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How to solve an integral of a gaussian distribution

  1. Aug 19, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_{-\infty}^\infty\! e ^{(x-a)^{2}}\, dx[/tex]

    2. Relevant equations

    [tex]\int \! e^u\, du = e^u + C[/tex]
    3. The attempt at a solution
    i just know that du = 2(x-a), but there is no x to make use of substitution, so I am confused on how to go about solving this since I cannot use substitution
  2. jcsd
  3. Aug 19, 2009 #2
    EDIT: I'm assuming you mean
    \int_{-\infty}^\infty\! e ^{-(x-a)^{2}}\, dx
    since what you wrote doesn't converge.

    Anyway, this is a very tricky integral that requires a trick. By far the best way starts as follows:

    First, letting I denote the original integral (after replacing x-a with x), notice that
    [tex]I = 2\int_{0}^\infty\! e ^{-x^{2}}\, dx[/tex]​
    by symmetry. Next comes the first trick:
    [tex]I^2 = 4 \int_{0}^\infty\! e ^{-x^{2}}\, dx \int_{0}^\infty\! e ^{-x^{2}}\, dx = 4 \int_{0}^\infty\! e ^{-x^{2}}\, dx \int_{0}^\infty\! e ^{-y^{2}}\, dy[/tex]​
    since the x in the integral is just a dummy variable. But then
    [tex]I^2 = 4\int_0^\infty\int_0^\infty\! e ^{-(x^{2}+y^{2})}\,dx\,dy.[/tex]​

    The traditional method is to at this point switch to polar coordinates, but a better way (I think) involves the substitution y=xu (so dy=xdu):
    [tex]I^2 = 4\int_0^\infty\int_0^\infty\! e^{-x^2(1+u^2)}x\,dx\,du.[/tex]​
    From here, I think you should be able to figure this out on your own.
  4. Aug 20, 2009 #3
    hmm my text book is certainly not nice as it leaves this as an exercise with just telling me "(look up any integrals you need)"
    and yes, i forgot to include the - sign for the exponential
  5. Aug 20, 2009 #4
    That's wierd.
  6. Aug 21, 2009 #5


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    Homework Helper

    On a wild hunch: this seems to be a traditional type problem in a first mathematical stat or calc-based probability course. If you've seen this (normal density integrating to 1)

    \int_{-\infty}^\infty \frac 1 {\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{\sigma^2}} \, dx = 1

    the "trick" is to identify your integrand as part of some normal density and go from there.

    of course, the item i have posted could be what the hint "look up any integrals needed" refers to as well.
  7. Aug 21, 2009 #6
    You mean
    \int_{-\infty}^\infty \frac 1 {\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \, dx = 1
    of course
  8. Aug 22, 2009 #7


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    Homework Helper

    Yes - many thanks for correcting my typing, and more for possibly clearing up a problem for the original poster.
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