$\int_0^\infty x^{-\frac{3}{2}}e^{-\frac{(x-1)^2}{x}}dx=\int_0^\infty x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx$ This is my third question following the previous post.

Prove that \begin{equation} \int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\frac{2}{3}\int_0^\infty x^{\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx \end{equation}

I am really having trouble to prove it. Please help me to compute the integrals. Every answer would be greatly appreciated. Thank you.

Examining the original integral, use the substitution $u=x^{-1}$ so that
$$\frac{du}{dx}=-x^{-2}\Rightarrow dx=-(u^{-2})du$$ and
$u\rightarrow 0$ when $x\rightarrow\infty$, $u\rightarrow \infty$ when $x\rightarrow0$ 
Therefore
$$\int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=-\int_{\infty}^0u^{\frac{3}{2}}e^{-\frac{(1-u)^2}{u}}(u^{-2})du\\=\int_{0}^{\infty}u^{-\frac{1}{2}}e^{-\frac{(1-u)^2}{u}}du\\=\int_{0}^{\infty}u^{-\frac{1}{2}}e^{-\frac{(u-1)^2}{u}}du$$
If we replace the variable $u$ by $x$ (which is equivalent to the trivial substitution $x=u$) we will end up with the desired result:-
$$\int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx$$  
Next we apply integration by parts to the integral in the RHS of the above equation, by setting $dv=x^{-\frac{1}{2}}$ and $u=e^{-\frac{(x-1)^2}{x}}$ so that $v=2x^{\frac{1}{2}}$ and $du=e^{-\frac{(x-1)^2}{x}}\left(\frac{1}{x^2}-1\right)$ and recalling that
$$\int_0^{\infty}(u)(dv)dx=[uv]_0^\infty-\int_0^{\infty}v(du)dx$$
leading to 
$$\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\left[2x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}\right]_0^{\infty}-2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}\left(\frac{1}{x^2}-1\right)dx\\=0-2\int_0^{\infty}x^{-\frac{3}{2}}e^{-\frac{(x-1)^2}{x}}dx+2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx\\=-2\int_0^{\infty}x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx+2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx \\\Rightarrow 3\int_0^{\infty}x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx=2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx\\\Rightarrow \int_0^{\infty}x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx=\frac{2}{3}\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx$$
Let $I_n$ denote the integral we wish to evaluate. Making use of the equality of the integrals in the first part of the question, we have
$$2I_n=\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx+\int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx$$ 
Using the substitution $u=x^{\frac{1}{2}}$ we have $\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}$ , resulting in
$$2I_n=2\int_0^\infty e^{-\large\left(u-\frac{1}{u}\right)^2}du+2\int_0^\infty\frac{1}{u^2} e^{-\large\left(u-\frac{1}{u}\right)^2}du \\\Rightarrow I_n=\int_0^\infty \left(1+\frac{1}{u^2}\right) e^{-\large\left(u-\frac{1}{u}\right)^2}du$$ 
If we use the substitution $s=u-\frac{1}{u}$ , we have $ds=du\left(1+\frac{1}{u^2}\right)$  so that the integral reduces to a Gaussian integral (note the change in limits, as $s\rightarrow-\infty$  when  $u\rightarrow 0$ and $s\rightarrow\infty$ when $u\rightarrow \infty$):-
$$I_n=\int_{-\infty}^\infty e^{-s^2}ds=\sqrt{\pi}$$   
