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Branch cuts of Ln(x)

  1. May 12, 2009 #1
    1.I am trying to evaluate the [tex]\int[/tex][tex]\frac{ln(x)}{(x^{2}+1)^{2}}[/tex] from 0 to inf


    2. I think I am having problems defining what my branch cut should be.

    3. Here is what I have tried to no avail. First set the branch cut from -infty to 0. Then changing it to a contour integral gives us ([tex]\int[/tex][tex]\frac{ln(x)}{(x^{2}+1)^{2}}[/tex]) = 2 [tex]\pi[/tex] i (a-1(i)+a-1(-i))=[tex]\frac{i \pi^{2}}{4}[/tex]

    I get somewhat lost after this. Please help. I'm studying for finals but i dont really understand this
     
    Last edited: May 12, 2009
  2. jcsd
  3. May 13, 2009 #2
    I'd use an upper half-washer with an indentation around the origin. The branch-cut can then be anywhere in the lower half-plane including on the real axis. Then consider the following contour integral:

    [tex]\mathop\oint\limits_{W_u}\frac{\text{Log}(z)}{(1+z^2)^2} dz[/tex]

    Assuming you can show the integral goes to zero over both large and small semi-circles of the contour, we're left with:

    [tex]\mathop\oint\limits_{W_u}\frac{\text{Log}(z)}{(1+z^2)^2} dz=2\int_0^{\infty}\frac{\ln(r)}{(1+r^2)^2}dr+\pi i\int_0^{\infty}\frac{1}{(1+r^2)^2}dr=2\pi i \frac{d}{dz} \frac{\log(z)}{(z+i)^2}\Bigg|_{z=i}[/tex]

    where I let [tex]z=r e^{0i}[/tex] on the leg from [tex]0[/tex] to [tex]\infty[/tex] and [tex]z=re^{\pi i}[/tex] on the leg from [tex]-\infty[/tex] to [tex]0[/tex] to arrive at that expression.
     
  4. May 13, 2009 #3
    Thanks so much. I worked it out as you have laid it out and received the correct answer. but how did you know to put the contour only in the upper half plane?
     
  5. May 13, 2009 #4
    You could use the lower half-plane and adjust the analysis accordingly. You just have to work a bunch of them to gain experience to analyze it and try things and be willing to fail at it from time to time. Remember, good cooks try again after failing a recipe, bad ones give up. :)
     
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