# contour integration and branch points

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 around a contour that consists of a straight line joining the points $i$ and $-i$ and the right half of the semicircle $|z|=1$.

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

The location of the branch cut is unclear to me.

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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

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{-\pi/2}^{\pi/2}\cos\pars{ax}\cos^{b}\pars{x}\,\dd x:\ {\large ?}\,,\qquad \pars{~a >\ b\ >\ -1~}}$.

\begin{align}&\color{#c00000}{\int_{\pi/2}^{\pi/2}\cos\pars{ax}\cos^{b}\pars{x}% \,\dd x} =\Re\int_{\pi/2}^{\pi/2}\expo{\ic ax}\cos^{b}\pars{x}\,\dd x \\[3mm]&=\Re \int_{\verts{z}\ =\ 1 \atop{\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi/2}} z^{a}\pars{z^{2} + 1 \over 2z}^{b}\,{\dd z \over \ic z} \\[3mm]&={1 \over 2^{b}}\,\Im\color{#00f}{% \int_{\verts{z}\ =\ 1 \atop{\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi/2}} z^{a - b - 1}\pars{z^{2} + 1}^{b}\,\dd z}\tag{1} \end{align}

If we 'close' the arc with the line $\ds{\braces{\pars{0,y}\ \mid \ -1 \leq y \leq 1}}$ wich include indents 'around' $\ds{z = \pm\ic}$ and $\ds{z = 0}$, the integral vanishes out. That means that the integral in question is equal to a minus the integral along the above mentioned segment, from $\ds{y = 1}$ t0 $\ds{y = -1}$, with a suitable handling of the indented points. It's summarized in the following expression: \begin{align}&\color{#00f}{% \int_{\verts{z}\ =\ 1 \atop{\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi/2}} z^{a - b - 1}\pars{z^{2} + 1}^{b}\,\dd z} =\lim_{\epsilon \to 0^{+}}\left\lbrace% -\ \overbrace{\left.\int_{0}^{-\pi/2}z^{a - b - 1}\pars{z^{2} + 1}^{b}\,\dd z\, \right\vert_{z\ \equiv\ \ic + \epsilon\expo{\ic\theta}}}^{\ds{\large\tt I}}\right. \\[3mm]&-\ \overbrace{% \int_{1 - \epsilon}^{\epsilon}y^{a - b - 1}\expo{\ic\pars{a - b - 1}\pi/2} \pars{-y^{2} + 1}^{b}\ \ic\,\dd y}^{\ds{\large\tt II}}\ -\ \overbrace{\left.\int_{\pi}^{-\pi}z^{a - b - 1}\pars{z^{2} + 1}^{b}\,\dd z\, \right\vert_{z\ \equiv\ \epsilon\expo{\ic\theta}}}^{\ds{\large\tt III}} \\[3mm]&-\ \overbrace{\int_{-\epsilon}^{-1 + \epsilon}\pars{-y}^{a - b - 1} \expo{-\ic\pars{a - b - 1}\pi/2} \pars{-y^{2} + 1}^{b}\ \ic\,\dd y}^{\ds{\large\tt IV}} \\[3mm]&\left.-\ \overbrace{% \left.\int_{\pi/2}^{0}z^{a - b - 1}\pars{z^{2} + 1}^{b}\,\dd z\, \right\vert_{z\ \equiv\ -\ic + \epsilon\expo{\ic\theta}}}^{\ds{\large\tt V}} \right\rbrace\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\pars{2} \end{align}

Now, we'll check the integral behaviours when $\ds{\epsilon \to 0^{+}}$. Lets examine the integral $\ds{\large\tt I}$: \begin{align} &\mbox{When}\ \epsilon \to 0^{+}\,,\quad \left.\int_{0}^{-\pi/2}z^{a - b - 1}\pars{z^{2} + 1}^{b}\,\dd z\, \right\vert_{z\ \equiv\ \ic + \epsilon\expo{\ic\theta}} \sim \epsilon^{b + 1} \to 0\quad\mbox{since}\quad b > -1. \\[3mm]&\mbox{Similarly, the integral}\ {\large\tt V} \to 0. \\[3mm]&\mbox{The integral}\ {\large\tt III}\ \mbox{behaves as}\ \epsilon^{a - b} \to 0\ \mbox{since}\ a > b. \\[3mm]&\mbox{Integrals}\ {\large\tt II}\ \mbox{and}\ {\large\tt IV}\ \mbox{converge when}\ \epsilon \to 0^{+}\ \mbox{since}\ a - b - 1>-1\ \mbox{and}\ b > -1. \end{align}

Then, we are left with $\ds{\pars{~\mbox{see expression}\ \pars{2}~}}$ \begin{align}&\color{#00f}{% \int_{\verts{z}\ =\ 1 \atop{\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi/2}} z^{a - b - 1}\pars{z^{2} + 1}^{b}\,\dd z} \\[3mm]&=-\expo{\ic\pars{a - b}\pi/2}\int_{1}^{0}y^{a - b - 1} \pars{1 - y^{2}}^{b}\,\dd y +\expo{-\ic\pars{a - b}\pi/2}\int_{0}^{-1}\pars{-y}^{a - b - 1} \pars{1 - y^{2}}^{b}\,\dd y \\[3mm]&=\expo{\ic\pars{a - b}\pi/2}\int_{0}^{1}y^{a - b - 1} \pars{1 - y^{2}}^{b}\,\dd y -\expo{-\ic\pars{a - b}\pi/2}\int_{0}^{1}y^{a - b - 1}\pars{1 - y^{2}}^{b}\,\dd y \\[3mm]&=\color{#00f}{2\ic\sin\pars{\bracks{a - b}\,{\pi \over 2}}\int_{0}^{1} y^{a - b - 1}\pars{1 - y^{2}}^{b}\,\dd y} \end{align}

Replacing this result in expression $\pars{1}$: \begin{align}&\color{#c00000}{\int_{\pi/2}^{\pi/2}\cos\pars{ax}\cos^{b}\pars{x} \,\dd x} ={\sin\pars{\bracks{a - b}\pi/2} \over 2^{b - 1}}\,\int_{0}^{1} y^{a - b - 1}\pars{1 - y^{2}}^{b}\,\dd y \\[3mm]&={\sin\pars{\bracks{a - b}\pi/2} \over 2^{b - 1}}\,\int_{0}^{1} y^{\pars{a - b - 1}/2}\pars{1 - y}^{b}\,\half\,y^{-1/2}\,\dd y \\[3mm]&={\sin\pars{\bracks{a - b}\pi/2} \over 2^{b}}\,\int_{0}^{1} y^{\pars{a - b}/2 - 1}\pars{1 - y}^{b}\,\dd y \\[3mm]&={\sin\pars{\bracks{a - b}\pi/2} \over 2^{b}}\, {\Gamma\pars{\bracks{a - b}/2}\Gamma\pars{b + 1}\over \Gamma\pars{\bracks{a + b}/2 + 1}} \end{align}

Here we used the Gamma Function Euler Reflection Identity: \begin{align}&\color{#66f}{\large\int_{\pi/2}^{\pi/2}\cos\pars{ax}\cos^{b}\pars{x}\,\dd x} \\[3mm]&=\color{#66f}{\large{\pi \over 2^{b}}\, {\Gamma\pars{b + 1}\over \Gamma\pars{1 + \bracks{b + a}/2}\Gamma\pars{1 + \bracks{b - a}/2}}} \end{align}

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Thanks. I'm aware of that particular evaluation, however. I asked this question 18 months ago when I was having a difficult time understanding branch cuts. – Random Variable Jul 18 '14 at 2:49
Technically I asked this question a little less than 17 months ago. :) – Random Variable Jul 18 '14 at 3:04
@RandomVariable I found it recently. I'm just one year old in M.SE. When I started to write the answer I was not aware it will be a long answer. It's always a good exercise to integrate along arcs where it is frequently spiced with indents. Thanks. – Felix Marin Jul 18 '14 at 3:17