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Thread: Convolution Of Exponential Function

  1. January 24th 2011, 09:41 AM #1
    mathematicalbagpiper
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    Convolution Of Exponential Function

    Let f(x)=e^{-x^2}. Compute f*f.

    So I get:

    f*f=\int_{-\infty}^{\infty}e^{-y^2}e^{-(x-y)^2}dy

    =\int_{-\infty}^{\infty}e^{-y^2}e^{-y^2+2xy-x^2}dy

    =\int_{-\infty}^{\infty}e^{-2y^2+2xy-x^2}dy

    =e^{-x^2}\int_{-\infty}^{\infty}e^{-2y^2+2xy}dy

    =f(x)\int_{-\infty}^{\infty}e^{-2(y^2-xy)}dy

    =f(x)\int_{-\infty}^{\infty}e^{-2(y-\frac{x}{2})^2+\frac{x^2}{2}}dy

    =f(x)e^{\frac{x^2}{2}}\int_{-\infty}^{\infty}e^{-2(y-\frac{x}{2})^2}dy

    ...and that's as far as I can get :-(
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  2. January 24th 2011, 09:58 AM #2
    Ackbeet
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    Looks good so far. Try the substitution u=\sqrt{2}\,(y-x/2). Where does that lead you?
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  3. January 25th 2011, 04:51 AM #3
    mathematicalbagpiper
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    Leads me to a nice result using the improper integral of e^{-u^2}du, and the resulting following calculation.

    Got it! Thanks! :-D
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  4. January 25th 2011, 05:39 AM #4
    Ackbeet
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    What was your final answer?
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  5. January 25th 2011, 05:46 AM #5
    mathematicalbagpiper
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    I ended up with \sqrt{\frac{\pi}{2}}\cdot f(\frac{x}{\sqrt{2}}) if I remember correctly.
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  6. January 25th 2011, 05:47 AM #6
    Ackbeet
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    Same here.
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