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Math Help - Solve the integral problem.......

  1. #1
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    Solve the integral problem.......

    Can anyone solve the problem given in the attachment? here a, b and 2pi are constants. Please show the steps while u solve it.

    Bye.
    Attached Thumbnails Attached Thumbnails Solve the integral problem.......-image-50.jpg  
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  2. #2
    Super Member 11rdc11's Avatar
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    What did you get so far and is this what it looks like?

    \frac{1}{a2\pi}\int_{-\infty}^{\infty} ye^{(\frac{-y^2}{b})}~dy

    \frac{-be^{(\frac{-y^2}{b})}}{a4\pi}\bigg|^{\infty}_{-\infty}
    Last edited by 11rdc11; September 28th 2008 at 05:29 PM.
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  3. #3
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    the function is odd (f(-x)=-f(x)), so the integral is zero.
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  4. #4
    Super Member Showcase_22's Avatar
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    Well I want to test the previous post. If it's true then i'm going to have a lot less work to do!

    \frac {1}{a(2 \pi)}\int_{-\infty}^{+\infty}ye^{ \frac {-y^2}{b}}dy

    u=\frac {-y^2}{b}

    \frac {du}{dy}=\frac{-2y}{b}

    dy= \frac {b du}{-2y}

    The limits remain the same. I do not think it is necessary to write that part of the working out.

    \frac{1}{a(2 \pi)(-2)}\int_{- \infty}^{\infty} \frac {y}{y}udu

     \frac{1}{-4a \pi}\int_{- \infty}^{\infty} udu

    \frac {1}{-4a \pi} [\frac{u^2}{2}]^{\infty}_{-\infty}

    \frac {1}{-8a \pi}[u^2]^{\infty}_{-\infty}

    \frac {1}{-8a \pi}((\infty)^{2}-(-\infty)^{2})

    \frac {1}{-8a \pi}(0)=0

    WOW! That is the find of the century for me!!!!

    (oh, and that's how I would integrate it.)
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  5. #5
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    showcase, you have two mistakes in your arguments -- namely, with "the limits stay the same", and after the change of variables you lost exponential for some reason.

    basically 11rdc11 wrote out the solution -- the primitive is \frac{-be^{(\frac{-y^2}{b})}}{a4\pi} and then the integral is \lim_{y\to\infty}\frac{-be^{(\frac{-y^2}{b})}}{a4\pi}-\lim_{y\to -\infty}\frac{-be^{(\frac{-y^2}{b})}}{a4\pi}=0.

    the short way of this, as I said, is to note that the function is odd, so the integral is zero immediately, though I am sloppy here, because for this argument to work, one needs to check that the integral is finite at first
    Last edited by choovuck; September 29th 2008 at 03:05 PM.
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  6. #6
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by choovuck View Post
    the function is odd (f(-x)=-f(x)), so the integral is zero.

    Lol I forgot about it being odd. It would have saved me some work
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  7. #7
    Super Member Showcase_22's Avatar
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    Good point!

    I was doing that late last night and still getting the hang of Latex. :s
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