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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?

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Hint: use u=x^2 substitution, then it should come out to look like the gamma distribution function. – student101 Sep 30 '12 at 4:21
You seem to be under the impression that $r^4e^{-r^2}\to\infty$ as $r\to\infty$. You might want to reconsider. – Gerry Myerson Sep 30 '12 at 4:26
up vote 2 down vote accepted


$$\Longrightarrow \int_0^\infty x^2e^{-x^2/b^2}dx=\left.-\frac{b^2x}{2}e^{-x^2/b^2}\right|_0^\infty+\frac{b^2}{2}\int^\infty_0e^{-x^2/b^2}dx=\frac{b^3}{2}\sqrt\frac{\pi}{2}$$

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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Of course, this takes $\int_0^{\infty}e^{-x^2/b^2}\,dx$ as known. Only OP knows whether that's legit. – Gerry Myerson Sep 30 '12 at 4:33
Indeed. I assumed he knows that as he's trying to make integration by parts, though he quickly goes into polar coordinates. Let's see if this is clear to him and if it is not then I shall try to give one of the very numerous proofs of this result (perhaps double integration...?) – DonAntonio Sep 30 '12 at 4:38
@DonAntonio: This comes from my answer to this question, which did not take $\int_0^{\infty}e^{-x^2/b^2}\,dx$ and suggested the trig substitution. – Brian M. Scott Sep 30 '12 at 6:55
I see @BrianM.Scott , thanks. – DonAntonio Sep 30 '12 at 10:33

There is an easier way than going through all the integration by parts. Note that $$\frac4{\pi b^2}x^2e^{-x^2/b^2} = - \frac{4}{\pi b^2} \frac{d}{d\lambda} e^{-\lambda x^2}$$ with $\lambda = b^{-2}$. Using this relation and exchanging differentiation and integration yields $$\int_0^\infty\!dx\,\frac4{\pi b^2}x^2e^{-x^2/b^2} =- \frac{4}{\pi b^2} \frac{d}{d\lambda} \int_0^\infty\!dx\,e^{-\lambda x^2} = - \frac{4}{\pi b^2}\frac{d}{d\lambda} \sqrt{\frac{\pi}{4\lambda}} $$ from which the final result can be obtained easily.

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