<s>[INST] How to prove $\frac{v}{\pi} \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx =\frac{1}{\sqrt{2\pi}} e^{-\frac{v^2}{2}}$ If $v>0$ why 
$$\frac{v}{\pi}  \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx
=\frac{1}{\sqrt{2\pi}} e^{-\frac{v^2}{2}}$$.
I saw it in 

I tried to solve it by $t=\frac{1}{x^2}-1$
$$\frac{v}{\pi}  \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx
=\frac{v}{\pi}\int_0^{\infty} \frac{1}{\frac{1}{t+1} \sqrt{1-\frac{1}{t+1}}} e^{\frac{-v^2}{2}(t+1)} \, \frac{1}{2}(t+1)^{\frac{-3}{2}}dt $$
$$=\frac{v e^{\frac{-v^2}{2}} }{2\pi}\int_0^{\infty}\frac{1}{\sqrt{t}}  e^{\frac{-v^2}{2}t} \, dt $$
Thanks in advance for any help you are able to provide.
 [/INST] First substitute $y=\frac{v^2}{2x^2}$. Then $x^2=\frac{v^2}{2y}$ and $\mathrm dx=\color{green}{-\frac{v}{2\sqrt 2} y^{-\frac32}}\,\mathrm dy$. Thus $$\int_0^1 \frac{1}{\color{blue}{x^2} \color{orange}{\sqrt{1-x^2}}} e^{\frac{-v^2}{2x^2}} \,\mathrm dx=\color{green}-\frac{\color{blue}2\color{green}v}{\color{green}{2\sqrt 2} \color{blue}{v^2}}\int_\infty^{\frac{v^2}2} \frac{\color{blue}y\exp(-y)}{\color{green}{\sqrt{y^3}}\color{orange}{\sqrt{1-\frac{v^2}{2y}}}}\,\mathrm dy=\frac{1}{v\sqrt 2}\int_{\frac{v^2}2}^\infty \frac{\exp(-y)}{\sqrt{y-\frac{v^2}2}}\,\mathrm dy.$$
Now use $z=y-\frac{v^2}2$ and this turns into $$\frac{1}{v\sqrt 2}\int_0^\infty \frac{\exp\left(-z-\frac{v^2}2\right)}{\sqrt{z}}\,\mathrm dz=\frac{\exp\left(-\frac{v^2}2\right)}{v\sqrt 2} \color{violet}{\int_0^\infty \frac{\exp(-z)}{\sqrt z}\,\mathrm dz}=\frac{\exp\left(-\frac{v^2}2\right)}{v\sqrt 2} \color{violet}{\Gamma\left(\frac12\right)}=\frac{\color{violet}{\sqrt \pi}\exp\left(-\frac{v^2}2\right)}{v\sqrt 2}.$$
Hence $$\frac{v}{\pi}\int_0^1 \frac{1}{{x^2} {\sqrt{1-x^2}}} e^{\frac{-v^2}{2x^2}} \,\mathrm dx=\frac{v}{\pi}\frac{{\sqrt \pi}\exp\left(-\frac{v^2}2\right)}{v\sqrt 2}=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{v^2}2\right).$$ QED.
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