TITLE: A contour integral with 3 branch points
QUESTION [6 upvotes]: I came across an exercise in a textbook that says to evaluate $$ \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos (ax) \cos^{b} (x) \, dx  \, , \quad a > b > -1$$ by integrating $f(z) = z^{a-1} (z+z^{-1})^{b}$  around a contour that consists of the line segment joining the points $i$ and $-i$, along with the right half of the unit circle. 
It also says that the contour should be indented at the points $0, i$, and $-i$.
But where should the branch cuts be placed?

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

\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}