Define $\binom{\sin x}{x}$ and try to calculate an approximation of $\int_0^{\pi}\binom{\sin x}{x}dx$ I was playing with Wolfram Alpha online calculator when I wondered how o define $$\binom{\sin x}{x}\tag{1}$$ in the integrand of $$\int_0^{\pi}\binom{\sin x}{x}dx,\tag{2}$$ and how to calculate an approximation of such definite integral (isn't required a very good approximation, only an idea of how to calculate an approximation).
There were my codes
Plot Binom(sin(x),x), from x=0 to pi
int Binom(sin(x),x)dx
that provide us a series expansion of the integral at $x=0$, and finally 
int Binom(sin(x),x)dx, from x=0 to pi

Question. Please define/expain me the integrand $(1)$ and calculate an approximation of $$\int_0^{\pi}\binom{\sin x}{x}dx.$$ 
  Many thanks.

 A: I'll start on the definition of the integrated function. It just requires a careful application of known facts:
$$\binom{\sin x}{x}=\frac{\Gamma(1+\sin x)}{\Gamma(1+x)\Gamma(1+\sin x-x)}$$
To deal with the denominator, we use the known reflection formula:
$$\frac{1}{\Gamma(1-z)\Gamma(z)}=\frac{\sin \pi z}{\pi}$$

$$\frac{1}{\Gamma(1+x)}=\frac{-\sin \pi x}{\pi} \Gamma(-x)$$
$$\frac{1}{\Gamma(1+\sin x-x)}=\frac{-\sin \pi (x-\sin x)}{\pi} \Gamma(x-\sin x)$$
Now we use the main definition of the Gamma function to show how all this works together:
$$\Gamma(z)=\int_0^\infty t^{z-1} e^{-t} dt$$


$$\binom{\sin x}{x}= -\frac{\sin \pi x ~\sin \pi (x-\sin x)}{\pi^2} \times \\ \times \int_0^\infty \int_0^\infty \int_0^\infty ~ \frac{w^{x}}{v^{x}}~ \frac{u^{\sin x}}{w^{\sin x}} \frac{e^{-u-v-w}}{v w}~ du~dv~dw $$

This is quite complicated, but now nothing prevents you from integrating w.r.t. $x$ now right inside this triple integral (unless some problems with convergence are involved).

Unrelated to the previous, but worthy of note. Changing $x \to \pi-x$ we have:
$$\int_0^{\pi}\binom{\sin x}{x}dx=\int_0^{\pi}\binom{\sin x}{\pi-x}dx$$
Another relation follows from the properties of binomial coefficients:
$$\int_0^{\pi}\binom{\sin x}{x}dx=\int_0^{\pi}\binom{1+\sin x}{1+x}dx-\int_0^{\pi}\binom{\sin x}{1+x}dx$$
We can continue this recursion if we want. This actually helps us deal with some of the numerical issues arising in the original integral when $x$ is close to $0$.
