# branch cuts and contour integration

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

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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 $-i$.

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So there is a branch cut from $i$ to $i \infty$, a branch cut along the negative imaginary axis, and a branch cut from $-i$ to $-i \infty$? And you're never crossing over any of them? –  Random Variable Feb 27 '13 at 5:07
@Random: we avoid crossing any cuts because the contour is indented to avoid the branch points. –  Ron Gordon Feb 27 '13 at 5:28
I didn't mean to use the term "cross." You can't cross a branch cut. What I meant to ask is if we're always staying on the original branches. –  Random Variable Feb 27 '13 at 5:34
So far as I can tell we are. Think of the branch along the imaginary axis as a branch of $\sqrt{1-z^2}$ between $(-1,1)$. As long as we stay within those bounds, then, yeah, we are on the original branch. Again, because we indent the contours, we make sure we are. –  Ron Gordon Feb 27 '13 at 12:48
I meant to ask if the second cut was along the negative REAL axis, not the negative imaginary axis. The direct evaluation of that line integral seems impossible. How did you get it into such a nice form? Did you bound it first? And I think the it should be a quarter circle indentation. –  Random Variable Feb 27 '13 at 21:38