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Integrating exp(x^2) like gaussian integral?

  1. Oct 30, 2009 #1
    Integrating exp(x^2) like gaussian integral??

    Hi,
    I can't solve this integral [tex]\int^{1}_{0}\\e^{x^2}\\dx[/tex]

    Can I solve this integral like gaussian integral?
    Please help me
    Thanks.
     
  2. jcsd
  3. Oct 30, 2009 #2
    Re: Integrating exp(x^2) like gaussian integral??

    The Gaussian integral you are referring to is usually being integrated from minus infinity to plus infinity. Integrals like yours can be expressed in terms of the so-called error function but not in terms of more elementary functions.

    The numerical value of the integral in your case is 1.46265.
     
  4. Oct 30, 2009 #3
    Re: Integrating exp(x^2) like gaussian integral??

    I found a solvation but I am not sure that the solvation is right.Firstly I write the integral in cartesian coordinates as surface integral after in polar coordinates.
    Where [tex]r=\sqrt{x^2+y^2}[/tex]

    [tex]\int_S {e^{r^2}}dA=\int_{0}^{1} \int_{0}^{1} e^{{\sqrt{x^2+y^2}}^2}dxdy=\int_{0}^{1} e^{x^2}dx\int_{0}^{1} e^{y^2}dy=[\int_{0}^{1}e^{x^2}]^2[/tex]

    [tex]\int_S {e^{r^2}}dA=\int_{0}^{\pi /4} \int_{0}^{1} e^{r^2}rdrd\theta=\frac{\pi }{8}{(e-1)}=[\int_{0}^{1}e^{x^2}]^2[/tex]

    I determine the limits from graph of [tex]e^{x^2}[/tex]

    [tex]\int_{0}^{1}e^{x^2}=\sqrt{\frac{\pi }{8}{(e-1)}}[/tex]

    I solve this problem like solvation of gaussian integral but I'm in doubt that it is right.
    Please help me and show me right solvation.
    Thanks.
     
    Last edited: Oct 30, 2009
  5. Oct 30, 2009 #4
    Re: Integrating exp(x^2) like gaussian integral??

    Your result is incorrect and the reason is that the rectangle [0,1] x [0,1] in Cartesian coordinates cannot easily be parametrized in polar coordinates, in particular it does not correspond to [0,pi/4] x [0,1] , which describes an 45 degree sector of the unit disc.
     
  6. Oct 30, 2009 #5

    Gib Z

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    Re: Integrating exp(x^2) like gaussian integral??

    If you don't mind getting a infinite series for an answer, replacing the integrand with its Taylor series and integrating term by term gives a relatively nice looking series:

    [tex]\sum_{n=0}^{\infty} \frac{1}{(2n+1) n!}[/tex]
     
  7. Nov 1, 2009 #6
    Re: Integrating exp(x^2) like gaussian integral??

    If that is so, then what can we do to solve this integral?I worked to solve this integral like the solvation of gaussian integral but I can't find the parameters of integrals.If this solution is right, what is the parameters of integrals else what is the right solution of this problem?
    I really wonder the solution.
    Please help me,
    Thanks in advance.
     
  8. Nov 2, 2009 #7

    Gib Z

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    Re: Integrating exp(x^2) like gaussian integral??

    It has no more elementary expression other than in terms of the Error Function, which is pretty much just shorthand notation for that integral anyway, or you could try the series I put up if you like that better. There is NO "nicer" solution in terms of things like pi and e.
     
  9. Nov 2, 2009 #8
    Re: Integrating exp(x^2) like gaussian integral??

    I gave this series to Maple, and the result it gave me involves the error function in the form erfc:

    [tex]\frac{i\sqrt {\pi }}{2} \left( -1+{\rm erfc} \left( i \right) \right) [/tex]
     
  10. Nov 3, 2009 #9

    Gib Z

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    Re: Integrating exp(x^2) like gaussian integral??

    Erfc is the complementary function of the error function, and is often defined to be 1-erf(x).

    Since [tex]erf(x)=\frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} dt[/tex] and through some formal manipulations, we first let x= i, and then change variables via u= it we can easily show how the expression maple gives is indeed [tex]\int^1_0 e^{t^2} dt[/tex]
     
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