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  1. #1
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    difficult improper integral

    integral (from 0 to infinite) of e^(-2x) / sqrt(x). i tried letting u = sqrt(x) so that x = u^2 and dx = 2u du and that turned my integral into 2 integral (from 0 to infinite) of e^(-2u^2) du. how would you do this integral? i checked wolfram alpha and it showed something about the error function but it didn't show the steps.
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  2. #2
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    Quote Originally Posted by oblixps View Post
    integral (from 0 to infinite) of e^(-2x) / sqrt(x). i tried letting u = sqrt(x) so that x = u^2 and dx = 2u du and that turned my integral into 2 integral (from 0 to infinite) of e^(-2u^2) du. how would you do this integral? i checked wolfram alpha and it showed something about the error function but it didn't show the steps.

    Its an unelementary integral.
    Post the whole problem. The problem asks you to find its value or just determine its convergence/divergence?
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  3. #3
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    Quote Originally Posted by oblixps View Post
    integral (from 0 to infinite) of e^(-2x) / sqrt(x). i tried letting u = sqrt(x) so that x = u^2 and dx = 2u du and that turned my integral into 2 integral (from 0 to infinite) of e^(-2u^2) du. how would you do this integral? i checked wolfram alpha and it showed something about the error function but it didn't show the steps.
    You have it. First recall \int_0^{\infty} e^{-w^2}dw=\frac{\sqrt{\pi}}{2}

    Now you have 2\int_0^{\infty}e^{-2u^2}du. How about letting w=\sqrt{2}u
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  4. #4
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    Quote Originally Posted by shawsend View Post
    You have it. First recall \int_0^{\infty} e^{-w^2}dw=\frac{\sqrt{\pi}}{2}

    Now you have 2\int_0^{\infty}e^{-2u^2}du. How about letting w=\sqrt{2}u
    There is a well known trick to evaluating this last integral, which is to consider:

    I^2=\left[\int_{-\infty}^{\infty}e^{-u^2}du\right]^2=\int_{-\infty}^{\infty}e^{-x^2}dx \int_{-\infty}^{\infty}e^{-y^2}dy=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy

    Now we can rewrite the last integral in polars instead of cartesians:

    \int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy=\int_{r=0}^{\infty}\int_{\theta=0}^  {2\pi}e^{-r^2} rd\theta dr

    which can be evaluated by elementary means.

    CB
    Last edited by CaptainBlack; March 7th 2010 at 09:05 PM.
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    Quote Originally Posted by CaptainBlack View Post
    There is a well known trick to evaluating this last integral, which is to consider:

    I^2=\left[\int_0^{\infty}e^{-u^2}du\right]^2=\int_0^{\infty}e^{-x^2}dx \int_0^{\infty}e^{-y^2}dy=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy

    Now we can rewrite the last integral in polars instead of cartesians:

    \int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy=\int_{r=0}^{\infty}\int_{\theta=0}^  {2\pi}e^{-r^2} rd\theta dr

    which can be evaluated by elementary means.

    CB
    would the limits on theta be from 0 to pi/2? since the original limits of e^(-x^2) are from 0 to infinite so we are considering the first quadrant.
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  6. #6
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    Quote Originally Posted by oblixps View Post
    would the limits on theta be from 0 to pi/2? since the original limits of e^(-x^2) are from 0 to infinite so we are considering the first quadrant.
    There is a mistake in my posting the first two integrals should be from -\infty to +\infty, then I want to leave it to the reader to divide I by 2 to get what you asked for. But changing the integral to be the first quadrant works as well (it shift where we divide by 2 to a different place but just as valid).

    CB
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