Prove that $\int_0^{\pi/2} \cos^{p+q-2}(\theta) \cos((p-q)\theta)d\theta = \frac{\pi}{(p+q-1)2^{p+q-1}B(p,q)}$

Question

Does anybody know how to prove this identity $?$:

\begin{align*} & \int_{0}^{\pi/2} \cos^{p + q - 2}\,\left(\,{\theta}\,\right) \cos\left(\,{\left[\,{p - q}\,\right] \theta}\,\right) \,{\rm d}\theta \\[2mm] = & \ \frac{\pi}{\left(\,{p + q - 1}\,\right) 2^{p + q - 1}\, \operatorname{B}\left(\,{p,q}\,\right)}\ , \qquad p+q>1,\ q<1 \end{align*}

where $\operatorname{B}\left(\,{x,y}\,\right)$ denotes Beta Function.

• I am confused because the result contains beta function in the denominator.
• I did it using Cauchy's Beta Integral but my friend says there's another method using contour integration: I can't figure it out.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\int_{0}^{\pi/2} \cos^{p + q - 2}\,\pars{\theta} \cos\pars{\bracks{p - q}\theta}\,\dd\theta} \\[5mm] = & \ \Re\int_{0}^{\pi/2}\cos^{p + q - 2}\,\pars{\theta} \expo{\ic\pars{p - q}\theta}\,\,\dd\theta \\[5mm] = & \ \left.\Re\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2} \pars{z^{2} + 1 \over 2z}^{p + q - 2}\,z^{p - q}\,\, {\dd z \over \ic z}\right\vert_{z\ =\ \exp\pars{\ic\theta}} \end{align} The last expression is an integration along an arc in the first quadrant of the complex plane. The integration can be extended along $\ds{\braces{z \mid \Re\pars{z} = 0,\ \Im\pars{z} \in \pars{0,1}}}$ and $\ds{\braces{z \mid \Re\pars{z} \in \pars{0,1},\ \Im\pars{z} = 0}}$. The last piece doesn't yield any contribution to the integration because the integrand is imaginary. The whole contour doesn't include any pole. Therefore, \begin{align} & \color{#44f}{\int_{0}^{\pi/2} \cos^{p + q - 2}\,\pars{\theta} \cos\pars{\bracks{p - q}\theta}\,\dd\theta}. \\[5mm] = & \ \left.{1 \over 2^{p + q - 2}}\,\,\Im\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2} \pars{1 + z^{2}}^{p + q - 2}\,\,z^{1 - 2q}\,\, \dd z\right\vert_{z\ =\ \exp\pars{\ic\theta}} \\[5mm] = & \ -\,{1 \over 2^{p + q - 2}}\,\,\Im\int_{1}^{0} \pars{1 - y^{2}}^{p + q - 2}\,\,y^{1 - 2q}\,\,\overbrace{\pars{\expo{\ic\pi/2}}^{1 - 2q}\,\,\,\ic}^{\ds{-\expo{-\ic\pi q}}} \,\dd y \\[5mm] = & \ {\sin\pars{\pi q} \over 2^{p + q - 2}} \int_{0}^{1}y^{1/2 - q}\,\,\pars{1 - y}^{p + q - 2} \,\,\,{1 \over 2}\,y^{-1/2}\,\,\dd y \\[5mm] = & \ {\sin\pars{\pi q} \over 2^{p + q - 1}} \int_{0}^{1}y^{-q}\,\,\pars{1 - y}^{\,p + q - 2} \,\,\,\dd y \\[5mm] = & \ {\sin\pars{\pi q} \over 2^{p + q - 1}} {\Gamma\pars{-q + 1}\Gamma\pars{p + q - 1} \over \Gamma\pars{p}} \\[5mm] = & \ {1 \over 2^{p + q - 1}}\ \overbrace{\bracks{\sin\pars{\pi q}\Gamma\pars{1 - q}}} ^{\ds{\pi/\Gamma\pars{q}}}\ {\Gamma\pars{p + q}/\pars{p + q - 1} \over \Gamma\pars{p}} \\[5mm] = & \ {\pi \over 2^{p + q - 1}\,\pars{p + q - 1}}{1 \over \Gamma\pars{p}\Gamma\pars{q}/\Gamma\pars{p + q}} \\[5mm] = & \ \bbx{\color{#44f}{{\pi \over \pars{p + q - 1}2^{p + q - 1}\,\on{B}\pars{p, q}}}} \\ & \end{align}

Answer 2

One solution using mostly real methods:

\begin{eqnarray*} \int_0^{\pi/2} \cos^{p+q-2}(\theta) \cos((p-q)\theta)d\theta &=& \frac{1}{2^{p+q-1}}\Re \int_0^{\pi/2} \left(e^{i\theta} + e^{-i\theta}\right)^{p+q-2}e^{i((p-q)\theta)} d\theta \\ &=& \frac{1}{2^{p+q-1}}\Re \int_0^{\pi/2} \left(1 + e^{-i\theta}\right)^{p+q-2}e^{i((2p-2)\theta)} d\theta \\ &=& \frac{1}{2^{p+q-1}}\Re \sum_{j=0}^{\infty} \binom{p+q-2}{j} (-1)^{j} \int_0^{\pi/2} e^{i((2p-2-2j)\theta)} d\theta \\ &=& \frac{1}{2^{p+q-1}} \sum_{j=0}^{\infty} \binom{p+q-2}{j} (-1)^{j} \int_0^{\pi/2} \cos((2p-2-2j)\theta)d\theta \\ &=& \frac{1}{2^{p+q-1}} \sum_{j=0}^{\infty} \binom{p+q-2}{j} \frac{(-1)^{j}\sin(\pi(p-j))}{2(j-p+1)}, \quad p-1\notin \mathbb{Z}_{\geq 0}\\ &=& \frac{\sin(\pi p)}{2^{p+q-1}} \sum_{j=0}^{\infty} \binom{p+q-2}{j} \frac{(-1)^j}{2(j-p+1)} \\ &=& \frac{\sin(\pi p)}{2^{p+q-1}(1-p)} \sum_{j=0}^{\infty} \frac{(-1)^j(p+q-1-j)_{j}(1-p)_{j}}{(2-p)_{j}}\frac{1}{j!}\\ &=& \frac{\sin(\pi p)}{2^{p+q-1}(1-p)} \sum_{j=0}^{\infty} \frac{(2-p-q)_{j}(1-p)_{j}}{(2-p)_{j}}\frac{1}{j!}\\ &=& \frac{\sin(\pi p)}{2^{p+q-1}(1-p)} {}_2F_{1}\left(2-p-q,1-p;2-p;1\right)\\ &=& \frac{\sin(\pi p)}{2^{p+q-1}(1-p)} \frac{\Gamma(1-p)\Gamma(q+p-1)}{\Gamma(q)} \quad(*)\\ &=& \frac{\pi }{2^{p+q-1}(q+p-1)} \frac{\Gamma(q+p)}{\Gamma(q)\Gamma(p)}; \quad p<1,\; \Gamma(p)\Gamma(1-p) = \frac{\pi}{\sin(\pi p)}\\ \end{eqnarray*}

Where we have used the following relation in $(*)$

$$ {}_2F_{1}(a,b;c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} \quad c>a+b$$

Answer 3

EDIT: I posted a more detailed answer here.

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The contour integration approach is to integrate the function $z^{p-q-1} \left(z-\frac{1}{z} \right)^{p+q-2}$ counterclockwise around a closed contour that consists of the right half of the circle $|z|=1$ and the vertical line segment from $z=i$ to $z=-i$.

Because the function has branch points at $z=0,i$, and $-i$, the contour needs to be indented at those points.

You need to check what conditions on the parameters are needed so that the contributions from the indentations vanish in the limit.