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Tough Integration question

  1. Oct 3, 2005 #1
    Anyone know how to solve this?
    [tex]\text 8\pi\int_{0}^{\infty}\frac{t^3}{(4+t^2)^\frac{5}{2}} dt[/tex]
     
  2. jcsd
  3. Oct 3, 2005 #2

    hotvette

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    Sure, make the following substitution:

    [tex]u = 4 + t^2[/itex]
     
  4. Oct 3, 2005 #3
    yeah, i tried that, but i can't go further coz i can't get rid of the t^3
     
  5. Oct 3, 2005 #4

    hotvette

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    No need to get rid of it. Take it further. I believe you'll end up with 2 integrals in fractional powers of u.
     
    Last edited: Oct 4, 2005
  6. Oct 3, 2005 #5
    can you help me with one more step? i really can't make two integrals with power of u..
     
  7. Oct 3, 2005 #6
    ok, i got it now, thanks for the help
     
  8. Oct 4, 2005 #7

    dextercioby

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    HINT:

    [tex] \sqrt{4+t^{2}}=u [/tex]

    [tex] t \ dt= u \ du [/tex]

    [tex] \int_{0}^{\infty} \frac{t^{3}+4t-4t}{\left(\sqrt{t^{2}+4}\right)^{5}} \ dt =\int_{0}^{\infty} \frac{t^{2}+4}{\left(\sqrt{t^{2}+4}\right)^{5}}t \ dt - 4\int_{0}^{\infty}\frac{t \ dt}{\left(\sqrt{t^{2}+4}\right)^{5}} [/tex]

    Daniel.
     
  9. Oct 4, 2005 #8
    can't it be done by contour integration?
     
  10. Oct 5, 2005 #9

    benorin

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    Let t^2=x-4 -> 2tdt=dx and 0<=t<=infinity -> 4<=x<=infinity

    so the integral becomes 4*Pi*Int((x-4)*x^(-5/2),x=4..infinity) = 8*Pi/3

    excuse my maple notation, I have yet to learn LaTeX.
     
  11. Oct 5, 2005 #10

    saltydog

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    Nice. Here we go: (just do a click on the equation and a pop-up window will display the LaTex commands)

    [tex]
    \begin{align*}
    8\pi\int_0^{\infty}\frac{t^3}{(4+t^2)^{5/2}}dt &=
    \frac{8\pi}{2}\int_4^{\infty}\frac{x-4}{x^{5/2}}dx \\ &=
    4\pi\int_4^{\infty}x^{-5/2}(x-4)dx \\ &=
    4\pi\int_4^{\infty}(x^{-3/2}-4x^{-5/2})dx \\ &=
    4\pi\left(8/3x^{-5/2}-2x^{-1/2}\right)_4^{\infty} \\ &=
    4\pi\left(-(1/3-1)\right) \\ &=
    \frac{8\pi}{3}
    \end{align}
    [/tex]
     
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