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Integral of a logarithm using residues

  1. Dec 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Hey everyone,

    So here's the problem, nice and simple. I have to find the following integral:
    [itex]\int_{0}^{\infty} \frac{x^{p}ln(x)}{x^{2}+1}dx, 0<p<1[/itex]


    2. Relevant equations

    The only thing relevant is the residue theorem:
    [itex]\oint_{c}f(z)=2\pi i \times[/itex] sum of residues enclosed

    3. The attempt at a solution
    So ive gone through it using a branch cut so that [itex]0<\theta<2\pi[/itex]. I then replaced x with [itex]z=re^{i(\theta + 2in\pi)}[/itex], setting n = 0 when integrating from 0 to infinity for Im(z)>0, then setting n = 1 when integrating from infinity back to 0 when Im(z)<0. Then I get this:

    [itex]I=-\pi^{2}/2 \frac{cos(\pi p/2)}{sin(\pi p)e^{i\pi p}}[/itex]

    Of course I can simplify it but its definitely wrong cos its negative, and there is a dependence on a complex exponential....so what have I done wrong? Am I even supposed to be using a branch cut or is it enough to do this with a contour indented at the singularity z = i in the upper half of the complex plane...? I dont see how that would work though, im pretty sure branch cut is the way to go, but I get this...
     
  2. jcsd
  3. Dec 6, 2013 #2
    Suppose we did some work and we ended up with the integral:

    [tex]\int_0^{\infty} \frac{r^p}{1+r^2} dr,\quad 0<p<1[/tex]

    Can you compute that by contour integration over a key-hole contour with the key-slot along the negative real axis?

    First though, solve:

    [tex]\int_0^{\infty} \frac{\sqrt{r}}{1+r^2} dr[/tex]

    then:

    [tex]\int_0^{\infty} \frac{r^{1/3}}{1+r^2} dr[/tex]

    then try to solve it for the general case, then we can go back and set up the original problem.
     
    Last edited: Dec 6, 2013
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