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Evaluate this integral

  1. Oct 10, 2013 #1


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    1. The problem statement, all variables and given/known data
    [itex]\displaystyle \int_0^{\infty} \dfrac{xlnx}{(1+x^2)^2} dx [/itex]

    2. Relevant equations

    3. The attempt at a solution
    Integrating by parts and using ILATE rule

    [itex]\left[ ln \dfrac{x}{\sqrt{1+x^2}} - \dfrac{lnx}{2(1+x^2)} \right] [/itex]

    Now I find the limit as x tends to infinity and get 0. But how do I evaluate limit when x tends to zero.
  2. jcsd
  3. Oct 10, 2013 #2


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    Your indefinite integration went off the rails somewhere. How did you ever get a SQRT(1+x^2) term?
  4. Oct 11, 2013 #3


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    I have checked my integrals thrice and it seems that I can't find any error in it. Anyways, I'm posting here for you

    [itex]lnx \displaystyle \int \dfrac{xdx}{(1+x^2)^2} - \int \dfrac{1}{x} \int \frac{xdx}{(1+x^2)^2}\\
    \frac{-lnx}{2(1+x^2)} + \frac{1}{2} \int \dfrac{dx}{x(1+x^2)} \\
    \dfrac{1}{2} \left[ lnx - ln \sqrt{1+x^2} - \dfrac{lnx}{1+x^2} \right] [/itex]

    If this is further simplified it turns out to be the same as posted earlier. I only missed that 1/2, but it makes no difference when limit is to be evaluated.
  5. Oct 12, 2013 #4

    Ray Vickson

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    First, check that your indefinite integral is correct, by differentiating it to see if you recover the integrand. That should always be your first step! I think your integral is incorrect; but do not take my word for it; check it for yourself.
  6. Oct 12, 2013 #5


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    OK, I made a silly mistake and now here's the correct one

    [itex]\dfrac{1}{4} \ln \dfrac{x^2}{1+x^2} - \dfrac{1}{2} \dfrac{\ln x}{1+x^2} [/itex]

    I can evaluate limit when x tends to infinity but can't get it for zero. Can you please help?
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