integral of a "sin-omial" coefficients=binomial

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.

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

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.