Show that $\int (f'(x))^2dx=\frac{1}{2\sqrt{\pi}}Var(Y)$

Question

Let $f(x)=\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})$, $x\in\mathbb{R}$. Show that $$\int (f'(x))^2dx=\frac{1}{2\sqrt{\pi}}Var(Y)$$ where $Y$ is a Normal random variable with $\mu=0$ and $\sigma^2=\frac{1}{2}$

I know that $f(x)$ is the standard normal density, $$f'(x)=-\frac{x}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})$$ and $$f'(x)^2=\frac{x^2}{2\pi}\exp(-x^2)$$ then $$\int (f'(x))^2dx=\int \frac{x^2}{2\pi}\exp(-x^2)dx$$

but I don't know what to do from here.

Answer 1

Hint. One may recall that $$ \int_0^\infty u^s\exp(-u)\:du=\Gamma(s+1),\quad s>-1. \tag1 $$ Then, by the change of variable $x=\sqrt{u}$, $u:=x^2$, $dx=\dfrac{du}{\sqrt{u}}$ in the following integrand $$ \int_0^\infty x^2\exp(-x^2)\:dx $$ one gets $$ \begin{align} \int_0^\infty (f'(x))^2dx&=\frac1{2\pi}\int_0^\infty x^2\exp(-x^2)\:dx \\\\&=\frac1{4\pi}\int_0^\infty u^{1/2}\exp(-u)\:du \\\\&=\frac1{4\pi}\Gamma\left(\frac32\right) \\\\&=\frac1{8\sqrt{\pi}}. \end{align} $$