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Integration of e^(-x^2)

  1. Apr 13, 2005 #1
    Hi everybody,
    Do you have any idea how this is solved?
    [tex]\int_{-\infty}^{+\infty} e^{-x^2} dx =?[/tex]
  2. jcsd
  3. Apr 13, 2005 #2

    matt grime

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    yes. first note the integral exists, let its value be T.

    By some theorem

    T^2 = [tex]\int_{-\infty}^{+\infty} e^{-x^2} dx\int_{-\infty}^{+\infty} e^{-y^2} dy[/tex]

    which equals

    [tex]\int_{-\infty}^{+\infty} e^{-x^2-y^2} dxdy[/tex]

    now put it into polars
  4. Apr 13, 2005 #3
    I=[tex]\int_{0}^{+\infty} e^{-x^2} dx[/tex], is solved using a trick, and integrating in polar coordinates, in which the Jacobian helps solving it. It is equal to [tex]\frac{\sqrt{\pi}}{2}[/tex]. Your integral is just 2I, I think.
  5. Apr 13, 2005 #4


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    Oggy,the "trick" applies for his integral.Yours,if u apply Fubini's theorem,will not lead to an integral over [itex]\mathbb{R}^{2} [/itex],but over the semiplane [itex] x\geq 0 [/itex] and it wouldn't be the same...

    To evaluate that integral (and to get the value u wrote),u need another "trick":the integrand is even,therefore

    [tex] \int_{0}^{+\infty} e^{-x^{2}} \ dx =\frac{1}{2}\int_{\mathbb{R}} e^{-x^{2}} \ dx [/tex]

  6. Apr 13, 2005 #5
    elaborating with what dexter had stated:

    [tex]\int_{0}^{\infty} e^{-x^2} dx[/tex]

    Convert to DI problem:

    [tex]{(\int_{0}^{\infty} e^{-x^2} dx)}^2[/tex]

    [tex]=(\int_{0}^{\infty} e^{-x^2} dx) (\int_{0}^{\infty} e^{-y^2} dy)[/tex]

    [tex]=\int_{0}^{\infty}\int_{0}^{\infty} e^{-x^2}e^{-y^2} dxdy[/tex]

    [tex]=\int_{0}^{\infty}\int_{0}^{\infty} e^{-{(x^2+y^2)}} dxdy[/tex]

    From here using a Change of Variables step or noticing the region we are integrating is simply a semicircular arc, we can use polar coordinates:

    [tex]=\int_{0}^{\frac{\pi}{2}}\int_{0}^{R} e^{-r^2}r drd\theta[/tex]

    Where we take the limit of this as R --> infinity and the extra r in the integrand is from the Jacobian when switching to polar coordinates (proven in the change of variables theorem)

    And NOW we can use a U-substitution to easily solve this iterated double integral:

    [tex]{(\int_{0}^{\infty} e^{-x^2} dx)}^2 = \frac{\pi}{4}[/tex]


    [tex]\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}[/tex]

    It seems like we can solve quite a few "definite" or "improper" integrals of non elementary functions using multivar. calculus. Does anyone know any other methods?
  7. Apr 13, 2005 #6


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    The theorem of residues of A.Cauchy never fails...:wink:

  8. Apr 14, 2005 #7
    That's exactly what I refered to.
  9. Apr 14, 2005 #8
    Well... you didn't explicitly state residue calculus now did you :wink:
  10. Apr 14, 2005 #9
    Thanks for your help. I am not very familiar with integration theorems but i wanted to calculate this integral because the normal distibution function has a term like this
    (e^(-x^2)) and the integral is 1. I checked it in Mathematica and the indefinite integral of the function g(x)=e^(-x^2) is (1/2)*[(pi)^(1/2)]*erf(x). Do u know what this erf(x) is? (a simple explanation as I am not very advanced in analysis)
  11. Apr 14, 2005 #10


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    The way i know it (it coincides with the definition swallowed by Mathematica and my Maple)

    [tex] \mbox{erf} (x)=:\frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^{2}} \ dt [/tex]

  12. Apr 14, 2005 #11


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