improper Integral exercise leads to Gaussian integral

Exercise:

Show that, $\int_{0}^{1}{\frac{dx}{\sqrt{-\ln x}}=\sqrt{\pi }}$

Solution: Let ${{u}^{2}}=-\ln x\Rightarrow -{{u}^{2}}=\ln x\Leftrightarrow x={{e}^{-{{u}^{2}}}}$ hence $dx=-2u{{e}^{-{{u}^{2}}}}du$

So $\int_{0}^{1}{\frac{dx}{\sqrt{-\ln x}}=\int_{u\left( 0 \right)}^{u\left( 1 \right)}{\frac{-2u{{e}^{-{{u}^{2}}}}du}{\sqrt{{{u}^{2}}}}}=-\int_{\infty }^{0}{\frac{2u{{e}^{-{{u}^{2}}}}du}{u}}=\int_{0}^{\infty }{2{{e}^{-{{u}^{2}}}}}du}=2\int_{0}^{\infty }{{{e}^{-{{u}^{2}}}}}du$

Notice that ${{e}^{-{{u}^{2}}}}$ is a Gaussian function over the entire real line which is even function

So this Gaussian integral  $I=\int_{-\infty }^{\infty }{{{e}^{-{{u}^{2}}}}}du=2\int_{0}^{\infty }{{{e}^{-{{u}^{2}}}}}du$

Hence ${{I}^{2}}=\int_{-\infty }^{\infty }{{{e}^{-{{u}^{2}}}}du}\int_{-\infty }^{\infty }{{{e}^{-{{v}^{2}}}}}dv=\int_{-\infty }^{\infty }{\int_{-\infty }^{\infty }{{{e}^{-{{u}^{2}}}}{{e}^{-{{v}^{2}}}}dudv}}=\int_{-\infty }^{\infty }{\int_{-\infty }^{\infty }{{{e}^{-\left( {{u}^{2}}+{{v}^{2}} \right)}}}dudv}$

By taking $u=r\cos \theta \,\,\And \,\,v=r\sin \theta \Rightarrow {{u}^{2}}+{{v}^{2}}={{r}^{2}}$

So $\int_{-\infty }^{\infty }{\int_{-\infty }^{\infty }{{{e}^{-\left( {{u}^{2}}+{{v}^{2}} \right)}}dudv}=\int_{0}^{2\pi }{\int_{0}^{\infty }{{{e}^{-{{r}^{2}}}}rdrd\theta }}}$

$=\int_{0}^{2\pi }{\left( \int_{0}^{\infty }{r{{e}^{-{{r}^{2}}}}}dr \right)d\theta =2\pi \int_{0}^{\infty }{r{{e}^{-{{r}^{2}}}}dr}}$

Take $w={{r}^{2}}$ $\Rightarrow dw=2rdr\Leftrightarrow \frac{dw}{2}=rdr$



So ${{I}^{2}}=\pi \int_{0}^{\infty }{{{e}^{-w}}dw}=\pi \left( -{{e}^{-w}} \right)_{0}^{\infty }=\pi \left( -{{e}^{-\infty }}+{{e}^{0}} \right)=\pi $

$\Leftrightarrow I=\sqrt{\pi }$   thus $\int_{-\infty }^{\infty }{{{e}^{-{{u}^{2}}}}}du=2\int_{0}^{\infty }{{{e}^{-{{u}^{2}}}}}du=\sqrt{\pi }$