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Integral Action

  1. Nov 26, 2005 #1
    Left my integral table at home, could someone tell me the value of these integrals?
    [tex]\int_0^{\infty} \frac{dx}{(x^2+b^2)^2}[/tex]

    [tex]\int_0^{\infty} \frac{x^2}{(x^2+b^2)^2} dx[/tex]
    and the same as the latter but with the denominator raised to the power 4. Thanks!
    Last edited: Nov 26, 2005
  2. jcsd
  3. Nov 26, 2005 #2
    Just solve them yourself, they're both pretty standard trig substitutions. You should always solve the integral forms from the table yourself, at least once, before you use them. They're a nice tool for saving yourself time and effort, not a crutch to avoid learning your integration techniques properly.
  4. Nov 27, 2005 #3


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  5. Nov 27, 2005 #4
    What do I do with the arctan at infinity? Do I just use [tex]\frac{\pi}{2}[/tex] or do I have to be saucy about it?
    Last edited: Nov 27, 2005
  6. Nov 27, 2005 #5


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    Well, I wouldn't recommend be "saucy" about homework! Technically, you should take the limit of arctan t as t goes to [itex]\infty[/itex] but that is, of course, [itex]\frac{\pi}{2}[/itex].
  7. Nov 27, 2005 #6


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    EDIT: I was going to post the below, but I just realized the it doesn't matter what [itex]\mbox{sgn}(b)[/itex] is, assuming [itex]b\neq 0[/itex], take it to be positive since b is only given by b2, we may assume it is positive.

    For [tex]\int_0^{\infty} \frac{dx}{(x^2+b^2)^2}=\lim_{M\rightarrow\infty} \frac{1}{2b^3} \left(\frac{bM}{b^2+M^2} + \tan^{-1}\left( \frac{M}{b}\right) -0\right) [/tex]
    [tex]=0+\lim_{M\rightarrow\infty} \frac{1}{2b^3} \tan^{-1}\left( \frac{M}{b}\right)=\mbox{sgn}(b) \frac{\pi}{4b^3}[/tex]
    Last edited: Nov 27, 2005
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