Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

up vote 2 down vote accepted

$$\begin{align*}u=&x&u'=&1\\v'=&xe^{-x^2/b^2}&v=&-\frac{b^2}{2}e^{-x^2/b^2}\end{align*}\;\;\;\;\;\;\Longrightarrow$$

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

And thus your integral equals

$$\frac{4}{\pi b^2}\frac{b^3}{2}\sqrt\frac{\pi}{2}=\sqrt\frac{2}{\pi}\,b$$

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.