgroup$
    – t.hu
    Commented May 23, 2017 at 10:03
  • $\begingroup$ Can I use Mathematica to get two following integrals by using Mathematica as well? $$\int_{-\infty}^{\infty} \frac{e^{ax}}{1+e^{x}} dx\; \; with\; a<0$$ $$\int_{-\infty}^{\infty} \frac{e^{ax}}{1+e^{x}} dx\; \; with\; 0<a<1$$ Thanks for any information!!! $\endgroup$
    – t.hu
    Commented May 23, 2017 at 10:07
  • $\begingroup$ It means the real part of $\text{a}$, but the problem is: the integral is defined outside the region $\Re\left(\text{a}\right)<0$, so when $\text{a}\ge0$ $\endgroup$
    – Jan Eerland
    Commented May 23, 2017 at 10:13
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    $\begingroup$ integrate this using contour looks way too complicated. try this instead: $$\frac{1}{1+e^x} + \frac{1}{1+e^{-x}} = 1\implies\int_{-\infty}^\infty \frac{e^{ax^2}}{1+e^x} dx = \frac12 \int_{-\infty}^{\infty} e^{ax^2}dx\implies\cdots$$ $\endgroup$
    – achille hui
    Commented May 23, 2017 at 11:25

2 Answers 2

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Indeed, the integral converges since the integrand behaves like $ e^{ax^2} $ as $x\to-\infty$ and like $ e^{x^2(a-1/x)} $ as $x\to+\infty$, and $a<0$.

Here is also a calculation which, however, does not use the contour method. Let us split the integral as $$ I=\int_{-\infty}^\infty \frac{e^{ax^2}}{1+e^x}dx = \int_{0}^\infty \frac{e^{ax^2}}{1+e^x}dx+ \int_{-\infty}^0\frac{e^{ax^2}}{1+e^x}dx=I_1+I_2\,. $$ Then, using the formula for the geometric series $$ I_1=\int_0^\infty \frac{e^{ax^2}}{1+e^{-x}}e^{-x}dx =\sum_{n=0}^\infty (-1)^n \int_0^\infty e^{ax^2-(n+1)x}dx=-\sum_{n=1}^\infty (-1)^n \int_0^\infty e^{ax^2-nx}dx $$ whereas $$ I_2= \int_{0}^\infty \frac{e^{ax^2}}{1+e^{-x}}dx =\sum_{n=0}^\infty (-1)^n \int_0^\infty e^{ax^2-nx}dx\,. $$ In the sum $I_1+I_2$ all terms cancel out except for $n=0$, thus $$ I=\int_0^\infty e^{ax^2}dx=\frac{1}{2}\int_{-\infty}^\infty e^{ax^2}dx=\frac{\sqrt{\pi}}{2\sqrt{-a}}\,, $$ for $a<0$.

EDIT: In fact, as pointed out by Achille Hui in the comments, there is no need for geometric series $$ I = \int_0^\infty e^{ax^2}\left( \frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right) dx =\int_0^\infty e^{ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{-a}}\,. $$

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On the path of Brightsun and Achille Hui (to achieve the computation)

$\begin{align}I&=\displaystyle \int_{-\infty}^{\infty} \frac{e^{ax^{2}}}{1+e^{x}} dx\\ &=\int_{-\infty}^{0} \frac{e^{ax^{2}}}{1+e^{x}} dx+\int_{0}^{\infty} \frac{e^{ax^{2}}}{1+e^{x}} dx \end{align}$

Perform the change of variable $y=-x$ in the first integral,

$\begin{align}I&=\int_{0}^{\infty} \frac{e^{ax^{2}}}{1+e^{-x}} dx+\int_{0}^{\infty} \frac{e^{ax^{2}}}{1+e^{x}} dx\\ &=\int_0^1\left(\dfrac{1}{1+e^{-x}}+\dfrac{1}{1+e^{x}}\right)e^{ax^{2}}dx\\ &=\int_{0}^{\infty} e^{ax^{2}}dx\\ \end{align}$

$a<0$, perform the change of variable $y=\sqrt{|a|}x$,

$\begin{align} I&=\dfrac{1}{\sqrt{|a|}}\int_{0}^{\infty} e^{-x^{2}}dx\\ &=\dfrac{\sqrt{\pi}}{2\sqrt{|a|}} \end{align}$

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