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Improper integrals and trig substitution

  1. Feb 8, 2011 #1


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    1. The problem statement, all variables and given/known data
    It's been a couple of years since I've done real math, so I'm kinda stuck on this one. This is actually part of a physics problem, not a math problem - but I'm stuck on the calculus part. I'm trying to solve this guy:

    \int \limits_{-\infty}^{\infty} \frac{x^2}{(x^2+a^2)^2}\textrm{d}x

    a is a constant

    2. Relevant equations

    [itex]\textrm{tan}^2 \theta + 1 = \textrm{sec}^2 \theta[/itex]

    3. The attempt at a solution
    I make the substitution

    [itex] x = a \textrm{tan} \theta[/itex]


    \textrm{d}x = a\textrm{sec}^2\theta\textrm{d}\theta

    giving me

    \int{\frac{a^2 \textrm{tan}^2 \theta a\textrm{sec}^2 \theta \textrm{d} \theta}{(a^2 \textrm{tan}^2 \theta + a^2)^2}}

    and eventually I get it to boil down to (using the aforementioned tangent identity and canceling terms)
    [itex] \frac{1}{a} \int \textrm{tan}^2 \theta \textrm{d} \theta [/itex]
    I thought I was supposed to change the limits to
    [itex] \pm\frac{\pi}{2} [/itex]
    , but when I solve the above simplified integral I get
    [itex] \textrm{tan}\theta - \theta[/itex]
    which is not convergent

    My problem is taking the limit for the tangent at [itex]\pm\frac{\pi}{2}[/itex]
    I'm probably screwing up with the limits of integration. What exactly am I supposed to do with a trig substitution and the limits when dealing with an improper integral? I was following an old calculus book of mine, but this doesn't seem exactly right...

    Thanks for your help!
    Last edited: Feb 9, 2011
  2. jcsd
  3. Feb 9, 2011 #2


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    Just realized that it shouldn't end at
    [itex]\frac{1}{a} \int \frac{\textrm{tan}^2\theta}{\textrm{sec}^2\theta}\textrm{d}\theta = \frac{1}{a} \int \textrm{sin}^2 \theta \textrm{d}\theta[/itex]
    I'm still unsure of what exactly to do with the limits...
  4. Feb 9, 2011 #3


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    I'm getting

    [tex] \int_{-\infty}^{\infty} \frac{x^2}{\left(x^{2}+a^{2}\right)^{2}} dx = \int_{0}^{\infty} x \, \frac{2x}{\left(x^{2}+a^{2}\right)^{2}} dx = ... [/tex]

    and now you can do part integration.
  5. Feb 9, 2011 #4
    To solve your resulting integral [itex] \int sin^2\theta \, d\theta [/itex], use the half-angle formula on [itex] sin^2 \theta [/itex]. As for the limits of integration, there are two possibilities. Once you've integrated, you can substitute back in to get expressions in x and then use the original limits. Or you can find the new limits for theta by solving
    \infty = x = a \tan\theta
    -\infty = x = a \tan\theta
    for theta. (Hint: To do the second method, think about where cosine is 0, and whether sine is positive or negative at this point.)
  6. Feb 9, 2011 #5


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    Thanks all for your help, I think I got it!
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