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?