1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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;


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook