<s>[INST] Help? I cannot do the integration by parts correctly?? Can someone please show me how to integrate
$$\int_0^\infty\frac4{\pi b^2}x^2e^{-x^2/b^2}dx\;?$$
please show steps how to integrate this problem.  This is what i have so far.
$$\frac4{\pi b^2}\int_0^\infty x^2 e^{-x^2/b^2}dx\;.$$
Then i take the property I^2 = $$\int_0^\infty\frac4{\pi b^2}x^2e^{-x^2/b^2}dx\;*\int_0^\infty\frac4{\pi b^2}y^2e^{-y^2/b^2}dx\;?$$
Then i substitue in x=rcos0  and y=rsin0  dxdy=rdrd0
Then I get:  $$\int_0^{\pi/2}\int_0^\infty r^5\cos^2\theta\sin^2\theta e^{(-r^2)/b^2}drd\theta=\int_0^{\pi/2}\cos^2\theta\sin^2\theta\int_0^\infty r^5e^{(-r^2)/(b^2)}drd\theta\;.$$
Then I do $$\int_0^\infty r^5e^{(-r^2)/(b^2)}drd\theta\;.$$ using integration by parts
 $u=r^4$, $dv=re^{-r^2}dr$, so that $du=4r^3dr$ and $v=-\frac12e^{-r^2/b^2}$. That will leave you with something of the form $r^4(-\frac12e^{-r^2/b^2})(from 0 to infinity)-\int_0^\infty r^3e^{-r^2/b^2}dr$ 
Then took the limit of r from 0 to $infty$ of $r^4(-\frac12e^{-r^2/b^2})$ I got infinity. So now my problem looks like $\infty-\int_0^\infty r^3e^{-r^2/b^2}dr$
Then I did integration by parts on $\int_0^\infty r^3e^{-r^2/b^2}dr$. I let $w=r^2$, $dz=re^{-r^2/b^2}dr$, so that $dw=2rdr$ and $z=-\frac12e^{-r^2/b^2}$. Then i have $\infty-(wz-\int_0^\infty r^2e^{-r^2/b^2}dr)$.
Then I do integration by parts one more time. But when i find w and z and take the limit i get inifinity again, so i get something that looks like $\infty-(infty-\int_0^\infty r^2e^{-r^2/b^2}dr)$ .  Can someone please tell me what I am doing wrong?
 [/INST] $$\begin{align*}u=&x&u'=&1\\v'=&xe^{-x^2/b^2}&v=&-\frac{b^2}{2}e^{-x^2/b^2}\end{align*}\;\;\;\;\;\;\Longrightarrow$$
$$\Longrightarrow \int_0^\infty x^2e^{-x^2/b^2}dx=\left.-\frac{b^2x}{2}e^{-x^2/b^2}\right|_0^\infty+\frac{b^2}{2}\int^\infty_0e^{-x^2/b^2}dx=\frac{b^3}{2}\sqrt\frac{\pi}{2}$$
And thus your integral equals
$$\frac{4}{\pi b^2}\frac{b^3}{2}\sqrt\frac{\pi}{2}=\sqrt\frac{2}{\pi}\,b$$
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