\(\int_{0}^{\infty}{{\rm{exp}}\bigl({-b^2\bigl({x^2+\frac{a^2}{x^2}}\bigr)}\bigr)\,dx}\)

[Grigorios Kostakos]

Prove that: \[\displaystyle \int_{0}^{+\infty}{{\rm{e}}^{-b^2\bigl({x^2+\frac{a^2}{x^2}}\bigr)}\,dx}=\frac{\sqrt{\pi}}{2|b|}{\rm{e}}^{-2b^2a}\,,\quad a\in(0,+\infty)\,,b\in{\mathbb{R}}^{*}\,.\]

[Papapetros Vaggelis]

Hello Grigoris. Here is a solution.

Firstly, for each \(\displaystyle{x>0}\) holds :

\(\displaystyle{\mathrm{exp}\left(-b^2\,\left(x^2+\dfrac{a^2}{x^2}\right)\right)>0}\)

and

\(\displaystyle{\mathrm{exp}\left(-b^2\,\left(x^2+\dfrac{a^2}{x^2}\right)\right)\leq e^{-b^2\,x^2}\,,x>0}\) , where

\(\displaystyle{\int_{0}^{\infty}e^{-b^2\,x^2}\,\mathrm{d}x=\dfrac{1}{\left|b\right|}\,\int_{0}^{\infty}e^{-(\left|b\right|\,x)^2}\,\mathrm{d}(\left|b\right|\,x)=\dfrac{\sqrt{\pi}}{\left|b\right|}}\)

Therefore,

\(\displaystyle{I=\int_{0}^{\infty}\mathrm{exp}\left(-b^2\,\left(x^2+\dfrac{a^2}{x^2}\right)\right)\,\mathrm{d}x<\infty}\)

If \(\displaystyle{x>0}\) , then \(\displaystyle{x^2+\dfrac{a^2}{x^2}=\left(x-\dfrac{a}{x}\right)^2+2\,a}\) , so :

\(\displaystyle{I=e^{-2\,a\,b^2}\,\int_{0}^{\infty}\mathrm{exp}\left(-b^2\,\left(x-\dfrac{a}{x}\right)^2\right)\,\mathrm{d}x}\) .

By substituting \(\displaystyle{x=\dfrac{a}{t}\,,t\in\left[0,+\infty\right)}\) , we get :

\(\displaystyle{\begin{aligned} I&=e^{-2\,a\,b^2}\,\int_{0}^{\infty}\dfrac{a}{t^2}\,\mathrm{exp}\left(-b^2\,\left(t-\dfrac{a}{t}\right)^2\right)\,\mathrm{d}t\\&=e^{-2\,a\,b^2}\,\int_{0}^{\infty}\dfrac{a}{x^2}\,\mathrm{exp}\left(-b^2\,\left(x-\dfrac{a}{x}\right)^2\right)\,\mathrm{d}x\end{aligned}}\)

and now :

\(\displaystyle{\begin{aligned} 2\,I &=e^{-2\,a\,b^2}\,\int_{0}^{\infty}\left(1+\dfrac{a}{x^2}\right)\,\mathrm{exp}\left(-b^2\,\left(x-\dfrac{a}{x}\right)^2\right)\,\mathrm{d}x\\&=\dfrac{e^{-2\,a\,b^2}}{\left|b\right|}\,\int_{0}^{\infty}\mathrm{exp}\left(-\left(\left|b\right|\,x-\dfrac{a\,\left|b\right|}{x}\right)^2\right)\,\mathrm{d}(\left|b\right|\,x-\dfrac{a\,\left|b\right|}{x})\\&=\dfrac{\sqrt{\pi}}{\left|b\right|}\,e^{-2\,a\,b^2}\end{aligned}}\)

Finally,

\(\displaystyle{I=\int_{0}^{\infty}\mathrm{exp}\left(-b^2\,\left(x^2+\dfrac{a^2}{x^2}\right)\right)\,\mathrm{d}x=\dfrac{\sqrt{\pi}}{2\,\left|b\right|}\,e^{-2\,a\,b^2}}\)

We are using the identity: \(\displaystyle{\Gamma\,\left(\dfrac{1}{2}\right)=\int_{0}^{\infty}\dfrac{e^{-t}}{\sqrt{t}}\,\mathrm{d}t\stackrel{t=y^2}{=}2\,\int_{0}^{\infty}e^{-y^2}\,\mathrm{d}y=\sqrt{\pi}}\) .