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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.

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    $\begingroup$ @Masacroso I think $I_n$ is the integral of something alike Fejer kernel, so $\sin((2n+1)x)$. $\endgroup$ – xbh Dec 29 '18 at 7:08
  • $\begingroup$ @Masacroso Edited!!! $\endgroup$ – Rohan Shinde Dec 29 '18 at 7:10
  • $\begingroup$ oeis.org/A201552 $\endgroup$ – James Arathoon Dec 29 '18 at 12:21
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    $\begingroup$ See here: math.stackexchange.com/a/2885887/515527 $\endgroup$ – Zacky Dec 29 '18 at 12:36
  • $\begingroup$ @James Arathoon The statement I guessed was from OEIS only and it also doesn't have any proofs. $\endgroup$ – Rohan Shinde Dec 29 '18 at 13:01
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$\def\b{\beta}$\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 $I_n(\b) = \frac{\pi}{2} T(\b,n)$.)

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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 ($m> \beta (n-1)$)

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  • $\begingroup$ 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"? $\endgroup$ – goblin Jan 1 at 23:46
  • $\begingroup$ Notice that the final sum is not necessarily real. $\endgroup$ – user26872 Jan 2 at 0:37
  • $\begingroup$ 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 $(\beta-1)!(z^2-1)^{-\beta}=((-2z)^{-1}\partial_z)^{\beta-1}(z^2-1)^{-1}$. $\endgroup$ – user26872 Jan 5 at 20:45

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