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Residue Calculus integrate Sqrt(x)ln(x)/(1+x^2)

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Integrate from zero to infinity;

    f(x)=\sqrt(x)log(x)/(x^2+1)

    2. Relevant equations

    Branch cut makes log(z)= ln|z|+i Arg(z)

    Poles are at +/- i and Res(z=i) is \pi/4 e^(i \pi/4)

    I'll need to close the contour; probably as an annullus in the top half of the plane with r<1, R>1


    3. The attempt at a solution
    I know that the total integral is 2\pi i Res(z=i) since that's the only pole enclosed in the annulus; but I can't see how to show that the semicircles go to zero as the radius goes to zero/infinity
    I worked out that (converting z=r' e^ix
    |∫f(z) dz|<=∫|fz|dz and |f(z)|= \sqrt((r')(ln|r'|^2+x^2))/(\sqrt(r'^2 cos^2(2x)+1)^2+r'^4 sin^2(2x))
    |f(z)|<=\sqrt((r')(ln|r'|^2+x^2))/(r'^2) (by taking out the +1 in the denominator
    But I don't see how this has a limit at zero as r->0 and R->\infty
    Help would be much appreciated.
     
  2. jcsd
  3. Oct 14, 2012 #2
    First consider the indentation around the origin. If we let [itex]z=\rho e^{it}[/itex], then the integral becomes:

    [tex]\lim_{\rho\to 0} \int_a^b \frac{\rho^{1/2} e^{it/2} \left(\ln(\rho)+it\right) \rho i e^{it}}{\rho^2 e^{2 i t}+1} dt[/tex]

    Now, what is:

    [tex]\lim_{\rho\to 0} \frac{\rho^{1/2} e^{it/2} \left(\ln(\rho)+it\right) \rho i e^{it}}{\rho^2 e^{2 i t}+1}[/tex]

    Well surely the denominator goes to just one right? Scrap everything in the numerator that's not a [itex]\rho[/itex] (we could just let them equal a big number for that matter and it won't affect the limit). And so you're left with:

    [tex]\lim_{\rho\to 0} \rho^{3/2} \ln(\rho)[/tex]

    Now you try and do the same type of analysis with [itex]R\to\infty[/itex] for the larger arc.
     
    Last edited: Oct 14, 2012
  4. Oct 16, 2012 #3
    Thanks thatreally helped. i got it out now
     
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