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I find the following averaged-integral amusing and intriguing, to say the least. Is there any proof?

For any pair of integers $n\geq k\geq0$, we have $$\frac1{\pi}\int_0^{\pi}\frac{\sin^n(x)}{\sin^k(\frac{kx}n)\sin^{n-k}\left(\frac{(n-k)x}n\right)}dx=\binom{n}k. \tag1$$

I also wonder if there's any reason to relate these with an MO question that I just noticed. Perhaps by inverting?

AN UPDATE. I'm extending the above to a stronger conjecture shown below.

For non-negative reals with $r\geq s$, a generalization is given by $$\frac1{\pi}\int_0^{\pi}\frac{\sin^r(x)}{\sin^s(\frac{sx}r)\sin^{r-s}\left(\frac{(r-s)x}r\right)}\,dx =\binom{r}{s}. \tag2$$

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    $\begingroup$ Does that mean you have a proof? If so, what is the question? If not, where does the equality come from? $\endgroup$ – Geoff Robinson Nov 2 '16 at 0:50
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    $\begingroup$ I do not have a proof. It comes from some experimentation and curiosity. $\endgroup$ – T. Amdeberhan Nov 2 '16 at 1:01
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    $\begingroup$ very interesting formula. I verified it for a few easy cases (e.g. ${2n}\choose{n}$) and also few not-so-trivial(e.g. ${3}\choose{1}$) with Mathematica. $\endgroup$ – BigM Nov 2 '16 at 4:32
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    $\begingroup$ @T.Amdeberhan I'm slightly occupied at the moment, but I worked out an argument on paper. I'll write it out tomorrow if it comes to fruition-- essentially it uses Ramanujan's master theorem in two variables. This is essentially an identity theorem on the naturals. if $|a_1(x+iy)|,|a_2(x+iy)| < C e^{(\pi-\delta)|y| + \rho|x|}$ for $x>0$ and $\rho$ arbitrary then $a_1\Big{|}_{\mathbb{N}} = a_2\Big{|}_{\mathbb{N}} \Rightarrow a_1 = a_2$. Therefore if $f(z)= \int_0^\pi \dfrac{\sin^{z}(x)}{\sin^{k}(\frac{k}{z}x)\sin^{z-k}(\frac{z-k}{z}x)}$ is bounded as such it necessarily equals $\dbinom{z}{k}$. $\endgroup$ – user78249 Nov 2 '16 at 22:45
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    $\begingroup$ My attempt led to a binomial-coefficient identity to prove ... see mathoverflow.net/q/253835/454 $\endgroup$ – Gerald Edgar Nov 3 '16 at 12:31
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Tonight I read here [the answer by esg to another your question] that $\frac1{2\pi}\int_{-\pi}^\pi e^{-ik t}(1+e^{it})^ndt=\binom{n}{k}$, which is, well, obvious at least when both $n$ and $k$ are positive integers: just expand the binomial $(1+e^{it})^n$ and integrate. Denoting $\alpha=k/n$ we may rewrite this as $\frac1{2\pi}\int_{-\pi}^\pi (f(t))^n dt=\binom{n}{\alpha n}$, where the function $f(t)=(1+e^{it})e^{-i\alpha t}$ is complex-valued. For making it real-valued, we change the path between the points $-\pi$ and $\pi$. The value of the integral does not change (since $f^n$ is analytic between two paths, for integer $n$ it is simply entire function.) On the second path $f$ takes real values. Namely, for $t\in (-\pi,\pi)$ we define $s(t)=\ln \frac{\sin (1-\alpha)t}{\sin \alpha t}$. It is straightforward (some elementary high school trigonometry) that $$f(t+is(t))=\frac{\sin t}{\sin^{\alpha} \alpha t\cdot \sin^{1-\alpha}(1-\alpha)t},$$ so we replace the path from $(-\pi,\pi)$ to $\{t+s(t)i:t\in (-\pi,\pi)\}$ (limit values of $s(t)$ at the endpoints are equal to 0) and take only the real part of the integral (this allows to replace $d(t+s(t)i)$ to $dt$ in the differential). We get $$ \frac1{2\pi}\int_{-\pi}^\pi \frac{\sin^n t}{\sin^{\alpha n} \alpha t\cdot \sin^{(1-\alpha)n}(1-\alpha)t}dt=\binom{n}{\alpha n} $$ as desired.

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    $\begingroup$ I cannot imagine a better explanation for this beautiful identity. $\endgroup$ – Mark Wildon Nov 19 '16 at 13:38
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    $\begingroup$ Bravo, nice proof! It works for any (incl complex) values of $n$ and $k$ too, not just integer. Consider $(1/2\pi)\int_{-\pi}^\pi e^{-ikt}(1+e^{it})^n dt$ for $\Re(k)<0$, $\Re(n)>0$, then the integrand is analytic for $-\pi<\Re(t)<\pi$. We can deform the contour to two the vertical strips $\pm\pi+iy$ for $y>0$. Then you get a Beta function and the integral becomes $-(1/\pi)\sin(k\pi)B(-k,n+1)$ which is $\binom{n}{k}$, using the reflection formula $\sin(k\pi)=\pi/(\Gamma(k)\Gamma(1-k))$. The derivation was valid for $\Re(k)<0$, but both sides are analytic, so continuation gives you all $k$. $\endgroup$ – Alex Selby Nov 19 '16 at 19:08
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    $\begingroup$ Simply beautiful! $\endgroup$ – esg Nov 19 '16 at 20:19
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    $\begingroup$ Something interesting about this proof is that it is "backwards" from the normal way of substituting. Writing $u(t)=t+is(t)$, you'd normally want to have $t$ expressed analytically in terms of $u$ to get from the sinomial expression with $\sin^{\alpha}(\alpha t)$ etc to $(e^{-i\alpha u}+e^{i(1-\alpha)u})^n$. But there is presumably no such simple expression of $t$ in terms of $u$, so you had to know to start from $(e^{-i\alpha u}+e^{i(1-\alpha)u})^n$ and work the other way, using magic to end up with the sinomial expression. (Slightly wondering if other integrals can be unlocked like this.) $\endgroup$ – Alex Selby Nov 22 '16 at 0:57
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    $\begingroup$ @AlexSelby I know a similar proof that $\sum 1/n^2=\pi^2/6$: take integral $\int_{-1}^1 \log(1+z)/z dz$ (which equal $1/2$ times $\sum 1/(2k+1)^2$ as follows from series expanding) and replace the contour to an arc of a unit circle. It is borrowed from D. Russel, Another Eulerian-type proof. Math. Mag. 1991 60, p.349. $\endgroup$ – Fedor Petrov Nov 24 '16 at 22:18
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(not an answer)

Denote $\alpha=k/n$, $f(x)=(\frac{\sin x}{\sin \alpha x})^\alpha (\frac{\sin x}{\sin (1-\alpha) x})^{1-\alpha}$. Then your claim may be rewritten as $\pi^{-1}\int_0^\pi f^n(x)dx=\frac{\Gamma(n+1)}{\Gamma(\alpha n+1)\Gamma((1-\alpha)n+1)}$, and it looks to be true without additional assumption that $\alpha n$ is integer (I checked for $\alpha=0.3;n=7$ or $n=7.4$ on WolframAlpha). We may multiply this by Beta-function $\int_0^1 t^{\alpha n}(1-t)^{(1-\alpha)n}dt=\frac{\Gamma(\alpha n+1)\Gamma((1-\alpha)n+1)}{\Gamma(n+2)}$, and we have to prove that $\int_0^\pi\int_0^1 h^n(t,x)dtdx=\frac{\pi}{n+1}$, where $h(t,x)=f(x)t^\alpha(1-t)^{1-\alpha}$. That is, our function $h$ on the rectangular $[0,\pi]\times [0,1]$ (with the normalized Lebesgue measure) should be equidistributed with the function $t$ on $[0,1]$. Another similar approach could be multiplying by two $\Gamma$-functions $\int_0^{\infty} y^{\alpha n}e^{-y}dy=\Gamma(\alpha n+1)$, $\int_0^{\infty} z^{(1-\alpha) n}e^{-z}dz=\Gamma((1-\alpha) n+1)$. On the probabilistic language, we get the following equivalent

Claim. Let EXP denote the exponential law (with density $e^{-t}dt$, $t>0$). Let $Y,Z$ be independent random variables distributed by EXP, and let $X$ be a third independent (of $Y,Z$) random variable distributed uniformly on $[0,\pi]$. Then for any fixed $\alpha\in (0,1)$ we have $$\left(Y\frac{\sin X}{\sin \alpha X}\right)^\alpha \left(Z\frac{\sin X}{\sin (1-\alpha) X}\right)^{1-\alpha}\in \text{EXP}.$$

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  • $\begingroup$ Interesting indeed. $\endgroup$ – T. Amdeberhan Nov 3 '16 at 12:34
  • $\begingroup$ Need $n$ to be an integer? $\endgroup$ – Ivan Izmestiev Nov 3 '16 at 14:12
  • $\begingroup$ @IvanIzmestiev No: if they are equidistributed, all moments are equal, including these for non-integer $n$. And calculations confirm this. $\endgroup$ – Fedor Petrov Nov 3 '16 at 15:11
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    $\begingroup$ C. L. Mallows, A formula for expected values, Amer. Math. Monthly 87 (1980), 584 gives probabilistic proof of integral $(2)$. $\endgroup$ – Nemo Nov 30 '17 at 14:52
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I've found the time and thought I should post this as I had a little breakthrough. This isn't an answer to the question but is an answer to a question posted in the comments. If the result holds, does it hold for complex values? I am being brief here and certainly not rigorous as I thought it would be a nice quip to add; nonetheless the result should follow if one wishes to fill in the gaps. If we assume the answer to the OP's question is yes, then

$$\frac{1}{\pi}\int_0^\pi \dfrac{\sin^{z}(t)}{\sin^{k}(\frac{k}{z}t)\sin^{z-k}(\frac{z-k}{z}t)}\,dt = \dbinom{z}{k}$$

This is rather involved (and would be too involved if I chose to make it rigorous) so pay close attention. Consider firstly a consequence of Ramanujan's master theorem

If $f_1(z)$ and $f_2(z)$ are holomorphic for $\Re(z) > 0$ and if $|f_{12}(x+iy)| < C e^{\tau|y|+\rho|x|}$ for $\tau < \pi$ and $\rho>0$ then

$$f_1 \Big{|}_{\mathbb{N}} = f_2\Big{|}_{\mathbb{N}} \Rightarrow f_1 = f_2$$

So essentially what we are going to do is show this in two steps. Firstly that

$$f_k(z) = \frac{1}{\pi}\int_0^\pi \dfrac{\sin^{z}(t)}{\sin^{k}(\frac{k}{z}t)\sin^{z-k}(\frac{z-k}{z}t)}\,dt$$

is bounded so that Ramanujan's master theorem will prevail and necessarily $f_k(z) = \dbinom{z}{k}$ since $\dbinom{z}{k}$ is equally so bounded.

Taking the function $g(z) = \sup_{t \in [0,\pi]} \Big{|}\dfrac{\sin^{z}(t)}{\sin^{k}(\frac{k}{z}t)\sin^{z-k}(\frac{z-s}{z}t)}\Big{|}$ for $\Re(z) > k$ we can show that this function is properly bounded. For each $t$ we know $\sin(t)^{z}$ is bounded as required as $y \to \infty$ for $\epsilon < t < \pi - \epsilon$; because this is exponentiation with a positive real value base--it is periodic. As $x \to \infty$ it just tends to $0$ so all good there. Now $\sin^{k-z}(\frac{z-s}{z}t)$ is exponentiation of a value which tends to $\sin(t)$. This is a little tricky but

$$\sin^{k}(t - \frac{k}{z}t)$$ is bounded and now all that's left is the troublesome

$$\sin^{-z}(t - \frac{k}{z}t)$$

which clearly grows like $\frac{1}{\sin^{x}(t)}$ as $\Re(z) = x \to \infty$. As $\Im(z) = y\to\infty$ it is not periodic, but it is eventually bounded by $\sin^{-z}(t\pm i\delta)$ though not exactly. This bound is of type $\tau < \pi$. This works for all $t\in [\epsilon,\pi-\epsilon]$ and so as $\epsilon \to 0$ it will follow taking close care to observe the end points tend to $1$ as $t \to 0,\pi$. Therefore $g(z) < Ce^{\tau|y| + \rho|x|}$, $f_k$ is of a Ramanujan bound for $\Re(z) > k$ and necessarily

$$f_k(z) = \frac{1}{\pi}\int_0^\pi \dfrac{\sin^{z}(t)}{\sin^{k}(\frac{k}{z}t)\sin^{z-k}(\frac{z-k}{z}t)}\,dt = \dbinom{z}{k}$$

This is all rather hand waivey because I don't want to take up too much space, the amount of epsilons and deltas is exhausting; plus this is more of an extended comment.

Taking $f_s(z)$ is much trickier. Performing the same procedure in the opposite direction is impossible, this is because $\dbinom{z}{s}$ is not bounded in $s$ in the sense described above. It grows like $\sin(\pi s)$ which isn't subject to Ramanujan's master theorem. I thought I could trick it into working but I've had no luck.

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    $\begingroup$ It's a good start. $\endgroup$ – T. Amdeberhan Nov 3 '16 at 0:31
  • $\begingroup$ @T.Amdeberhan Again, I just did this really fast on paper. I have the quasi more rigorous arguments, but I think this is a few pages. For $s$, the lower argument, I do have a good idea, but it's even longer. And I don't want to over extend my reach at the moment, it's more of a longshot. $\endgroup$ – user78249 Nov 3 '16 at 0:39
  • $\begingroup$ The question about interpolating for non integer n and k is another case concerned by mathoverflow.net/questions/181943/…. $\endgroup$ – Wolfgang Nov 9 '16 at 3:44
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Here is another way to prove it. Surprisingly, for $n$ integral and $k$ real, the integral in question can be written down as an indefinite integral. This gives a direct proof for non-integer $k$, though obviously less clear than the contour method. (In fact, it is convenient to avoid integer $k$ in this method, and extend to integer $k$ by continuity.)

Writing $y=x/n$ and $l=n-k$, we have for example for $n=2$: $$\int\frac{\sin^2(2y)}{\sin^k(ky)\sin^l(ly)}dy= \frac{\frac{2}{l-k}\sin((l-k)y)+\sin(2y)} {kl\sin^{k-1}(ky)\sin^{l-1}(ly)}$$

In general for $n$ even ($n$ odd is similar with cosines): $$I_{n,k}(y)=\int\frac{\sin^n(ny)}{\sin^k(ky)\sin^l(ly)}dy= \frac{\sum_{r=0}^{n-1}\sum_{s=0}^{n-1}\lambda_{r,s}\sin(((n-1-2r)k+(n-1-2s)l)y)} {kl\sin^{k-1}(ky)\sin^{l-1}(ly)}$$

where $$\lambda_{r,s}=\begin{cases} \displaystyle \frac{(-1)^r(n-1)^{\underline{r}}}{(r-l)^{\underline{\smash{r-s}}}s!}\lambda_{0,0},\;\;r\ge s\\ \displaystyle \frac{(-1)^s(n-1)^{\underline{s}}}{(s-k)^{\underline{\smash{s-r}}}r!}\lambda_{0,0},\;\;s\ge r \end{cases} $$ $$\lambda_{0,0}=(-1)^{n/2+1}2^{1-n},$$ and $x^{\underline{r}}$ denotes the falling power $x(x-1)\ldots(x-r+1)$.

It is easy to check the derivative, $I_{n,k}'(y)$ is correct by considering the coefficient of $\cos((ak+bl)y)/(\sin^k(ky)\sin^l(ly))$ for each $a$, $b$. If $a\neq b$ then you get zero, otherwise for $a=b=n-2r$, $(ak+bl)y=(n-2r)ny$ and you get $\frac{1}{2}(-1)^r\binom{n}{r}\lambda_{0,0}\cos((n-2r)ny)$. Then $\sin^n(ny)$ arises from the binomial expansion: $$(-1)^{n/2}2^{-n}\sum_{r=0}^n (-1)^r\binom{n}{r}\cos((n-2r)ny)=\sin^n(ny).$$

Note that $I_{n,k}(0)=0$ because, being the integral of something well-behaved at $0$, $I_{n,k}(y)$ must be continuous at $0$, so its numerator must vanish to order $n-2$ like its denominator. Using L'H\^{o}pital, taking $n-2$ derivatives of the numerator gives only sines, which themselves vanish at 0. To evaluate $I_{n,k}(\pi/n)$, note that $\sin(k\pi/n)=\sin(l\pi/n)$ and $\sin(((n-1-2r)k+(n-1-2s)l)y)=\sin(2(r-s)k\pi/n)$. Conditioning on $r-s=d>0$, you get (e.g., by considering partial fractions in $k$) $$\sum_{s=0}^{n-1-d}\lambda_{s+d,s}= \frac{(-1)^d(n-1)!\binom{n-2}{d-1}}{(k-1)^{\underline{\smash{n-1}}}}\lambda_{0,0},$$ and similarly for $d<0$ with the opposite sign, and using $-d$ in place of $d$. Using the binomial expansion of $(1-e^{2\pi ik/n})^{n-2}$, you get $$\sum_{d=1}^{n-1}(-1)^d\binom{n-2}{d-1}\sin\left(\frac{2dk\pi}{n}\right)= (-1)^{n/2}2^{n-2}\sin^{n-2}\left(\frac{k\pi}{n}\right)\sin(k\pi)$$ So finally, putting the pieces together, $$\frac{n}{\pi}I_{n,k}\left(\frac{\pi}{n}\right)=\frac{n\sum_{r=0}^{n-1}\sum_{s=0}^{n-1}\lambda_{r,s}\sin\left(\frac{2(r-s)k\pi}{n}\right)} {\pi kl\sin^{k-1}\left(\frac{k\pi}{n}\right)\sin^{l-1}\left(\frac{l\pi}{n}\right)}=\frac{n!\sin(k\pi)}{\pi k^{\underline{\smash{n+1}}}}$$ which (for even $n$) we recognise as $\binom{n}{k}$ by the reflection formula for factorials.

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(not an answer)

I've found a way to convert the integral evaluation to a binomial sum identity. Incidentally, this gives the details leading up to the follow-up question here.

I have highlighted (in bold) where we need some potential rigor to make this argument complete.

Let $\zeta=e^{\pmb{i}x/n}, \pmb{i}=\sqrt{-1}$. Equation (1) becomes an integral along an arc on the unit circle $$\frac{n}{\pmb{i}\pi}\int_1^{e^{\pmb{i}\pi/n}}\frac{(\zeta^n-\zeta^{-n})^n} {(\zeta^k-\zeta^{-k})^k(\zeta^{n-k}-\zeta^{-(n-k)})^{n-k}}\frac{d\zeta}{\zeta}=\binom{n}k. \tag3$$ Define the rational complex functions (with a pole at the origin) $f_m(z)=(z^m-z^{-m})^m$ and $$F_{n,k}(z)= \frac{f_n(z)}{f_k(z)f_{n-k}(z)}=\frac{(1-z^{2n})^nz^{-2k(n-k)}}{(1-z^{2k})^k(1-z^{2n-2k})^{n-k}} =\frac{(1-z^{2n})^n}{(1-z^{2k})^k(z^{2k}-z^{2n})^{n-k}}.$$ To verify (3), compute a contour integral around the unit circle $\mathcal{C}$ (oriented positively) $$\frac{n}{\pmb{i}\pi}\int_1^{e^{\pmb{i}\pi/n}}F_{n,k}(z)\frac{dz}z= \pmb{\frac{2n}{2\pmb{i}\pi}\int_1^{e^{\pmb{i}\pi/n}}F_{n,k}(z)\frac{dz}z= \frac1{2\pmb{i}\pi}\int_{\mathcal{C}}F_{n,k}(z)\frac{dz}z}=\text{Res}(F_{n,k}(z);0).$$ This is equivalent to determining the constant term in $F_{n,k}(z)$, which in turn reduces to the identity $$\sum (-1)^a\binom{n}a\binom{k+b-1}b\binom{n-k+c-1}c=\binom{n}k\tag4$$ where the sum runs through $a,b,c\geq0$ such that $(a+c)n+(b-c)k=k(n-k)$.

It remains to prove (4).

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  • $\begingroup$ This looks like the direction that motivated this: mathoverflow.net/questions/253835/… $\endgroup$ – Suvrit Nov 5 '16 at 1:08
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    $\begingroup$ $F_{n,k}(z)$ doesn't have the symmetry you need to extend the integral from the arc to the full unit circle. (In fact, it has singularities on the unit circle.) $\endgroup$ – Alex Selby Nov 5 '16 at 2:00
  • $\begingroup$ True. But, you may avoid singularities by small half-circles "bumps." $\endgroup$ – T. Amdeberhan Nov 5 '16 at 2:04
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    $\begingroup$ You can do that, but you haven't explained (or I haven't understood) why $2n$ times the integral along the arc is equal to the integral around the circle (deformed or otherwise). $\endgroup$ – Alex Selby Nov 5 '16 at 2:15

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