Integral of a logarithm using residues

1. Dec 5, 2013

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:
$\int_{0}^{\infty} \frac{x^{p}ln(x)}{x^{2}+1}dx, 0<p<1$

2. Relevant equations

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

3. The attempt at a solution
So ive gone through it using a branch cut so that $0<\theta<2\pi$. I then replaced x with $z=re^{i(\theta + 2in\pi)}$, 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:

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

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. Dec 6, 2013

jackmell

Suppose we did some work and we ended up with the integral:

$$\int_0^{\infty} \frac{r^p}{1+r^2} dr,\quad 0<p<1$$

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

First though, solve:

$$\int_0^{\infty} \frac{\sqrt{r}}{1+r^2} dr$$

then:

$$\int_0^{\infty} \frac{r^{1/3}}{1+r^2} dr$$

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