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.

To arrive to a gaussian integral 
$$I= \int \frac{e^{\frac{-v^2}{2x^2}}}{x^2 \sqrt{1-x^2}}\,  dx$$
Let
$$x=\frac{1}{\sqrt{t^2+1}}\implies dx=-\frac{t}{\left(t^2+1\right)^{3/2}}\,dt$$ This makes
$$I=-\int e^{-\frac{1}{2} \left(t^2+1\right) v^2}\,dt=-\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{v^2}{2}} \text{erf}\left(\frac{t
   v}{\sqrt{2}}\right)}{v}$$
$$J=-\int_0^\infty e^{-\frac{1}{2} \left(t^2+1\right) v^2}\,dt=-\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{v^2}{2}}}{\sqrt{v^2}}$$
