1. ## Complex Analysis, Integration

Derive $\int_0^\infty \frac {(x^{a-1})} {(1+x^2)} dx= \frac {\pi} {(2\sin(\frac {\pi a} {2}))}, 0< a < 2$

Evaluate $\int_0^{2\pi} \frac {\cos \theta} {a-\cos \theta} d\theta, a>1$

Evaluate $\frac {1} {2\pi i} \int_C \frac {z^{n-1}} {3z^n -1} dz$ where n is a positive integer, and C is the unit circle with the coutnerclockwise orientation.

2. Originally Posted by terr13
Derive $\int_0^\infty \frac {(x^{a-1})} {(1+x^2)} dx= \frac {\pi} {(2\sin(\frac {\pi a} {2}))}, 0< a < 2$
I think you mean $0 < a < 1$ anyway if you use keyhole contour you get:

$\left( 1 - e^{2\pi i (a-1)} \right) \int_0^{\infty} \frac{x^{a-1}}{x^2+1}dx = 2\pi i \left( \text{res}(f,i) + \text{res}(f,-i) \right)$ where $f(z) = \frac{z^{a-1}}{z^2+1}$

3. Originally Posted by terr13
Evaluate $\int_0^{2\pi} \frac {\cos \theta} {a-\cos \theta} d\theta, a>1$
Notice that $\int_0^{2\pi} \frac{\cos \theta}{a - \cos \theta} d\theta = \oint_{|z|=1} \frac{\frac{1}{2}(z+z^{-1})}{a - \frac{1}{2}(z+z^{-1})}\cdot \frac{1}{iz} dz$.

Simplify, $\frac{1}{i}\oint_{|z|=1} \frac{z^2+1}{z(2az - z^2 - 1)} dz$

Let $f(z) = \frac{z^2+1}{z(2az - z^2 - 1)}$ then the poles are when $z(2az-z^2-1) = 0$. One is easy $z=0$ the other is when $z^2 - 2az + 1 = 0$ and that happens for $z_{1,2}= a \pm \sqrt{a^2-1}$. Since we require $|z|<1$ (we are inside the disk) it means $z_1 = a - \sqrt{a^2-1}$.

Now $f(z) = \frac{g(z)}{h(z)}$ with $h'(z_1)\not = 0$ so $\text{res}(f,z_1) = \frac{g(z_1)}{h'(z_1)} =\frac{z_1^2 + 1}{(2az_1-z_1^2-1) + z_1(2a - 2z_1)}$.

Thus, $\text{res}(f,z_1) = \frac{2az_1}{2z_1(a-z_1)} = \frac{a}{a-z_1} = \frac{a}{\sqrt{a^2-1}}$.

While, $\text{res}(f,0) = \frac{g(0)}{h'(0)} = -1$.

Thus, the integral is $2\pi \left( \frac{a}{\sqrt{a^2-1}} - 1 \right)$