Evaluate the definite-integrals $\int_0^{\pi} \frac{\sin nx}{\sin x}dx$

Question

Evaluate the definite-integrals $\int_0^{\pi} \dfrac{\sin nx}{\sin x}dx$

my teacher say that, using the formula : $\sin nt=\dfrac{e^{ni}-e^{-ni}}{2i}$, but i can't :(.

Answer 1

If $n$ is even then the function $f(z)=\frac{\sin(nz)}{\sin(z)}$ is symmetric around $z=\frac{\pi}{2}$, so the integral is simply zero. On the other hand, if $n=2k+1$ we have:

$$\frac{\sin((2k+1)z)}{\sin z} = \frac{e^{(2k+1)iz}-e^{-(2k+1)iz}}{e^{iz}-e^{-iz}} \\= e^{2kiz}+e^{(2k-2)iz}+\ldots+e^{2iz}+1+e^{-2iz}+\ldots+e^{-2kiz}$$ and all the terms of the previous sum vanish when integrated between $0$ and $\pi$, except $1$.

This gives: $$ \int_{0}^{\pi}\frac{\sin(nz)}{\sin z}\,dx = \frac{1-(-1)^n}{2} \pi.$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large \int_{0}^{\pi}\frac{\sin\pars{nx}}{\sin\pars{x}}\,\dd x} =\,{\rm sgn}\pars{n}\Im\int_{0}^{\pi} \frac{\exp\pars{\ic\verts{n}x} - 1 - x\bracks{\pars{-1}^{n} - 1}/\pi}{\sin\pars{x}}\,\dd x \\[5mm]&=\left.\,{\rm sgn}\pars{n}\,\Im\oint_{\verts{z}=1} \frac{z^{\verts{n}} - 1 + \ic\bracks{\pars{-1}^{n} - 1}\ln\pars{z}/\pi} {\pars{z^{2} - 1}/\pars{2\ic z}}\,\frac{\dd z}{\ic z} \,\right\vert_{\, 0< \,{\rm Arg}\pars{z} < \pi} \\[5mm]&=\left.2\,{\rm sgn}\pars{n}\,\Im\oint_{\verts{z}=1} \frac{z^{\verts{n}} - 1 + \ic\bracks{\pars{-1}^{n} - 1}\ln\pars{z}/\pi}{z^{2} - 1} \,\dd z\,\right\vert_{\, 0< \,{\rm Arg}\pars{z} < \pi} \\[1cm]&=-2\,{\rm sgn}\pars{n}\,\Im\int_{-1}^{0} \frac{\pars{-x}^{\verts{n}}\expo{\ic\pi\verts{n}} - 1 + \ic\bracks{\pars{-1}^{n} - 1}\bracks{\ln\pars{-x} + \ic\pi}/\pi}{x^{2} - 1} \,\dd x \\[5mm]&\phantom{=}-2\,{\rm sgn}\pars{n}\,\Im\int_{0}^{1} \frac{x^{\verts{n}} - 1 + \ic\bracks{\pars{-1}^{n} - 1}\ln\pars{x}/\pi}{x^{2} - 1}\,\dd x \\[5mm]&=-\,\frac{4}{\pi}\,\,{\rm sgn}\pars{n}\bracks{\pars{-1}^{n} - 1}\ \underbrace{\int_{0}^{1}\frac{\ln\pars{x}}{x^{2} - 1}\,\dd x} _{\dsc{\frac{\pi^{2}}{8}}} =\color{#66f}{\large% \,{\rm sgn}\pars{n}\bracks{1 - \pars{-1}^{n}}\frac{\pi}{2}} \end{align} We set the $\ds{\ln}$-branch cut: $$ \ln\pars{z}=\ln\pars{\verts{z}} + \,{\rm Arg}\pars{z}\ic\,;\qquad -\,\frac{\pi}{2} < \,{\rm Arg}\pars{z} < \frac{3\pi}{2}\,,\quad z \not=0. $$

---

The last integral $\ds{\pars{~\mbox{the logarithmic one}~}}$ is easily evaluated as follows: \begin{align}&\dsc{\int_{0}^{1}\frac{\ln\pars{x}}{x^{2} - 1}\,\dd x} =-\,\half\int_{0}^{1}\frac{\ln\pars{x}}{1 - x}\,\dd x -\half\int_{0}^{1}\frac{\ln\pars{x}}{1 + x}\,\dd x \\[5mm]&=-\,\half\int_{0}^{1}\frac{\ln\pars{1 - x}}{x}\,\dd x +\half\int_{0}^{-1}\frac{\ln\pars{-x}}{1 - x}\,\dd x \\[5mm]&=-\,\half\int_{0}^{1}\frac{\ln\pars{1 - x}}{x}\,\dd x +\half\int_{0}^{-1}\frac{\ln\pars{1 - x}}{x}\,\dd x =-\,\half\int_{-1}^{1}\frac{\ln\pars{1 - x}}{x}\,\dd x \\[5mm]&=\half\int_{-1}^{1}\Li{2}'\pars{x}\,\dd x =\frac{\Li{2}\pars{1} - \Li{2}\pars{-1}}{2} =\frac{\pi^{2}/6 - \pars{-\pi^{2}/12}}{2}=\dsc{\frac{\pi^{2}}{8}} \end{align}