Prove that $\int_{0}^{\frac{\pi}{2}}{\frac{\sin(2n+1)x}{\sin(x)}dx}=\frac{\pi}{2}$ for $n\ge0$

Question

Prove that
$$\int_{0}^{\frac{\pi}{2}}{\frac{\sin(2n+1)x}{\sin(x)}dx}=\frac{\pi}{2}$$ for $n\ge0$

I am not able to proceed with the integral. For the case $k+1$ please guide me through the problem.

Thanks in advance!

Answer 1

Hint: $$ \frac{1}{2} + \cos y + \cdots +\cos (ny) = \frac{\sin (n+1/2)y}{2 \sin (y/2)}. $$ This is a useful identity that can be used to prove the convergence of Fourier series. It is proved by induction on $n \in \mathbb{N}$.

Answer 2

$$\frac{\sin((2n+1)x)}{\sin(x)} = \frac{e^{(2n+1)ix}-e^{-(2n+1)ix}}{e^{ix}-e^{-ix}}=\sum_{k=-n}^{n}e^{2ikx} $$ and for every $k\in\mathbb{Z}\setminus\{0\}$, $$ \int_{0}^{\pi/2} e^{2ikx}\,dx \in i\mathbb{R},$$ but the integral $\int_{0}^{\pi/2}\frac{\sin((2n+1)x)}{\sin(x)}\,dx$ has to be real, hence it equals the contribute given by $k=0$ only.

Answer 3

Using Prosthaphaeresis Formula,

$$\sin(m+2)x-\sin mx=2\sin x\cdot\cos(m+1)x$$

So, if $\displaystyle I_m=\int_0^{\pi/2}\dfrac{\sin mx}{\sin x}dx,$

$$I_{m+2}-I_m=2\int_0^{\pi/2}\cos(m+1)x\ dx=\dfrac2{m+1}\cdot\left\{\sin(m+1)\dfrac\pi2-\sin0\right\}$$

If $\displaystyle m+1$ is even $\displaystyle=2r$(say), $$I_{2r+1}=I_{2r-1}$$

What is $I_1?$

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ ${\large\mbox{With}\ m = 1, 3, 5, \ldots}$:

\begin{align} &\color{#f00}{\int_{0}^{\pi/2}{\sin\pars{mx} \over \sin\pars{x}}\,\dd x} = \Im\int_{0}^{\pi/2}{\expo{\ic mx} - 1 \over \pars{\expo{\ic x} - \expo{-\ic x}}/\pars{2\ic}}\,\dd x = \Im\int_{z\ =\ \exp\pars{\ic x} \atop {\vphantom{\Large A}0\ <\ x\ <\ \pi/2}} \quad{z^{m} - 1 \over \pars{z^{2} - 1}/\pars{2\ic z}}\,{\dd z \over \ic z} \\[3mm] = &\ 2\,\Im\int_{z\ =\ \exp\pars{\ic x} \atop {\vphantom{\Large A}0\ <\ x\ <\ \pi/2}} \quad{1 - z^{m} \over 1 - z^{2}}\,\dd z = 2\,\Im\pars{-\int_{1}^{0}{1 - y^{m}\expo{\ic m\pi/2} \over 1 + y^{2}}\,\ic\,\dd y} = 2\int_{0}^{1}{1 - y^{m}\ \overbrace{\cos\pars{m\pi/2}}^{\ds{=\ 0}\ } \over 1 + y^{2}}\,\dd y \\[3mm] = &\ 2\int_{0}^{1}{\dd y \over 1 + y^{2}} = \color{#f00}{{\pi \over 2}} \end{align}