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? $\endgroup$ Nov 2 '16 at 0:50
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    $\begingroup$ I do not have a proof. It comes from some experimentation and curiosity. $\endgroup$ Nov 2 '16 at 1:01
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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 $\endgroup$ Nov 3 '16 at 12:31
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    $\begingroup$ Formula $(2)$ is due to C. L. Mallows, A formula for expected values, Amer. Math. Monthly 87 (1980), 584. Also see D. Stanton, R. Evans, M.E.H. Ismail, Coefficients in expansions of certain rational functions, Canad. J. Math. 34 (1982) 1011–1024. $\endgroup$
    – Nemo
    Nov 30 '17 at 14:01

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. $\endgroup$ 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. $\endgroup$ Nov 24 '16 at 22:18

(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 $$\left(Y\frac{\sin X}{\sin \alpha X}\right)^\alpha \left(Z\frac{\sin X}{\sin (1-\alpha) X}\right)^{1-\alpha}\in \text{EXP}.$$

  • $\begingroup$ Interesting indeed. $\endgroup$ Nov 3 '16 at 12:34
  • $\begingroup$ Need $n$ to be an integer? $\endgroup$ 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. $\endgroup$ 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

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. $\endgroup$ 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

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.


(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).

  • $\begingroup$ 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." $\endgroup$ 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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