Solving the integral $\int_0^{\pi/2} \frac{\sin((2n+1)t)}{\sin t} \mathrm{d}t$

Question

I really don't know how to solve this integral $$\int_0^{\pi/2} \frac{\sin((2n+1)t)}{\sin t} \mathrm{d}t$$ Should I use firstly a formula of $\sin(a+b)$?

Answer 1

This is known as the Dirichlet integral and the integrand sin(2n+1)t/sint = 1 + 2cos2t + 2cos4t + ...+ 2cos(2nt). You can use this to integrate term by term and the answer is: pi/2

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#00f}{\large\int_{0}^{\pi/2}{\sin\pars{\bracks{2n + 1}t} \over \sin\pars{t}}\,\dd t} =\int_{0}^{\pi/2}{\expo{\ic\pars{2n + 1}t} - \expo{-\ic\pars{2n + 1}t} \over \expo{\ic t} - \expo{-\ic t}}\,\dd t \\[3mm] = & \ \int_{0}^{\pi/2}\expo{-2nt\,\ic}{\expo{\ic\pars{2n + 1}2t} - 1 \over \expo{2\ic t} - 1}\,\dd t =\int_{0}^{\pi/2}\expo{-2nt\,\ic}\sum_{k = 0}^{2n}\expo{2\ic k t}\,\dd t \\[3mm] = & \ \sum_{k = -n}^{n}\int_{0}^{\pi/2}\expo{2kt\,\ic}\,\dd t = \sum_{k = -n}^{-1}\int_{0}^{\pi/2}\expo{2kt\,\ic}\,\dd t + {\pi \over 2} + \sum_{k = 1}^{n}\int_{0}^{\pi/2}\expo{2kt\,\ic}\,\dd t \\[3mm] = &\ {\pi \over 2} + 2\sum_{k = 1}^{n}\ \overbrace{\int_{0}^{\pi/2}\cos\pars{2kt}\,\dd t}^{\ds{=\ 0}} = \color{#00f}{\Large{\pi \over 2}} \end{align}

Answer 3

It is a general very useful fact that $$D_n(t)=\sum_{k=-n}^ne^{ikt}=1+2\sum_{k=1}^n\cos kt=\frac{\sin\left(n+\frac 1 2\right)t}{\sin \frac t 2}$$

You can go head and try to prove it. $D_n(t)$ is known as the Dirichlet kernel, and is widely used in Fourier Theory. In particular, if $S_n(t)$ is the $n$-th partial sum of $f$'s Fourier series, $$S_n(t)=\frac{1}{2\pi}\int_0^{2\pi}f(u)D_n(u-t)dt=(f\ast D_n)(t)$$

Answer 4

If $\displaystyle I_n=\int_0^{\dfrac\pi2} \frac{\sin((2n+1)t)}{\sin t} \mathrm{d}t$

$\displaystyle I_n-I_{n-1}=\int_0^{\dfrac\pi2} \frac{\sin((2n+1)t)-\sin((2n-1)t)}{\sin t} \mathrm{d}t$

Using Prosthaphaeresis Formulas, $\displaystyle \sin((2n+1)t)-\sin((2n-1)t)=2\sin t\cos2nt $

$\displaystyle\implies I_n-I_{n-1}=2\int_0^{\dfrac\pi2}\cos2nt\mathrm{d}t=\cdots =0$

Now, $\displaystyle I_0=\int_0^{\dfrac\pi2} \frac{\sin((2\cdot+1)t)}{\sin t} \mathrm{d}t=\int_0^{\dfrac\pi2}\mathrm{d}t=\cdots$