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Math Help - Integration Help

  1. #1
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    Integration Help

    Hello, this is my first post. I was wondering if somebody could shed some light on how to solve the following integral:

    \int_0^{\infty} \sqrt{x}e^{-(1/2)x^2} dx

    I have tried all sorts: parts, reduction formulae and cannot seem to find a suitable substitution. Please Help!
    Last edited by Electric; April 2nd 2010 at 02:21 PM.
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  2. #2
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    Quote Originally Posted by Electric View Post
    Hello, this is my first post. I was wondering if somebody could shed some light on how to solve the following integral:

    int_0^infinity x^(1/2)e^[-(1/2)x^2]dx

    I have tried all sorts: parts, reduction formulae and cannot seem to find a suitable substitution. Please Help!
    Wolfram Mathematica Online Integrator

    It looks as if this has no "closed form" solution. You can solve it using the exponential integral.

    I'm not sure of this though, and there are a few amazing integrators on this forum. Maybe one of them can find a solution.
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  3. #3
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    Yeah, it is a tricky one! I have a feeling there is a sneaky substitution and then the error or gamma function has to be used somewhere however, I still cannot find a solution
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    Quote Originally Posted by Electric View Post
    Yeah, it is a tricky one! I have a feeling there is a sneaky substitution and then the error or gamma function has to be used somewhere however, I still cannot find a solution
    Maybe pm Krizalid and see if he can help.
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  5. #5
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Electric View Post
    Hello, this is my first post. I was wondering if somebody could shed some light on how to solve the following integral:

    \int_0^{\infty} x^{1/2}e^{-(1/2)x^2} dx

    I have tried all sorts: parts, reduction formulae and cannot seem to find a suitable substitution. Please Help!
    Yeah \infty is weird, right? When I'm typing teX I say in my head "integral from zero to inf-T"
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  6. #6
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    put x=\sqrt2\sqrt t and that'd turn the integral into a Gamma function one.
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  7. #7
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    Ah yes, thanks Krizalid. Overlooked that substitution For anyone interested to know the solution, here it is:




    x=\sqrt{2t}, dx=\frac{1}{\sqrt{2t}} dt

    =\int_0^{\infty} {2t}^{-(1/4)}e^{-t} dt

     since\;\;\; \Gamma{(n)}=\int_0^{\infty} t^{n-1}e^{-t} dt,

    =>\frac{1}{2^{1/4}} \int_0^{\infty} t^{-(1/4)} e^{-t} dt

    =\frac{1}{2^{1/4}}*\Gamma{(3/4)}

    =0.841*1.225

    =1.030
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