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Help with integration

  1. Jul 30, 2006 #1
    I'm having difficulty with this, trigonometic substitution won't work, neither would integration by parts...
    [tex]\int_{0}^{\infty} \frac{y^2}{1+y^4} dy[/tex]

    ETA:
    doing trigonometric substitution with [tex] y^2 = tan\theta[/tex], I would get
    [tex]\frac{1}{2} \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{tan \theta} d\theta [/tex]
     
    Last edited: Jul 30, 2006
  2. jcsd
  3. Jul 31, 2006 #2

    siddharth

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    Don't make the trig sub.

    Instead, divide numerator and denominator by [itex]y^2[/itex].
    You'll have
    [tex] \int \frac{dy}{y^2 + 1/y^2} [/tex]

    There's a tricky step (more like manipulation) to solve this. Here's a hint:
    [tex] y^2 + 1/y^2 = (y+1/y)^2 - 2 [/tex]
    [tex] y^2 + 1/y^2 = (y-1/y)^2 + 2 [/tex]

    Can you play around for a while and take it from here?
     
  4. Aug 1, 2006 #3
    hey, thanks for the help siddharth...yeah, i'll try to play around this form and fina a solution.

    thanks again
     
  5. Aug 8, 2006 #4

    dextercioby

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    Contour integration is the solution.

    Daniel.
     
  6. Aug 8, 2006 #5

    dextercioby

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    Also, u might try for a simple fraction expansion.

    [tex] y^{4}+1 =\left(y^{2}+\sqrt{2}y+1\right)\left(y^{2}-\sqrt{2}y+1\right) [/tex]

    The result is [itex]\frac{1}{4}\pi \sqrt{2} [/itex]

    Daniel.
     
  7. Aug 12, 2006 #6

    siddharth

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    You don't need contour integration to solve this.

    The trick in integrating
    [tex] \int \frac{dy}{y^2 + 1/y^2} [/tex]

    is to write it as

    [tex](1/2) \int \left( \frac{1-1/y^2}{(y+1/y)^2 - 2} + \frac{1+1/y^2}{(y-1/y)^2 + 2} \right) dy [/tex]

    This is very easy to integrate.
     
    Last edited: Aug 12, 2006
  8. Aug 12, 2006 #7

    benorin

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    This is very nice :)

     
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