1. ## Difficult integral

Evalutate

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

I started by letting

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

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

Thank you

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

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

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

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

Notice that

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

Lets find integral

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

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

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

3. Originally Posted by zzzoak

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

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

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

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

Notice that

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

Lets find integral

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

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

$\displaystyle = \int_{0}^{\infty}\ e^{\frac{-x^{2}}{2\theta^{2}}}dx \:$
Not sure I follow the last part. Could you possible expand on this part

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

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

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

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

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

5. Originally Posted by Danneedshelp
Evalutate

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

I started by letting

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

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 $\displaystyle \theta$ is a constant? If not what is the relationship between $\displaystyle x~and~\theta$ ?

6. Originally Posted by Sudharaka
Dear Danneedshelp,

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

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

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

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

Thus,

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

The correct answer is $\displaystyle \theta\sqrt{\frac{\pi}{2}}$. 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 $\displaystyle \pi$ with respect to $\displaystyle \theta$, but I don't thank that would be proper.

Note: The polar coordinates make things a little confusing, because the $\displaystyle \theta$ in the integrand is a constant and not related to $\displaystyle d\theta$.

8. You asserted, originally, that the problem was to integrate
$\displaystyle \int_0^\infty \frac{x^2}{\theta^2}e^{-\frac{x^2}{2\theta}}dx$

I suspect you have accidently switched to finding
$\displaystyle \int_{-\infty}^\infty \frac{x^2}{\theta^2}e^{-\frac{x^2}{2\theta}}dx$

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

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

and
$\displaystyle u =x \: \: \: v=e^{\frac{-x^{2}}{2\theta^{2}}}$
$\displaystyle \int udv = uv - \int v du$
I'm sorry to keep bringing this up, but I dont see how

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 $\displaystyle d(\frac{-x^{2}}{2\theta^{2}})$. I just don't see the equality posted above.

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

Thank you again.

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

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 $\displaystyle d(\frac{-x^{2}}{2\theta^{2}})$. 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...

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

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

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

so we can write...

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

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

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

and

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

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

And then apply $\displaystyle uv = \int vu'$

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

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

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

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

so we can write...

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

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

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

and

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

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

And then apply $\displaystyle uv = \int vu'$

Lol now I see what zzzoak was doing...

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

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

and hence

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

$\displaystyle u=x$,

$\displaystyle u' = 1$,

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