# Difficult integral

• Apr 14th 2010, 01:26 PM
Danneedshelp
Difficult integral
Evalutate

$\int_{0}^{\infty}\frac{x^{2}}{\theta^{2}}e^{\frac{-x^{2}}{2\theta^{2}}}dx$.

I started by letting

$u=x^{2}\Rightarrow\\du=2x$
and $dv=e^{\frac{-x^{2}}{2\theta^{2}}}\Rightarrow\\$?

I believe I need to use polar coordinates, but I not sure how. Any help would be greatly appreciated.

Thank you
• Apr 14th 2010, 03:02 PM
zzzoak

$I=\int_{-\infty}^{\infty} e^{-x^2} \: dx$

$I^2 \:=\:\int_{-\infty}^{\infty} e^{-x^2} \:dx\int_{-\infty}^{\infty} e^{-y^2} \: dy \:= \:\int_{0}^{\infty} e^{-r^2}2 \pi r dr \:=\:- \pi \int_{0}^{\infty} e^{-r^2} \: d(-r^2) \: =$

$= \:-\pi \:e^{-r^2} |_{0}^{\infty} \:=\pi$

$I \: = \: \sqrt{\pi}$.

Notice that

$\int_{-\infty}^{\infty} e^{-x^2} \: dx \: = \: 2 \:\int_{0}^{\infty} e^{-x^2} \: dx$.

Lets find integral

$
\int_{0}^{\infty}\frac{x^{2}}{\theta^{2}}e^{\frac{-x^{2}}{2\theta^{2}}}dx \:= \:
-\int_{0}^{\infty}\: x \: e^{\frac{-x^{2}}{2\theta^{2}}} \: d({\frac{-x^{2}}{2\theta^{2}}}) \:=
$

$= -x \: e^{\frac{-x^{2}}{2\theta^{2}}} \: |_{0}^{\infty} \:+\int_{0}^{\infty}\ e^{\frac{-x^{2}}{2\theta^{2}}}dx \: =$

$= \int_{0}^{\infty}\ e^{\frac{-x^{2}}{2\theta^{2}}}dx \:$
• Apr 14th 2010, 04:58 PM
Danneedshelp
Quote:

Originally Posted by zzzoak

$I=\int_{-\infty}^{\infty} e^{-x^2} \: dx$

$I^2 \:=\:\int_{-\infty}^{\infty} e^{-x^2} \:dx\int_{-\infty}^{\infty} e^{-y^2} \: dy \:= \:\int_{0}^{\infty} e^{-r^2}2 \pi r dr \:=\:- \pi \int_{0}^{\infty} e^{-r^2} \: d(-r^2) \: =$

$= \:-\pi \:e^{-r^2} |_{0}^{\infty} \:=\pi$

$I \: = \: \sqrt{\pi}$.

Notice that

$\int_{-\infty}^{\infty} e^{-x^2} \: dx \: = \: 2 \:\int_{0}^{\infty} e^{-x^2} \: dx$.

Lets find integral

$
\int_{0}^{\infty}\frac{x^{2}}{\theta^{2}}e^{\frac{-x^{2}}{2\theta^{2}}}dx \:= \:
-\int_{0}^{\infty}\: x \: e^{\frac{-x^{2}}{2\theta^{2}}} \: d({\frac{-x^{2}}{2\theta^{2}}}) \:=
$

$= -x \: e^{\frac{-x^{2}}{2\theta^{2}}} \: |_{0}^{\infty} \:+\int_{0}^{\infty}\ e^{\frac{-x^{2}}{2\theta^{2}}}dx \: =$

$= \int_{0}^{\infty}\ e^{\frac{-x^{2}}{2\theta^{2}}}dx \:$

Not sure I follow the last part. Could you possible expand on this part

$-\int_{0}^{\infty}\: x \: e^{\frac{-x^{2}}{2\theta^{2}}} \: d({\frac{-x^{2}}{2\theta^{2}}}) \:=
$

$= -x \: e^{\frac{-x^{2}}{2\theta^{2}}} \: |_{0}^{\infty} \:+\int_{0}^{\infty}\ e^{\frac{-x^{2}}{2\theta^{2}}}dx \: =\int_{0}^{\infty}\ e^{\frac{-x^{2}}{2\theta^{2}}}dx \:$
• Apr 14th 2010, 05:19 PM
zzzoak
$-\int_{0}^{\infty}\: x \: e^{\frac{-x^{2}}{2\theta^{2}}} \: d({\frac{-x^{2}}{2\theta^{2}}}) \:=
$

$
= \: -\int_{0}^{\infty}\: x \: d( \:e^{\frac{-x^{2}}{2\theta^{2}}} \:)
$

and
$u =x \: \: \: v=e^{\frac{-x^{2}}{2\theta^{2}}}$
$\int udv = uv - \int v du$
• Apr 14th 2010, 07:13 PM
Sudharaka
Quote:

Originally Posted by Danneedshelp
Evalutate

$\int_{0}^{\infty}\frac{x^{2}}{\theta^{2}}e^{\frac{-x^{2}}{2\theta^{2}}}dx$.

I started by letting

$u=x^{2}\Rightarrow\\du=2x$
and $dv=e^{\frac{-x^{2}}{2\theta^{2}}}\Rightarrow\\$?

I believe I need to use polar coordinates, but I not sure how. Any help would be greatly appreciated.

Thank you

Dear Danneedshelp,

Can you please tell me whether $\theta$ is a constant? If not what is the relationship between $x~and~\theta$ ?
• Apr 14th 2010, 09:55 PM
Danneedshelp
Quote:

Originally Posted by Sudharaka
Dear Danneedshelp,

Can you please tell me whether $\theta$ is a constant? If not what is the relationship between $x~and~\theta$ ?

Yes, $\theta$ is a constant.
• Apr 14th 2010, 10:29 PM
Danneedshelp
Quote:

Originally Posted by zzzoak
$-\int_{0}^{\infty}\: x \: e^{\frac{-x^{2}}{2\theta^{2}}} \: d({\frac{-x^{2}}{2\theta^{2}}}) \:=
$

$
= \: -\int_{0}^{\infty}\: x \: d( \:e^{\frac{-x^{2}}{2\theta^{2}}} \:)
$

and
$u =x \: \: \: v=e^{\frac{-x^{2}}{2\theta^{2}}}$
$\int udv = uv - \int v du$

$I^{2}$ $=\int_{\mathbb{R^{2}}}exp(\frac{-(x^{2}+y^{2})}{2\theta^{2}})dA$ $=
\int_{0}^{2\pi}\int_{0}^{\infty}exp(\frac{-r^{2}}{2\theta^{2}})rdrd\theta$
$=
2\pi\int_{0}^{\infty}exp(\frac{-r^{2}}{2\theta^{2}})rdr$
. Let $u=r^{2}$ and $\frac{du}{2}=rdr$, then
$\pi\int_{0}^{\infty}exp(\frac{-u}{2\theta^{2}})du=\pi(0-(-2\theta^{2}))=2\pi\theta^2$.

Thus,

$\sqrt{I^{2}}=I=\theta\sqrt{2\pi}$.

The correct answer is $\theta\sqrt{\frac{\pi}{2}}$. So, I am still stuck on this problem. I am not sure what I am doing wrong. I would have the correct answer if I integrated from 0 to $\pi$ with respect to $\theta$, but I don't thank that would be proper.

Note: The polar coordinates make things a little confusing, because the $\theta$ in the integrand is a constant and not related to $d\theta$.
• Apr 15th 2010, 12:18 AM
HallsofIvy
You asserted, originally, that the problem was to integrate
$\int_0^\infty \frac{x^2}{\theta^2}e^{-\frac{x^2}{2\theta}}dx$

I suspect you have accidently switched to finding
$\int_{-\infty}^\infty \frac{x^2}{\theta^2}e^{-\frac{x^2}{2\theta}}dx$
• Apr 15th 2010, 10:45 AM
Danneedshelp
Quote:

Originally Posted by zzzoak
$-\int_{0}^{\infty}\: x \: e^{\frac{-x^{2}}{2\theta^{2}}} \: d({\frac{-x^{2}}{2\theta^{2}}}) \:=
$

$
= \: -\int_{0}^{\infty}\: x \: d( \:e^{\frac{-x^{2}}{2\theta^{2}}} \:)
$

and
$u =x \: \: \: v=e^{\frac{-x^{2}}{2\theta^{2}}}$
$\int udv = uv - \int v du$

I'm sorry to keep bringing this up, but I dont see how

http://www.mathhelpforum.com/math-he...2ecbd45d-1.gif

and the rest of the expression above. I don't really understand your notation. I do see how you are using integration by parts, but I don't understand $d(\frac{-x^{2}}{2\theta^{2}})$. I just don't see the equality posted above.

Sorry for beating a dead horse, but I want to understand this.

Thank you again.
• Apr 15th 2010, 12:03 PM
Quote:

Originally Posted by Danneedshelp
I'm sorry to keep bringing this up, but I dont see how

http://www.mathhelpforum.com/math-he...2ecbd45d-1.gif

and the rest of the expression above. I don't really understand your notation. I do see how you are using integration by parts, but I don't understand $d(\frac{-x^{2}}{2\theta^{2}})$. I just don't see the equality posted above.

Sorry for beating a dead horse, but I want to understand this.

Thank you again.

I presume this is what's happening but here's another way of writing your integral...

$\int_0^{\infty} \frac{x^2}{\theta^2}e^{\frac{-x^2}{2\theta^2}}dx = \int_0^{\infty} \frac{x}{\theta^2}xe^{\frac{x^2}{2\theta^2}}dx$

Now $d(\frac{-x^2}{2\theta^2})$ means the derivative of $\frac{-x^2}{2\theta^2}$ which is...

$\frac{-2x}{2\theta^2} = -\frac{x}{\theta^2}$

so we can write...

$\int_0^{\infty} \frac{x}{\theta^2}xe^{\frac{-x^2}{2\theta^2}}dx = -\int_0^{\infty} xe^{\frac{-x^2}{2\theta^2}}d(\frac{x^2}{2\theta^2})dx$

However... I'm not following his working 100% so what I would have done is set...

$u=\frac{x}{\theta^2}$, $u'=\frac{1}{\theta^2}$

and

$v = -\theta^2 e^{\frac{-x^2}{2\theta^2}}$

(since $v' = xe^{\frac{-x^2}{2\theta^2}}$)

And then apply $uv = \int vu'$
• Apr 15th 2010, 12:33 PM
Quote:

I presume this is what's happening but here's another way of writing your integral...

$\int_0^{\infty} \frac{x^2}{\theta^2}e^{\frac{-x^2}{2\theta^2}}dx = \int_0^{\infty} \frac{x}{\theta^2}xe^{\frac{x^2}{2\theta^2}}dx$

Now $d(\frac{-x^2}{2\theta^2})$ means the derivative of $\frac{-x^2}{2\theta^2}$ which is...

$\frac{-2x}{2\theta^2} = -\frac{x}{\theta^2}$

so we can write...

$\int_0^{\infty} \frac{x}{\theta^2}xe^{\frac{-x^2}{2\theta^2}}dx = -\int_0^{\infty} xe^{\frac{-x^2}{2\theta^2}}d(\frac{x^2}{2\theta^2})dx$

However... I'm not following his working 100% so what I would have done is set...

$u=\frac{x}{\theta^2}$, $u'=\frac{1}{\theta^2}$

and

$v = -\theta^2 e^{\frac{-x^2}{2\theta^2}}$

(since $v' = xe^{\frac{-x^2}{2\theta^2}}$)

And then apply $uv = \int vu'$

Lol now I see what zzzoak was doing...

$\int_0^{\infty} \frac{x}{\theta^2}xe^{\frac{-x^2}{2\theta^2}}dx = \int_0^{\infty} x\frac{x}{\theta^2}e^{\frac{-x^2}{2\theta^2}}dx$

And taking $u=x$ and $v'=\frac{x}{\theta^2}e^{\frac{-x^2}{2\theta^2}}$

and hence

$v = \int \frac{x}{\theta^2}e^{\frac{-x^2}{2\theta^2}} dx = -e^{\frac{-x^2}{2\theta^2}}$

$u=x$,

$u' = 1$,

$v = -e^{\frac{-x^2}{2\theta^2}}$.