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.

  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ 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

$\def\b{\beta}$\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 ($m> \beta (n-1)$)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.