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

Question

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!

Answer 1

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 $$ I = \int_0^\infty e^{ax^2}\left( \frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right) dx =\int_0^\infty e^{ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{-a}}\,. $$

Answer 2

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 $y=\sqrt{|a|}x$,

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