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Homework Help: Guassian Intergral doesn't seem to sum to sqrt(pi)

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    According to

    3. The attempt at a solution
    I have got as far as the integral at:

    [tex]e^{-\frac{w^{2}}{4} } \int e^{-(t+ \frac{jw}{2})^{2}}dt[/tex]

    Apparently this makes the correct answer but I don't see how the integral evaluates to sqrt(pi)

    Can somebody please explain

    Last edited: Apr 9, 2012
  2. jcsd
  3. Apr 9, 2012 #2

    Ray Vickson

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    Homework Helper

    Your integral is wrong; you have written
    [tex]e^{-\frac{w^{2}}{4}\int e^{-(t^{2}+j\frac{w}{2})^{2}}},[/tex]
    whiich is not what the cited article wrote.

  4. Apr 9, 2012 #3
  5. Apr 9, 2012 #4
    Last edited: Apr 10, 2012
  6. Apr 10, 2012 #5
    Maybe i wasn't very clear. I cannot see how

    [tex]\int e^{-(t+ \frac{jw}{2})^{2}}dt[/tex]

    Becomes sqrt(pi). I can't be just sqrt(pi), as we don't just have the integral of e^(t^2))
  7. Apr 10, 2012 #6


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    [tex]\int_{-\infty}^\infty e^{-(t+ j\omega/2)^2}dt[/tex]

    Let [itex]u= t+ j\omega^2/2[/itex]. [itex]du= dt[/itex]. The integral becomes
    [tex]\int_{-\infty}^\infty e^{-u^2}{du}[/tex]

    That's a well known integral. If you haven't seen it before, let
    [tex]I= \int_{-\infty}^\infty e^{-u^2}{du}[/tex]
    so that, because of the symmetry,
    [tex]I/2= \int_0^\infty e^{-u^2}{du}[/tex]
    Then it is also true that
    [tex]I/2= \int_0^\infty e^{-v^2}dv[/tex]
    and, multiplying them
    [tex]I^2/4= \left(\int_0^\infty e^{-u^2}du\right)\left(\int_0^\infty e^{-v^2}dv\right)= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-u^2-v^2}dudv[/tex]

    Now change to polar coordinates, r and [itex]\theta[/itex], with [itex]u= r cos(\theta)[/itex] and [itex]v= r sin(\theta)[/itex]. The uv-integral takes each variable from 0 to [itex]\infty[/itex], the first quadrant, which, in terms of polar coordinates, has r from 0 to [itex]\infty[/itex] and [itex]\theta[/itex] from 0 to [itex]\pi/2[/itex]. Of course, [itex]-(u^2+v^2)=-r^2[/itex] and [itex]dudv= r drd\theta[/itex]

    The integral becomes
    [tex]\int_{\theta= 0}^{\pi/2}\int_{r=0}^\infty e^{-r^2}rdrd\theta= \frac{\pi}{2}\int_{r=0}^\infty e^{-r^2}rdr[/tex]

    To do that last integral, let [itex]x= r^2[/itex] so that dx= 2rdr and rdr= (1/2)dx.
    Now we have
    [tex]I^2/4= \frac{\pi}{4}\int_0^\infty e^{-x}dx[/tex]

    That last integral is, of course, 1 so we have [itex]I^2/4= \pi/4[/itex] and [itex]I= \sqrt{\pi}[/itex].
  8. Apr 14, 2012 #7
    nice going HallOfIvy, very good :)
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