Integration of the function $f\left( x \right) = \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}}$

Question

Let f be the function on $[-\pi,\pi]$ given by $f(0)=9$ and $f\left( x \right) = \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}}$ for $x\ne 0$. Find the value of $\frac{2}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} $

My approach is as follow. The function is continuous at $x=0$. Hence it is contionuos for $[-\pi,\pi]$

$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\left( {\frac{{9x}}{2}} \right)}} \times 9 \times \frac{{\left( {\frac{x}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}} = 9$

$\frac{x}{2}=\theta$

$\frac{2}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} $

$\int\limits_{ - \pi }^\pi {\frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\left( {\frac{{9x}}{2}} \right)}}dx} \Rightarrow 2\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\sin \left( {9\theta } \right)}}{{\sin \left( \theta \right)}}d\theta } \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {9\theta } \right)}}{{\sin \left( \theta \right)}}d\theta } $

$ \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {8\theta + \theta } \right)}}{{\sin \left( \theta \right)}}d\theta } \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {8\theta } \right)\cos \left( \theta \right) + \cos \left( {8\theta } \right)\sin \left( \theta \right)}}{{\sin \left( \theta \right)}}d\theta } $

$ \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{\sin \left( {8\theta } \right)\cos \left( \theta \right)}}{{\sin \left( \theta \right)}} + \cos \left( {8\theta } \right)} \right)d\theta } $

$\int\limits_0^{\frac{\pi }{2}} {\cos \left( {8\theta } \right)d\theta } = 0$

How so I proceed form here as expansion of $sin(8\theta)$ is cumbersome.

Is their any formula of converting $sin(n\theta)$ in terms of $sin(\theta)$

Answer 1

Via $$\cos t={e^{it}+e^{-it}\over2},\qquad\sin t={e^{it}-e^{-it}\over2i}$$ you can easily prove that $${\sin(9t)\over\sin t}=1+2\bigl(\cos(2t)+\cos(4t)+\cos(6t)+\cos(8t)\bigr)\ .$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{{2 \over \pi}\int_{-\pi}^{\pi} {\sin\pars{9x/2} \over \sin\pars{x/2}}\,\dd x} = {8 \over \pi}\int_{0}^{\pi/2} {\sin\pars{9x} \over \sin\pars{x}}\,\dd x \\[5mm] = &\ {8 \over \pi}\,\Im\int_{0}^{\pi/2} {\expo{9x\ic} - 1\over \sin\pars{x}}\,\dd x \\[5mm] = &\ \left.{8 \over \pi}\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2} {z^{9} - 1 \over \pars{z - 1/z}/\pars{2\ic}} \,{\dd z \over \ic z} \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left.{16 \over \pi}\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\ {1 - z^{9} \over 1 - z^{2}} \,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \end{align} I'll "close" the integration around a quarter circle, of radius $\ds{1}$, in the complex plane first quadrant: \begin{align} &\bbox[5px,#ffd]{{2 \over \pi}\int_{-\pi}^{\pi} {\sin\pars{9x/2} \over \sin\pars{x/2}}\,\dd x} = -{16 \over \pi}\,\Im\int_{1}^{0}\ {1 - \ic y^{9} \over 1 + y^{2}}\,\ic\,\dd y \\[5mm] = &\ {16 \over \pi}\int_{0}^{1}{\dd y \over 1 + y^{2}} = {16 \over \pi}\,{\pi \over 4} = \bbx{4} \\ & \end{align}