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Math Help - what is the value of this integral

  1. #1
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    what is the value of this integral

    \int_{-\epsilon}^{\epsilon}{\frac{1}{\sqrt{\pi}}e^{-\frac{x^{2}}{t^{2}+1}}(t^{2}+1)^{-\frac{1}{2}}dx}

    epsilon is infinitesimal
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by szpengchao View Post
    \int_{-\epsilon}^{\epsilon}{\frac{1}{\sqrt{\pi}}e^{-\frac{x^{2}}{t^{2}+1}}(t^{2}+1)^{-\frac{1}{2}}dx}

    epsilon is infinitesimal
    Note that \int_{-\epsilon}^{\epsilon}{\frac{1}{\sqrt{\pi}}e^{-\frac{x^{2}}{t^{2}+1}}(t^{2}+1)^{-\frac{1}{2}}dx}=\int_{-\epsilon}^{\epsilon}{\frac{1}{\sqrt{\pi}}e^{-\left(\frac{x}{\sqrt{t^{2}+1}}\right)^2}(t^{2}+1)^  {-\frac{1}{2}}dx}

    Let u=\frac{x}{\sqrt{t^2+1}}\implies \,du=\frac{\,dx}{\sqrt{t^2+1}}

    Thus,

    \int_{-\epsilon}^{\epsilon}{\frac{1}{\sqrt{\pi}}e^{-\left(\frac{x}{\sqrt{t^{2}+1}}\right)^2}(t^{2}+1)^  {-\frac{1}{2}}dx}=\frac{1}{2}\int_{-\frac{\epsilon}{\sqrt{t^2+1}}}^{\frac{\epsilon}{\s  qrt{t^2+1}}}\frac{2}{\sqrt{\pi}}e^{-u^2}\,du=\left.\frac{1}{2}\,\text{erf}\left(u\righ  t)\right|_{-\frac{\epsilon}{\sqrt{t^2+1}}}^{\frac{\epsilon}{\s  qrt{t^2+1}}}

    =\frac{1}{2}\left[\text{erf}\left(\frac{\epsilon}{\sqrt{t^2+1}}\righ  t)-\text{erf}\left(-\frac{\epsilon}{\sqrt{t^2+1}}\right)\right]=\frac{1}{2}\left[2\,\text{erf}\left(\frac{\epsilon}{\sqrt{t^2+1}}\r  ight)\right]=\color{red}\boxed{\text{erf}\left(\frac{\epsilon}  {\sqrt{t^2+1}}\right)}

    Does this make sense?

    --Chris
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  3. #3
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    what

    what is that erf there?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by szpengchao View Post
    what is that erf there?
    Its the error function. See here.
    Last edited by Chris L T521; October 14th 2008 at 08:41 PM.
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