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Straightforward integration

  1. Oct 3, 2010 #1
    Ok, so I need to evaluate this integral: [tex] \int_0^{\infty} \frac{s^3}{(s^2 + a^2)^3} ds [/tex]
    It seems evident that a substitution ought to do the trick. Let's try [itex] u = s^2 + a^2 [/itex], so [itex] du = 2sds [/itex]. Hence

    [tex] \int_0^{\infty} \frac{s^3}{(s^2 + a^2)^3} ds
    = \frac{1}{2} \int_{a^2}^{\infty} \frac{u - a^2}{u^3} du
    = \frac{1}{2} \int_{a^2}^{\infty} \frac{1}{u^2} - \frac{a^2}{u^3} ds
    = \frac{1}{2} \left(-\frac{1}{u} + \frac{1}{2} \frac{a^2}{u^2}\right)\right|_{a^2}^{\infty}
    = \frac{1}{2} \left(\frac{1}{a^2} - \frac{1}{2}\right) [/tex]

    This isn't right. In fact I know the answer must be [itex] \frac{1}{4a^2} [/tex]. (I came acorss this integral doing Griffiths E&M problem 8.4; for physical reasons the integral MUST have the value I indicated.) Indeed that's exactly what WolframAlpha gives; see here. Incindentally, WolframAlpha doesn't know what to do with this integral: [itex] \int_{a^2}^{\infty} \frac{u - a^2}{u^3} du [/itex], (which occurs after my substitution) see here, which I find to be strange and amusing. What's going on here? I don't understand, and I feel rather silly that this straightforward integral is causing me problems.
     
    Last edited: Oct 3, 2010
  2. jcsd
  3. Oct 3, 2010 #2

    CompuChip

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    If [itex]du = 2 s ds[/itex] then
    [tex]\frac{1}{2} \frac{s^2}{u^3} (2 s ds) = \frac{1}{2} \frac{u^2 - a^2}{u^3}[/tex]
    Note that this differs from your expression by a square on the u.
     
  4. Oct 3, 2010 #3
    I believe you messed up your substitution in the numerator/differential. I personally would use the sub: s=atanθ. This will give you an integral that is easier to deal with, After that sub integration by parts twice should get you your answer.
     
    Last edited: Oct 3, 2010
  5. Oct 3, 2010 #4
    [tex]
    \int_0^{\infty} \frac{s^3}{(s^2 + a^2)^3} ds
    = \frac{1}{2}\int_{a^2}^{\infty} \frac{s^2}{u^3} du
    = \frac{1}{2}\int_{a^2}^{\infty} \frac{u - a^2}{u^3} du
    [/tex]
    right, since [tex] s^2 = u - a^2 [/tex] by how I defined u.
     
  6. Oct 3, 2010 #5
    Wow, sorry I wasn’t paying attention at all since I’m so used to squared terms in the denominator being a trig sub.

    Your error is when you evaluate the integral. I get 1/2(-1/u + a^2/2u^2) evaluated at inf and a^2 which works out to your desired result.
     
  7. Oct 3, 2010 #6
    Yes, I made a typo. I have corrected it. There is still a problem, however.
     
  8. Oct 3, 2010 #7

    CompuChip

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    Isn't that the same as what fluxions wrote (apart from a typo in the second-to-last identity, where there is u instead of 1/u, but which is corrected afterwards)?

    Making such stupid mistakes, I refrain from replying more, and will get some sleep :)
     
  9. Oct 3, 2010 #8

    CompuChip

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  10. Oct 3, 2010 #9

    CompuChip

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    I just spotted one of the mistakes...
    [tex] \left( \frac{1}{u} - \frac{1}{2} \frac{a^2}{u^2} \right)_{u = a^2} = \frac{1}{a^2} - \frac{a^2}{2 (a^2)^2} = \frac{1}{a^2} - \frac{1}{2} \frac{1}{a^2} = \frac{1}{2} \frac{1}{a^2} [/tex]

    I completely overlooked that you square the square in the second term.
     
  11. Oct 3, 2010 #10
  12. Oct 3, 2010 #11
    Indeed. Problem solved. Thanks.
     
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