Integral of a logarithm using residues

[Dixanadu]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

So I've gone through it using a branch cut so that 0&lt;\theta&lt;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 don't see how that would work though, I am pretty sure branch cut is the way to go, but I get this...

 

[jackmell]

Homework Statement

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&lt;p&lt;1

Homework Equations

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

The Attempt at a Solution

So I've gone through it using a branch cut so that 0&lt;\theta&lt;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 don't see how that would work though, I am pretty sure branch cut is the way to go, but I get this...

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

\int_0^{\infty} \frac{r^p}{1+r^2} dr,\quad 0&lt;p&lt;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.