Integration Problem: $\frac{1}{\sqrt{2\pi}}\int^{\infty}_0\sqrt{w}e^{-w-\frac{wx^2}{2}}dw$

Question

I know that:$$\frac{1}{\sqrt{2\pi}}\int^{\infty}_0\sqrt{w}e^{-w-\frac{wx^2}{2}}dw=\frac{1}{{(x^2+2)}^{3/2}}$$ I've attempted to solve this with integration by parts $\int{fdg}=fg-\int{dfg}$

I let

$$\begin{array} ff=\sqrt{w} & dg=e^{-w-\frac{wx^2}{2}}\\df=\frac{1}{2\sqrt{w}}& g=\frac{-2e^{-w-\frac{wx^2}{2}}}{2+x^2}{}\end{array}$$

Which is:$$\frac{1}{2\pi}\bigr[\frac{-2\sqrt{w}e^{-w-\frac{wx^2}{2}}}{2+x^2}\Bigr|^{\infty}_{0}+\frac{1}{2+x^2}\int^{\infty}_0\frac{e^{-w-\frac{wx^2}{2}}}{\sqrt{w}}dw\bigr]=\frac{1}{2\pi}\bigr[\frac{1}{2+x^2}\int^{\infty}_0\frac{e^{-w-\frac{wx^2}{2}}}{\sqrt{w}}dw\bigr]$$

I have to integrate by parts again, this time I have:

$$\begin{array} ff=e^{-w-\frac{wx^2}{2}} & dg=\frac{1}{\sqrt{w}}\\df=\frac{(x^2+2)^2e^{-w-\frac{wx^2}{2}}}{2}& g=2\sqrt{w}\end{array}$$ Which is: $$\frac{1}{\sqrt{2\pi}}\bigr[\frac{1}{x^2+2}\bigr(2\sqrt{w}e^{-w-\frac{wx^2}{2}}\Bigr|^{\infty}_{0}+(x^2+2)\int^{\infty}_{0}\sqrt{w}e^{-w-\frac{wx^2}{2}}dw\bigr)\bigr]=\frac{1}{\sqrt{2\pi}}\int^{\infty}_0\sqrt{w}e^{-w-\frac{wx^2}{2}}dw$$

So, I'm back to where I started. Can somebody spot my mistake?

Answer 1

Hint

you have nearly with $w=u^2 \implies dw=2udu$ this:

$$I=\int \sqrt w e^{-w}dw=2\int u^2e^{-u^2}du$$ $$I=2\int u\times ue^{-u^2}du=\int u\times (-e^{-u^2})'du$$ Integrate by part once... $$\int_0^{\infty} u\times (-e^{-u^2})'du=(u\times (-e^{-u^2}))|_0^{\infty}-\int_0^{\infty} (-e^{-u^2})du$$ And use gaussian integral to get the value of the last integral

$$\int_0^{\infty} e^{-u^2}du=\frac {\sqrt {\pi}}2$$