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Thread: Difficult integral

  1. #1
    Senior Member Danneedshelp's Avatar
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    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
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  2. #2
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    Please use this

    $\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 \: $
    Last edited by zzzoak; Apr 14th 2010 at 02:26 PM.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by zzzoak View Post
    Please use this

    $\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 \: $
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  4. #4
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    $\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}}} $
    $\displaystyle \int udv = uv - \int v du$
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  5. #5
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    Quote Originally Posted by Danneedshelp View Post
    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 $\displaystyle x~and~\theta$ ?
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  6. #6
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Sudharaka View Post
    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.
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  7. #7
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by zzzoak View Post
    $\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 $\displaystyle d\theta$.
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  8. #8
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    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
    $\displaystyle \int_{-\infty}^\infty \frac{x^2}{\theta^2}e^{-\frac{x^2}{2\theta}}dx$
    and so your answer needs to be divided by 2.
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  9. #9
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by zzzoak View Post
    $\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.
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  10. #10
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    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 $\displaystyle uv = \int vu'$
    Last edited by Deadstar; Apr 15th 2010 at 11:06 AM. Reason: missed minus signs...
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  11. #11
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Deadstar View Post
    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

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


    $\displaystyle u=x$,

    $\displaystyle u' = 1$,

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