branch cuts and contour integration

Question

An exercise in a textbook says to evaluate $\displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos (ax) \cos^{b} (x) \ d x \ (a > b > -1)$ by letting $\displaystyle f(z) = z^{a-1} (z+z^{-1})^{b}$ and integrating along a contour that consists of a straight line joining the points $i$ and $-i$ and the right half of the semi-ciricle $|z|=1$.

But it also says that the contour should be indented at the points $0, i$, and $-i$.

The indentations at $i$ and $-i$ suggest that there are two branch cuts along the imaginary axis. Could someone explain to me exactly what's going on?

EDIT:

$ \displaystyle \Big|\int_{C_{i}}f(z)\ dz\Big|\le\int_{0}^{-\frac{\pi}{2}}\Big|\big(i+re^{it}\big)^{a-1}\big(i+re^{it}+(i+re^{it})^{-1}\big)^{b}\ ire^{it}dt\Big| $

$ \displaystyle =\int_{0}^{-\frac{\pi}{2}}\Big|i+re^{it}\Big|^{a-1}\Big|i+re^{it}+(i+re^{it})^{-1}\Big|^{b}\ r\ dt $

$ \displaystyle =\int_{0}^{-\frac{\pi}{2}}\Big|i+re^{it}\Big|^{a-1}\Big|2re^{it}+O(r^{2})\Big|^{b}\ r\ dt $

$ \displaystyle \approx\int^{-\frac{\pi}{2}}_{0}2^{b}r^{b+1}\ dt $ for small $r$

$ \displaystyle = \pi \ 2^{b-1} r^{b+1}$ (I removed the negative sign)

Therefore $\displaystyle \lim_{r \to 0}\Big|\int_{C_{i}}f(z)\ dz\Big| \le 0$ (since $b > -1$) $ \displaystyle\implies \lim_{r \to 0} \int_{C_{i}} f(z)\ dz = 0$

Answer 1

All that's happening is that you are avoiding the branch points with your contour so you can apply Cauchy's theorem. You have branch points where the argument any non-integral power is zero, in your case, at $z=0$ and $\pm i$.

Cauchy's theorem states that

$$\oint_C dz \: f(z)=0$$

So the original integral is going to equal the integral along the line joining $i$ to $-i$ plus any contribution from the indentations, if any. Is there any? Let's take the one at $z=i$ Let $z=i + \epsilon e^{i \phi}$ and the contribution is

$$\lim_{\epsilon \rightarrow 0}\: i \epsilon^{1+b} \int_0^{-\pi/2} d \phi \: e^{i (a+b) \phi}$$

This is zero because $-1>b$. (Whew!) Same for the other branch point at $z=-i$.

At $z=0$, we let $z= \epsilon e^{i \phi}$, and we will see that this contribution goes to zero because $a>b$.

So the integral you seek is simply the real part of the integral of $f(z)$ along the straight line from $i$ to $-i$.