Nice Exercise Solved by using substitution and Gaussian integral

Exercise:

Compute, $\int_{0}^{\infty }{{{e}^{-\left( {{x}^{2}}+\frac{4}{{{x}^{2}}} \right)}}}dx$

Solution: Let $x=\frac{2}{t}\Leftrightarrow dx=-\frac{2}{{{t}^{2}}}dt$ and put $I=\int_{0}^{\infty }{{{e}^{-\left( {{x}^{2}}+\frac{4}{{{x}^{2}}} \right)}}}dx$

Remark that $t=\frac{2}{x}$ as $x\to 0\,\,,\,\,t\to \infty $ and as $x\to \infty \,\,,\,\,t\to 0$

So $I=-\int_{\infty }^{0}{\frac{2}{{{t}^{2}}}{{e}^{-\left( \frac{4}{{{t}^{2}}}+{{t}^{2}} \right)}}}dt=\int_{0}^{\infty }{\frac{2}{{{t}^{2}}}{{e}^{-\left( {{t}^{2}}+\frac{4}{{{t}^{2}}} \right)}}dt}$

Hence $2I=\int_{0}^{\infty }{{{e}^{-\left( {{x}^{2}}+\frac{4}{{{x}^{2}}} \right)}}}dx+\int_{0}^{\infty }{\frac{2}{{{x}^{2}}}{{e}^{-\left( {{x}^{2}}+\frac{4}{{{x}^{2}}} \right)}}dx}=\int_{0}^{\infty }{\left( 1+\frac{2}{{{x}^{2}}} \right){{e}^{-\left( {{x}^{2}}+\frac{4}{{{x}^{2}}} \right)}}dx}$

Let $w=x-\frac{2}{x}\Leftrightarrow dw=\left( 1-\frac{2}{{{x}^{2}}} \right)dx$    as $x\to 0\,\,\,,\,\,w\to -\infty \,\,\,\And \,\,\,x\to \infty \,\,\,,w\to \infty $

And ${{w}^{2}}={{x}^{2}}+\frac{4}{{{x}^{2}}}-2\left( x \right)\left( \frac{2}{x} \right)={{x}^{2}}+\frac{4}{{{x}^{2}}}-4$ $\Leftrightarrow {{w}^{2}}+4={{x}^{2}}+\frac{4}{{{x}^{2}}}$

So $2I=\int_{-\infty }^{\infty }{{{e}^{-{{w}^{2}}-4}}dw}={{e}^{-4}}\int_{-\infty }^{\infty }{{{e}^{-{{w}^{2}}}}dw}$

Observe that $\int_{-\infty }^{\infty }{{{e}^{-{{w}^{2}}}}dw}$ is a Gaussian integral

To prove $\int_{-\infty }^{\infty }{{{e}^{-{{w}^{2}}}}dw=\sqrt{\pi }}$  we need to put $J=\int_{-\infty }^{\infty }{{{e}^{-{{w}^{2}}}}dw}$

So \[{{J}^{2}}={{\left( \int_{-\infty }^{\infty }{{{e}^{-{{w}^{2}}}}dw} \right)}^{2}}=\left( \int_{-\infty }^{\infty }{{{e}^{-{{x}^{2}}}}dx} \right)\left( \int_{-\infty }^{\infty }{{{e}^{-{{y}^{2}}}}dy} \right)=\int_{-\infty }^{\infty }{\int_{-\infty }^{\infty }{{{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}}dxdy}}\]

Note that ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ whenever $x=r\cos \theta \,\,\,\And \,\,\,y=r\sin \theta $

 So ${{J}^{2}}=\int_{0}^{\infty }{\int_{0}^{2\pi }{{{e}^{-{{r}^{2}}}}rdr}d\theta }=\int_{0}^{\infty }{2\pi r{{e}^{-{{r}^{2}}}}dr=\pi \int_{0}^{\infty }{2r{{e}^{-{{r}^{2}}}}dr=\pi \int_{0}^{\infty }{{{e}^{-{{r}^{2}}}}d\left( {{r}^{2}} \right)=-\pi \left[ {{e}^{-{{r}^{2}}}} \right]_{0}^{\infty }}}}$

$\Rightarrow {{J}^{2}}=-\pi \left( {{e}^{-\infty }}-{{e}^{0}} \right)=\pi \Leftrightarrow J=\sqrt{\pi }$  so $\int_{-\infty }^{\infty }{{{e}^{-{{w}^{2}}}}dw}=\sqrt{\pi }$

So $2I={{e}^{-4}}\sqrt{\pi }\Rightarrow I=\frac{\sqrt{\pi }}{2{{e}^{4}}}$