# Conjecture:$\int_0^{\pi/2} \left(\frac{\sin(2n+1)x}{\sin x}\right)^\beta dx$ is integer multiple of $\pi/2$,for integer $\beta>2$

I was solving the integral $$I_n=\int_0^{\frac {\pi}{2}} \left(\frac {\sin ((2n+1)x)}{\sin x}\right)^2 dx$$

With $$n\ge 0$$ And $$n\in \mathbb{N}$$

On solving, I got $$I_n =\frac {(2n+1)\pi}{2}$$

But, due to curiosity, I started investigating the family of integrals as

$$I_n(\beta) =\int_0^{\frac {\pi}{2}} \left(\frac {\sin (2n+1)x}{\sin x}\right)^{\beta} dx$$

On trying various values of $$\beta\gt 2$$ and $$\beta\in \mathbb{N}$$, I conjectured that $$I_n(\beta) =c_{\beta} \frac{\pi}{2}$$ where $$c_{\beta}$$ denotes "Number of arrays of $$\beta$$ integers in $$-n$$ to $$n$$ with sum $$0$$"

But, on trying a lot, I couldn't prove this statement. Also, I suppose that the statement could be proved with help of Dirichlet kernel, but I couldn't get the way out through it.

Any help and hints to prove/disprove the conjecture are greatly appreciated.

• @Masacroso I think $I_n$ is the integral of something alike Fejer kernel, so $\sin((2n+1)x)$. – xbh Dec 29 '18 at 7:08
• @Masacroso Edited!!! – Rohan Shinde Dec 29 '18 at 7:10
• oeis.org/A201552 – James Arathoon Dec 29 '18 at 12:21
• – Zacky Dec 29 '18 at 12:36
• @James Arathoon The statement I guessed was from OEIS only and it also doesn't have any proofs. – Rohan Shinde Dec 29 '18 at 13:01

$$\def\b{\beta}$$\begin{align*} \newcommand\cmt[1]{{\small\textrm{#1}}} I_n(\b) &= \int_0^{\pi/2} \left(\frac{\sin (2n+1)x}{\sin x}\right)^\b dx \\ &= \frac 1 4 \int_0^{2\pi} \left(\frac{\sin (2n+1)x}{\sin x}\right)^\b dx & \cmt{begin similar to user630708} \\ &= \frac{1}{4i} \oint_\gamma \left(\frac{z^{4n+2}-1}{z^2-1}\right)^\b \frac{dz}{z^{2n\b+1}} & \cmt{let z=e^{ix}} \\ &= \frac{1}{4i} \oint_\gamma \left(\sum_{k=0}^{2n}z^{2k}\right)^\b \frac{dz}{z^{2n\b+1}} & \cmt{partial sum of geometric series} \\ &= \left.\frac{1}{4i} \frac{2\pi i}{(2n\b)!} \left(\frac{d}{dz}\right)^{2n\b} \left(\sum_{k=0}^{2n}z^{2k}\right)^\b \right|_{z=0} & \cmt{Cauchy integral formula} \\ &= \left.\frac{\pi}{2} \frac{1}{(2n\b)!} \left(\frac{d}{dz}\right)^{2n\b} \sum_{\sum x_k=\b} \frac{\b!}{\prod x_k!} \prod (z^{2k})^{x_k} \right|_{z=0} & \cmt{multinomial expansion, k=0,1,\ldots,2n} \\ &= \left.\frac{\pi}{2} \frac{1}{(2n\b)!} \left(\frac{d}{dz}\right)^{2n\b} \sum_{\sum x_k=\b} \frac{\b!}{\prod x_k!} z^{2\sum k x_k} \right|_{z=0} \\ &= \frac{\pi}{2} \sum_{\sum x_k=\b \atop \sum k x_k = n\b} \frac{\b!}{\prod x_k!} & \cmt{only surviving terms have \sum k x_k = n\b} \\ &= \frac{\pi}{2} \sum_{\sum x_k=\b \atop \sum (n-k) x_k = 0} \frac{\b!}{\prod x_k!} \end{align*} In the last line note that $$\sum_{k=0}^{2n} n x_k=n\b$$ and so $$\sum_{k=0}^{2n} (n-k)x_k = 0$$. By inspection one can see that $$\sum_{\sum_{k=0}^{2n} x_k=\b \atop \sum_{k=0}^{2n} (n-k) x_k = 0} \frac{\b!}{\prod x_k!} = \textrm{number of arrays of \b integers in -n,\ldots,n with sum equal to 0,}$$ i.e., $$I_n(\b) = \frac{\pi}{2} T(\b,n),$$ where $$T(\b,n)$$ is OEIS A201552, as pointed out by James Arathoon in the comments. (On that page we also find an integral form of $$T(\b,n)$$ which, after a simple substitution, gives $$I_n(\b) = \frac{\pi}{2} T(\b,n)$$.)

This is easy using Residue Theory:

Note that by symmetry $$\int_{0}^{\pi/2}...dx=1/4\int_{-\pi}^{\pi}…dx$$ (use parity and a sub $$y=\pi-x$$ to Show that).

employing $$z=e^{ix}$$ we get

$$4 I_{n,\beta}=\oint_C \left[\frac{z^{4n+2}-1}{z^2-1}\right]^{\beta}\frac{dz}{i z^{2\beta n+1}}$$

where $$C$$ denotes the unit circle in the comlex plane. By the residue theorem (there is one pole inside the contour at $$z=0$$, using f.e. the geometric series you can Show that the Points $$z=\pm i$$ are removable singularities). We have

$$4 I_{n,\beta}=2\pi \text{Res}(\left[\frac{z^{4n+2}-1}{z^2-1}\right]^{\beta}\frac{1}{z^{2\beta n+1}} ,z=0)$$

Using $$\beta!(z^2-1)^{-\beta}=((2z)^{-1}\partial_z)^{\beta-1}(z^2-1)^{-1}$$ we get

$$(1-z^2)^{-\beta}=\frac{1}{2^{\beta-1}}\sum_{m\geq0}\binom{m+\beta-1}{\beta-1}z^{2m}\\ (1-z^{4n+2})^{\beta}=z^{2\beta}\sum_{k\geq0}(-1)^k\binom{\beta}{k}z^{4k}$$

which means that we have the condition $$4k+2(m+\beta)-2\beta n-1=-1$$ (since we are interested in $$a_{-1}$$ coefficent of the Laurent expansion) which essenitally kills one of the sums, and we end up with

$$I_{n,\beta}=\frac{\pi}{2^{\beta}}\sum_{m\geq0}(-1)^{\beta n /2-(\beta+m)/2}\binom{m+\beta-1}{\beta-1}\binom{\beta}{\beta n /2-(\beta+m)/2}$$

which is a finite sum, since the second binomial becomes zero when $$m$$ is large enough ($$m> \beta (n-1)$$)

• This seems cool, but where does the $\beta \neq 4$ exception come from? Is it from that line about how we "kill one of the sums"? – goblin Jan 1 at 23:46
• Notice that the final sum is not necessarily real. – user26872 Jan 2 at 0:37
• Things seem to go off the rails with $\beta!(z^2-1)^{-\beta}=((2z)^{-1}\partial_z)^{\beta-1}(z^2-1)^{-1}$, which is false for $\beta>1$. It is true that $(\beta-1)!(z^2-1)^{-\beta}=((-2z)^{-1}\partial_z)^{\beta-1}(z^2-1)^{-1}$. – user26872 Jan 5 at 20:45