
Residue theorem troubles
Hi. I am trying to evaluate the following integral:
I = $\displaystyle \int_{0}^{\infty } \frac{x^{p}}{12x\cos{\theta}+x^{2} }$
where 1 < p < 1 and 0 < $\displaystyle \theta$ < 2*Pi. One can easily show the following:
$\displaystyle 12x\cos{\theta}+x^{2}=(xe^{i \theta } )(x  e^{i \theta })$
When we transfer this over to the complex plane this means we have a branch point at x = 0 and a couple of simple poles at x = $\displaystyle e^{i \theta }$ and x = $\displaystyle e^{i \theta }$
I am attempting to apply the pacman or keyhole contour here where the branch cut is made along the positive real axis.
I end up with the following:
A = $\displaystyle \oint_{pacman} \frac{z^{p}}{12z\cos{\theta}+z^{2} }dz$ = $\displaystyle \int_{0}^{\infty } \frac{r^{p}}{12r\cos{\theta}+r^{2} }dr  \int_{0}^{\infty } \frac{e^{i2\pi p}r^{p}}{12r\cos{\theta}+r^{2} }dr$ = $\displaystyle (1  e^{i2 \pi p})$ I = $\displaystyle e^{i \pi p}2i \sin (\pi p)$ I
But by the residue theorem we have:
A = $\displaystyle 2 \pi i \sum Residues = 2 \pi i \frac{ \sin (\theta p)}{\sin (\theta )}$
which implies:
I = $\displaystyle \int_{0}^{\infty } \frac{x^{p}}{12x\cos{\theta}+x^{2} }$ = $\displaystyle \frac {e^{i \pi p} \pi \sin (\theta p)}{ \sin (\theta ) \sin (\pi p) }$
but this cannot be right since we are integrating a real valued function over the real line... it shouldnt have any imaginary components. What have I done wrong?

I didn't check all the details, but you probably know that $\displaystyle e^{i\pi}=1$?

yes, but p is a fraction or irrational or zero so making that substitution doesnt help really. Youll have some fractional or irrational root of 1.

I believe you're using the wrong residue for the pole at $\displaystyle e^{i\theta}$ based on the branch you're using: You define:
$\displaystyle z^{p}=e^{p\log(z)}=e^{p(\ln(r)+i\arg(z))},\quad 0\leq \arg(z)\leq 2\pi$
Then for $\displaystyle z=e^{i\theta}$
$\displaystyle \arg(z)=2\pi\theta$
Also, try it first for $\displaystyle p=1/4$ and $\displaystyle \theta=\pi/4$ or something easy like this. Do it numerically first in Mathematica, then do it numerically via the Residue Theorem:
Code:
In[56]:=
p = 1/4;
t = Pi/4;
NIntegrate[1/(x^p*(1  2*x*Cos[t] +
x^2)), {x, 0, Infinity}]
r1 = Exp[7*Pi*(I/16)]/
(Exp[(Pi)*(I/4)]  Exp[Pi*(I/4)]);
r2 = Exp[(Pi)*(I/16)]/(Exp[Pi*(I/4)] 
Exp[(Pi)*(I/4)]);
N[((2*Pi*I)/(1  Exp[2*Pi*I*p]))*
(r1 + r2)]
Out[58]=
3.4907507252185646
Out[61]=
3.4907507252151033 +
3.487868498008632*^16*I

Correct correct! My understanding of branch cuts was flawed. The final answer comes out to be:
$\displaystyle \frac {\pi \sin (p (\pi  \theta ))}{\sin (\pi p) \sin (\theta )}$