# Showing that $\int_0^\pi \frac{\sin{(kx)}}{\sin(x)} d x = \pi$ for odd integer $k$

I am trying to show that

$$\int_0^\pi \frac{\sin{(kx)}}{\sin(x)} d x = \pi$$ for odd integer $$k$$.

It seems like this could be done using multiple angle formulae, but I'm stuck.

I can get to $$\int_0^\pi \frac{\sin{((2N+1)x)}}{\sin(x)} d x$$ $$= \sum_{\ell = 0} ^N (-1)^\ell \frac{(2N+1)!}{(2\ell+1)!(2(N-\ell))!} \int_0 ^\pi (\cos(x))^{2(N-\ell)} (\sin(x))^{2\ell} d x,$$ using a multiple angle formula due to Viete (from the trig formulae Wikipedia page), but I don't know what to do from here. Is there a simpler way?

If we let $$I_n=\int \frac{\sin{((2n+1)x)}}{\sin{(x)}} dx$$ One can quite easily show that $$I_n=I_{n-1}+\frac{\sin{(2nx)}}{n}$$ Solving this recurrence relation gives $$I_n=\sum_{k=1}^n \frac{\sin{(2kx)}}{k}+x+c$$ So, for the integral in question, $$\int_0^\pi \frac{\sin{((2n+1)x)}}{\sin{(x)}}dx=[\sum_{k=1}^n \frac{\sin{(2kx)}}{k}+x]_0^\pi = \pi$$ As $$\sin{(2k\pi)}=0$$ for all $$k\in\mathbb{Z}$$.

• This is very clever. Commented Apr 1, 2019 at 23:40

Let $$k=2n+1$$ then $$I=\int_0^{\pi} \frac {\sin ((2n+1)x)}{\sin x} dx=\int_0^{\pi} \frac {\sin \left(\left(n+\frac 12\right)2x\right)}{\sin x} dx$$

Now substituting $$x=\frac t2$$ we get $$I=\frac 12\int_0^{2\pi} \frac {\sin \left(\left(n+\frac 12\right)t\right)}{\sin \left(\frac t2\right)} dt= \frac 12\int_0^{2\pi} D_n(t) dt$$

Where $$D_n(t)$$ represents the Dirichlet Kernel. And using its property that $$D_n(t)=1+2\sum_{k=0}^n \cos(kt)$$

We get $$I=\frac 12\int_0^{2\pi} \left(1+2\sum_{k=0}^n \cos(kt)\right)dt=\pi+ \sum_{k=0}^n \int_0^{2\pi} \cos (kt) dt$$

From where you notice that the last integral is $$0$$ for all integer $$k$$ from $$0$$ to $$n$$

Q.E.D

Use the addition formulas for $$\sin$$ and $$\cos$$ to show that $$\frac{\sin ((k+2)x)}{\sin(x)} = \frac{\sin (k x)}{\sin(x)} + 2\cos ((k+1)x)$$ From this formula the result you are after follows directly by using induction and the fact that $$\int_0^\pi \cos(nx){\rm d}x = 0$$ for all integers $$n>0$$. It also follows that the integral is $$0$$ for even $$k$$ although this can be shown more easily by using the fact that the integrand is odd about $$x = \pi/2$$.

Note that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ Then $$\frac{\sin(kx)}{\sin(x)}=\frac{e^{ikx}-e^{-ikx}}{e^{ix}-e^{-ix}}=\frac{(e^{ix})^k-(e^{-ix})^k}{e^{ix}-e^{-ix}}$$ Using (n is odd)$$a^n-b^n=(a-b) \sum _{j=0}^{n-1} a^j b^{n-1-j}$$ We have $$\frac{\sin(kx)}{\sin(x)}=\sum _{j=0}^{k-1} (e^{ix})^j (e^{-ix})^{k-1-j}=\sum _{j=0}^{k-1} e^{i(2j-(k-1))x}$$ $$\sum _{j=0}^{k-1} e^{i(2j-(k-1))x}=\sum _{j=0}^{k-1} \frac{e^{i(2j-(k-1))x}+ e^{-i(2j-(k-1))x}}{2}=\sum _{j=0}^{k-1}\cos((2j-(k-1))x)$$ It should be easy to move on now

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\bbox[10px,#ffd]{\left.\int_{0}^{\pi}{\sin\pars{kx} \over \sin\pars{x}}\,\dd x\,\right\vert_{\ k\ =\ \pm 1,\ \pm 3,\ \pm 5,\ldots}} = 2\,\mrm{sgn}\pars{k}\Im\int_{0}^{\pi/2}{\expo{\ic\verts{k}x} - 1 \over \sin\pars{x}}\,\dd x \\[5mm] = &\ \left.2\,\mrm{sgn}\pars{k}\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}{z^\verts{k} - 1 \over \pars{1 - z^{2}}\ic/\pars{2z}}\,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left.4\,\mrm{sgn}\pars{k}\Im\int_{x\ =\ 0}^{x\ =\ \pi/2} {1 - z^\verts{k} \over 1 - z^{2}}\,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] =& -4\,\mrm{sgn}\pars{k}\Im\int_{1}^{0} {1 - y^\verts{k}\expo{\ic\verts{k}\pi/2} \over 1 + y^{2}}\,\ic\,\dd y = 4\,\mrm{sgn}\pars{k}\int_{0}^{1}{\dd y \over 1 + y^{2}} \\[5mm] = & \bbx{\pi\,\mrm{sgn}\pars{k}\,\qquad k = \pm 1,\ \pm 3,\ \pm 5\ldots} \end{align}

Note that $$\ds{\left.\cos\pars{k\,{\pi \over 2}} \,\right\vert_{\ k\ \mrm{odd}} = \color{red}{\large 0}}$$.