# Can we simplify $\int_0^{\pi}\left(\frac{\sin nx}{\sin x}\right)^mdx$?

Letting $m,n$ be natural numbers, can we simplify the following definite integral?

$$\int_0^{\pi}\left(\frac{\sin nx}{\sin x}\right)^mdx$$

I've known the followings: $$\int_0^{\pi}\frac{\sin 2kx}{\sin x}dx=0, \int_0^{\pi}\frac{\sin (2k-1)x}{\sin x}dx=\pi, \int_0^{\pi}\left(\frac{\sin nx}{\sin x}\right)^2dx=n\pi.$$

Proof: Letting $I_n=\int_0^{\pi}\frac{\sin nx}{\sin x}dx$, we get $$I_n-I_{n-2}=\int_0^{\pi}\frac{\sin{nx}-\sin{(n-2)x}}{\sin x}dx=2\int_0^{\pi}\cos(n-1)xdx=2\left[\frac{\sin{(n-1)x}}{n-1}\right]_0^{\pi}=0$$ for $n\ge 3$. Then, we get $$I_{2k}=I_2=0, I_{2k-1}=I_1=\pi.$$

Letting $J_n=\int_0^{\pi}\left(\frac{\sin nx}{\sin x}\right)^2dx$, we get $$J_n-J_{n-1}=\frac{-1}{2}\int_0^{\pi}\frac{\cos{2nx}-\cos{(2n-2)x}}{\sin^2x}=\frac{-1}{2}\int_0^{\pi}\frac{-2\sin\left(\frac{4n-2}{2}x\right)\sin\left(\frac{2x}{2}\right)}{\sin^2}$$$$=\int_0^{\pi}\frac{\sin{(2n-1)x}}{\sin x}=I_{2n-1}=\pi.$$ Then, we get $$J_n=J_{n-1}+\pi=J_{n-2}+2\pi=\cdots=J_1+(n-1)\pi=n\pi.$$ Now the proof is completed.

I'm interested in the generalization of these definite integrals. Here is my question.

Question: Letting $m,n$ be natural numbers, can we simplify the following definite integral?

$$\int_0^{\pi}\left(\frac{\sin nx}{\sin x}\right)^mdx$$ I don't have any good idea. Could you show me how to simplify this?

• If $m$ is odd and $n$ is even then the integral is zero. Also it seems that $Q(n,n) = \pi \cdot \text{A077045}$ where oeis.org/A077045 Commented Aug 28, 2013 at 12:26
• May be the relation to the Chebyshev polynomials of the second kind $$\frac{\sin{n x}}{\sin x } = U_{n-1}(\cos x )$$ may help. Commented Aug 28, 2013 at 12:43
• i think you can use induction @mathlove Commented Aug 28, 2013 at 13:08
• The Maple command $$[seq(int((sin(k*x)/sin(x))^5, x = 0 .. Pi), k = 1 .. 10) ]$$ produces $$[\pi ,0,51\,\pi ,0,381\,\pi ,0,1451\,\pi ,0,3951\,\pi ,0]$$ and $$[seq(int((sin(k*x)/sin(x))^6, x = 0 .. Pi), k = 1 .. 10) ]$$ gives $$[\pi ,20\,\pi ,141\,\pi ,580\,\pi ,1751\,\pi ,4332\,\pi ,9331\,\pi , 18152\,\pi ,32661\,\pi ,55252\,\pi ] .$$ Commented Aug 28, 2013 at 13:46
• @user64494 - An interesting cross-link for your results: The first sequence $(a_n)$ without the zero (1, 51, 381, 1451, 3951, 8801, 17151) is the start of oeis.org/A083669, which lists two formulas $$a_n= 1 + \frac{5}{12} n (n+1) (23 n^2 + 23 n +14),$$ $$a_n= \int_0^{\pi}\frac{sin((n+\frac{1}{2})x)}{sin(\frac{x}{2})^5} dx$$ The second (1, 20, 141, 580, 1751, 4332, 9331, 18152, 32661, 55252) refers to oeis.org/A071816. Commented Aug 29, 2013 at 9:13

An explicit sum form may be derived. Consider first the case of odd $n$. Take $n=2N+1$ to be an odd integer and note that: \begin{aligned} \int_0^{\pi}\left(\frac{\sin nx}{\sin x}\right)^mdy&=\frac{1}{2}\int_0^{2\pi}\left(\frac{\sin (N+\frac{1}{2})y}{\sin \frac{y}{2}}\right)^mdy\\ &=\frac{1}{2}\int_0^{2\pi}\left(\sum_{k=-N}^{N}e^{Iky}\right)^mdy\\ &=\sum_{k_1,k_2,..,k_m=-N}^{N}\frac{1}{2}\int_0^{2\pi}e^{I\left(k_1+k_2+...k_m\right)x}dy\\ &=\sum_{k_1,k_2,..,k_m=-N}^{N}\pi\delta_{k_1+k_2+..k_m,0}=\pi L\\ \end{aligned}\tag{1} where we have used the Dirichlet Kernel and the integral representation of the Kronecker Delta.

We see straight away that the result is an integer multiple of $\pi$. I am hoping someone better schooled in combinatorics will be able to spot that number $L$ straight away (or perhaps name it if it has a name); it is the number of ways to add $m$ integers between $-N$ and $N$ to make $0$. If this doesn't happen, it is still possible to obtain an explicit sum for $L$ using an ordinary generating function by noting that $L$ is the coefficient of $x^0$ in: $$\left(\sum_{j=-N}^{N}x^j\right)^m=\left( {\frac {{x}^{N+1}-{x}^{-N}}{x-1}} \right) ^{m}\tag{2}$$ Unfortunately we can't put $x=0$ in there because that contains negative powers of $x$. Instead we multiply $(2)$ by $x^{Nm}$ and take $L$ as the coefficient of $x^{Nm}$ to get: $$\int_0^{\pi}\left(\frac{\sin (2N+1)x}{\sin x}\right)^mdx=\frac{\pi}{(Nm)!}\lim_{x=0} \dfrac{d^{Nm}}{dx^{Nm}}\left( {\frac {{x}^{2N+1}-1}{x-1}}\right)^m\quad :N\in\mathbb{Z}^+ \tag{3}$$ If we then apply the generalized Leibniz rule for the ${Nm}^{th}$ derivative to $(3)$ the RHS becomes:$$\frac{\pi}{(Nm)!}\sum_{j=0}^{Nm}{Nm \choose j}\left(\lim_{x=0} \dfrac{d^{j}}{dx^{j}}\left( {x}^{2 N+1}-1\right)^m\right)\left(\lim_{x=0} \dfrac{d^{Nm-j}}{dx^{Nm-j}} \dfrac{1}{\left(x-1\right)^m}\right)\tag{4}$$  we then note that:   \begin{aligned} \lim_{x=0} \dfrac{d^{Nm-j}}{dx^{Nm-j}} \dfrac{1}{\left(x-1\right)^m}&=\lim_{x=0} \dfrac{d^{Nm-j}}{dx^{Nm-j}} \left(\dfrac{(-1)^{m-1}}{(m-1)!}\dfrac{d^{m-1}}{dx^{m-1}} \dfrac{1}{\left(x-1\right)}\right)\\ &=\dfrac{(-1)^{m-1}}{(m-1)!} \lim_{x=0}\dfrac{d^{(N+1)m-j}}{dx^{(N+1)m-j}} \dfrac{1}{\left(x-1\right)}\\ &={\frac { \left( -1 \right) ^{m} \left( \left( N+1 \right) m-j-1 \right) !}{ \left( m-1 \right) !}} \end{aligned}\tag{5} and that: \begin{aligned} \lim_{x=0} \dfrac{d^{j}}{dx^{j}}\left( {x}^{2 N+1}-1\right)^m&=(-1)^m\lim_{x=0} \dfrac{d^{j}}{dx^{j}}\sum_{l=0}^{m}{m \choose l}(-1)^l{x}^{l(2 N+1)}\\ &=\cases{0&j\neq l \left( 2\,N+1 \right) \cr \left( -1 \right) ^{m+l}{m\choose l} \left( l \left( 2\,N+1 \right) \right) !&j=l \left( 2\,N+1 \right) \cr} \end{aligned}\tag{6} where we read $(6)$ as meaning that the result is zero unless $j$ is a multiple $l$ times $2N+1$ in which case we insert $l$ into the non-zero formula. After using $(5)$ and $(6)$ in $(4)$, summing $j$ only over multiples of $2N+1$ and simplifying the factorials and binomial coefficients we obtain an explicit sum form for $(3)$: $$\int_0^{\pi}\left(\frac{\sin (2N+1)x}{\sin x}\right)^mdx=\pi\sum_{l=0}^{\left\lfloor\tfrac{Nm}{2N+1}\right\rfloor}\left( -1 \right) ^{l}{m\choose l} { \left( N+1 \right) m-l \left( 2N+1 \right) -1\choose m-1}\tag{7}$$ where $\lfloor\, \rfloor$ denotes the floor function.

In the case of even $n=2N$ and even $m=2M$ we would use: $${{\rm e}^{ix}}\sum _{k=-N}^{N-1}{{\rm e}^{2\,ikx}}={\frac {\sin \left( 2\,xN \right) }{\sin \left( x \right) }}\tag{8}$$ to show: \begin{aligned} \int _{0}^{\pi }\! \left( {\frac {\sin \left( 2xN \right) }{\sin \left( x \right) }} \right) ^{2M}{dx}&=\frac{\pi}{(M(2N-1))!}\lim_{x=0} \dfrac{d^{M(2N-1)}}{dx^{M(2N-1)}}\left( {\frac {{x}^{2N}-1}{x-1}}\right)^{2M}\\ &=\pi \sum _{l=0}^{\left\lfloor {\tfrac{M\left(2N-1\right)}{2N}} \right\rfloor }\left( -1 \right) ^{l}{2M\choose l}{M \left( 2N+1\right) -l2N-1\choose 2M-1} \end{aligned}\tag{9} The integral is zero by symmetry when $n$ is even and $m$ is odd; otherwise, combining $(7)$ and $(9)$ gives: $$\int _{0}^{\pi }\! \left( {\frac {\sin \left( nx \right) }{\sin \left( x \right) }} \right)^{m}{dx}=\pi \sum _{l=0}^{\left\lfloor {\tfrac{m\left(n-1\right)}{2n}} \right\rfloor }\left( -1 \right) ^{l}{m\choose l}{\frac{m}{2}\left( n+1\right) -ln-1\choose m-1}\tag{10}$$

• Using \dfrac on the upper limit of a sum is just... off. And you know how to use \left and \right; you should have used it on your floor brackets. Commented Jul 10, 2016 at 10:33

$\displaystyle{\large% {\cal I}_{mn} \equiv \int_{0}^{\pi} \left\lbrack{\sin\left(nx\right) \over \sin\left(x\right)}\right\rbrack^{m}\,{\rm d}x \ = \ ?\,, \quad m, n = 0,1,2,\ldots}$

$$\lim_{m \to 0}\lim_{n \to 0}{\cal I}_{mn} = {\cal I}_{10} = 0\,, \qquad \lim_{n \to 0}\lim_{m \to 0}{\cal I}_{mn} = {\cal I}_{01} = {\cal I}_{11} = \pi$$

Let's consider $m > 1$ and $n > 1$: \begin{align} {\cal I}_{mn} &= {1 \over 2}\int_{0}^{2\pi}{\rm e}^{-{\rm i}m\left(n - 1\right)x/2} \left( {{\rm e}^{{\rm i}nx} - 1 \over {\rm e}^{{\rm i}x} - 1} \right)^{m}\,{\rm d}x = {1 \over 2}\oint_{\left\vert z\right\vert = 1}{1 \over z^{m\left(n - 1\right)/2}} \left(% {z^{n} - 1 \over z - 1} \right)^{m}\,{{\rm d}z \over {\rm i}z} \\[3mm]&= \pi\oint_{\left\vert z\right\vert = 1}{{\rm d}z \over 2\pi{\rm i}}\,{1 \over z}\, {1 \over z^{m\left(n - 1\right)/2}} \left({1 - z^{n} \over 1 - z}\right)^{m} \end{align}

Then, $\left.{\cal I}_{mn}\vphantom{\Large A}\right\vert_{m\ >\ 1 \atop n\ >\ 1} = 0$ whenever $\phantom{A}m\left(n - 1\right)\phantom{A}$ is odd. That means $\left.{\cal I}_{mn}\vphantom{\Large A}\right\vert_{m > 1 \atop n > 1} = 0$ when $m$ is odd and $n$ is even.

\begin{align} \left(1 - z^{n} \over 1 - z\right)^{m} &= \sum_{k = 0}^{m}{m \choose k}\left(-1\right)^{k}z^{nk} \sum_{k' = 0}^{\infty}{m + k'\choose k'}z^{k'} \sum_{\ell = 0}^{\infty}\delta_{\ell, nk + k'} \\[3mm]&= \sum_{\ell = 0}^{\infty}z^{\ell} \sum_{k = 0}^{\leq\ M \atop}\left(-1\right)^{k}{m \choose k} {m + \ell - nk \choose \ell - nk} \\[3mm]&= \sum_{\ell = 0}^{\infty}z^{\ell} \sum_{k = 0}^{\leq\ M \atop}\left(-1\right)^{k} {\left(m + \ell -nk\right)! \over k!\left(m - k\right)!\left(\ell - nk\right)!}\,, \quad M \equiv \min\left\lbrace m, {\ell \over n}\right\rbrace \\[1cm]& \end{align}

\begin{align} \int_{0}^{\pi} \left\lbrack{\sin\left(nx\right) \over \sin\left(x\right)}\right\rbrack^{m}\,{\rm d}x &= \pi\sum_{k = 0}^{\leq\ M\atop}\left(-1\right)^{k} {\left\lbrack m\left(n + 1\right)/2 -nk\right\rbrack! \over k!\left(m - k\right)!\left\lbrack m\left(n - 1\right)/2 - nk\right\rbrack!} \\[2mm]& \mbox{where}\quad M \equiv \min\left\lbrace m, {1 \over 2}\,m\left(1 - {1 \over n}\right)\right\rbrace = {1 \over 2}\,m\left(1 - {1 \over n}\right) \end{align}

Examples:

\begin{align} {\cal I}_{1n} &= \int_{0}^{\pi}{\sin\left(nx\right) \over \sin\left(x\right)}\,{\rm d}x = \pi {\left\lbrack \left(n + 1\right)/2\right\rbrack! \over \left\lbrack \left(n - 1\right)/2\right\rbrack!} = {1 \over 2}\,\left(n + 1\right)\pi\,, \qquad n = 3, 5, 7, \ldots \\[3mm] {\cal I}_{2n} &= \int_{0}^{\pi}\left\lbrack{\sin\left(nx\right) \over \sin\left(x\right)}\right\rbrack^{2}\,{\rm d}x = \pi\,{\left(n + 1\right)! \over \left(n - 1\right)!} = n\left(n + 1\right)\pi\,, \qquad n = 2, 3, 4, \ldots \\[6mm] {\cal I}_{3n} &= \int_{0}^{\pi}\left\lbrack{\sin\left(nx\right) \over \sin\left(x\right)}\right\rbrack^{3}\,{\rm d}x = \pi\, {\left\lbrack 3\left(n + 1\right)/2\right\rbrack! \over 3!\left\lbrack 3\left(n - 1\right)/2\right\rbrack!} - \pi\, {\left\lbrack 3\left(n + 1\right)/2 - n\right\rbrack! \over 2!\left\lbrack 3\left(n - 1\right)/2 -n\right\rbrack!} \\[3mm]&= {\pi \over 6}\,{3n + 3 \over 2}\,{3n + 1 \over 2} - {\pi \over 2}\,{n + 3 \over 2}\,{n + 1 \over 2} = {1 \over 4}\,\left(n^{2} - 1\right)\pi\,, \qquad n = 3, 5, 7, \ldots \end{align}