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Math Help - Heat diffusion, Gaussian

  1. #1
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    Heat diffusion, Gaussian

    Hi,
    I need to evaluate intergral (-inf, inf) of S(x-y, t)dx where y element R and where s(x, t) = (1/2sqrt(pi*k*t))e^((-x^2)/4kt). You are supposed to use the fact that intergral (-inf, inf) e^-ax^2dx = sqrt(pi/a). I know there is some trick you need to do at the beginning but not sure what so that it can be simplified easily.
    Thanks
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tbyou87 View Post
    Hi,
    I need to evaluate intergral (-inf, inf) of S(x-y, t)dx where y element R and where s(x, t) = (1/2sqrt(pi*k*t))e^((-x^2)/4kt). You are supposed to use the fact that intergral (-inf, inf) e^-ax^2dx = sqrt(pi/a). I know there is some trick you need to do at the beginning but not sure what so that it can be simplified easily.
    Thanks
    Let me attempt to translate this into LaTeX:
    \int_{-\infty}^{\infty}\frac{1}{2}\sqrt{\pi k t}~e^{-(x - y)^2/4kt}dx

    using
    \int_{\infty}^{\infty}e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    Let me attempt to translate this into LaTeX:
    \int_{-\infty}^{\infty}\frac{1}{2}\sqrt{\pi k t}~e^{-(x - y)^2/4kt}dx

    using
    \int_{\infty}^{\infty}e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}

    -Dan
    It looks to me like you can do the following:
    \int_{-\infty}^{\infty}\frac{1}{2}\sqrt{\pi k t}~e^{-(x - y)^2/4kt}dx = \frac{1}{2}\sqrt{\pi k t} \int_{-\infty}^{\infty}e^{-(x - y)^2/4kt}dx
    (since t is constant with respect to the integral.)

    Now let u = x - y \implies du = dx
    = \frac{1}{2}\sqrt{\pi k t} \int_{-\infty}^{\infty}e^{-u^2/4kt}du

    and temporarily set a = \frac{1}{4kt}:
    = \frac{1}{2}\sqrt{\pi k t} \int_{-\infty}^{\infty}e^{-au^2}du

    = \frac{1}{2}\sqrt{\pi k t} \int_{-\infty}^{\infty}e^{-au^2}du

    = \frac{1}{2}\sqrt{\pi k t} \sqrt{4 \pi kt}

    = \pi kt

    But it worries me slightly that this is no longer a function of y.

    -Dan
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