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Math Help - Integration by parts

  1. #1
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    Integration by parts

    Integrate x* 2*alpha*x*e^(-alpha*x^2) dx with limits infinity and zero?
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  2. #2
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    Re: Integration by parts

    Quote Originally Posted by JDAWES View Post
    Integrate x* 2*alpha*x*e^(-alpha*x^2) dx with limits infinity and zero?
    Hi JDAWES!
    Your question is not clear. Is it

    \int_0^\infty 2\alpha x^{2 }e^{-\alpha x^2}dx

    or

    \int_0^\infty x^{2\alpha}e^{-\alpha x^2}dx
    ?
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  3. #3
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    Re: Integration by parts

    \int_0^\infty 2\alpha x^{2 }e^{-\alpha x^2}dx

    we can here use the substitution t= x^2 \,\, \, dx = \frac{1}{2}t^{\frac{-1}{2}}\, dt

    so we get the following : \alpha \int^{\infty}_0 \, t^{1/2}e^{-\alpha t }\, dt

    Now this is solved by the Laplace transform or gamma function which converges for \alpha >0

    \alpha \int^{\infty}_0 \, t^{{1/2}}e^{-\alpha t }\, dt  = \alpha \frac{\Gamma {(\frac{3}{2})}}{\alpha^{\frac{3}{2}}}= \frac{\Gamma(\frac{1}{2})}{2\sqrt{\alpha }}= \frac{\sqrt{\pi}}{2\sqrt{\alpha }}

    ---------------------------------------------------------------

    Now for the integrand : \int_0^\infty x^{2\alpha}e^{-\alpha x^2}dx

    here if we use the same sub we get :

    \int^{\infty}_0 \, t^{\alpha -\frac{1}{2}}e^{-\alpha t }\, dt  = \frac{\Gamma {(\alpha +\frac{1}{2})}}{2\alpha^{\alpha +\frac{1}{2}}}
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