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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.

$\frac4{\pi b^2}\int_0^\infty x^2 e^{-x^2/b^2}dx\;.$

I know i need to use integration by parts. let $u = x$ and $dv = xe^{-x^2/b^2}$, then $du=dx$ and $v=\int xe^{-x^2/b^2}dx$. But this is where i get stuck.

I know i will need to somehow get $\int xe^{-x^2/b^2}dx$ by itself to use polar coordinates but not sure how to get it by itself and then put everything back together. I appreciate any help!!!

3 Answers 3

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These types of integrals are widely used in statistics, and one of the most practical approaches to tackle them is to exploit the gamma function. Recalling the definition of the gamma function

$ \Gamma(s) = \int_{0}^{\infty} t^{s-1} {\rm e}^{-t}dt\,. $

Making the change of variables $y=\frac{x^2}{b^2} $ casts the integral to the gamma function

$\int_0^\infty\frac4{\pi b^2}x^2e^{-x^2/b^2}dx\;= \frac{2b}{\pi}\int _{0}^{\infty }\sqrt {y}\,{{\rm e}^{-y}}{dy} = \frac{2b}{\pi} \Gamma(\frac{3}{2}) = \frac{b}{\pi}\Gamma(\frac{1}{2})= \frac{b}{\sqrt{\pi}}\,. $

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    @user43126:if you integrate your function from $-\infty$ to $\infty$ you do not get 1.2012-10-01
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This one requires a trick. Let $I$ be the value of the integral. Then

$\begin{align*} I^2&=\left(\int_0^\infty\frac4{\pi b^2}x^2e^{-x^2/b^2}dx\right)\left(\int_0^\infty\frac4{\pi b^2}y^2e^{-y^2/b^2}dy\right)\\\\ &=\frac{16}{\pi^2b^4}\int_0^\infty\int_0^\infty x^2y^2e^{-(x^2+y^2)/b^2}dydx\;. \end{align*}$

Now convert to polar coordinates: $x=r\cos\theta$, $y=r\sin\theta$, etc. You’re integrating over the first quadrant, so you want your double integral in polar coordinates to have $0\le\theta\le\frac{\pi}2$ and $0\le r<\infty$. When you’ve completed the integration, you’ll have $I^2$, from which you can easily get $I$.

Added: Ignoring the various constants, you have essentially something like $\int_0^{\pi/2}\int_0^\infty r^5\cos^2\theta\sin^2\theta e^{-r^2}drd\theta=\int_0^{\pi/2}\cos^2\theta\sin^2\theta\int_0^\infty r^5e^{-r^2}drd\theta\;.$

The inner integral (with respect to $r$) can be done by repeated integration by parts; for the first one let $u=r^4$, $dv=re^{-r^2}dr$, so that $du=4r^3dr$ and $v=-\frac12e^{-r^2}$. That will leave you with something of the form $\int_0^\infty r^3e^{-r^2}dr$ to deal with. Repeat the process, and you’ll have something of the form $\int_0^\infty re^{-r^2}dr$, which you can integrate outright.

At that point you’ll be integrating some multiple of $\cos^2\theta\sin^2\theta$. One way is to use the double angle formula for the sine to rewrite this as $\frac12\sin^22\theta$, then use the half-angle formula to rewrite $\sin^22\theta$ as $\frac12(1-\cos 4\theta)$, which you can integrate.

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    Brian, when I got to the part that you said i could integrate right out. i used the part i knew that Josh wrote earlier. But when I finished the integration by parts when i started substituting back in i got something with a bunch of infinities that all but one canceled out. I will add this in above2012-09-30
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I actually would use a different trick. I'm assuming that you know that \begin{equation} \int^\infty_0 dx~e^{-a x^2} = {1\over2} \sqrt{\pi\over a}~. \end{equation} The derivative of the right-hand side with respect to $a$ is a trivial calculation and the derivative of the left-hand side is proportional to the integral you want to solve for (after setting $a=b^{-2}$).

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    You may want to use$a$combination of our approaches. Using the trick Brian advocated is easier if done with the simple exponential integral in my response than with the full-fledged integral that you are trying to solve. Once you verify the equation that I wrote, you can follow the procedure I listed to get the answer to your integral.2012-09-30