I am interested in the following integral related to the Chebyshev polynomials

$$I_{n,m}:= \int_0^\pi \left(\frac {\sin nx}{\sin x}\right)^{m} dx,$$

where $n,m\in \mathbb{Z}^+.$

It is easy to see the following result.

For an even number $n \in \mathbb{Z}^+$ and an odd number $m \in \mathbb{N}$, we have

$$I_{n,m}=0.$$

The nonzero $I_{n,m}$ are the so called central multinomial coefficients , the largest coefficient of $(1+x+x^2\cdots +x^{n-1})^m$.

I conjecture that if $I_{n,m}\neq 0$, then the $I_{n,m}$ is a polynomial $P(n)$ of degree $m-1$.

For example, the following results have been proved :

$\displaystyle I_{n,1} = \pi\enspace$ for odd $\,n\,$, otherwise $\,0\,$;

$\displaystyle I_{n,2} = \pi n$;

$\displaystyle I_{n,3} =\frac{\pi}{4}(1+3n^2)\enspace$ for odd $\,n\,$, otherwise $\,0\,$;

$\displaystyle I_{n,4}= \frac{\pi n}{3}(1+2n^2)$.

But in the general case, I have no idea about the proof of the conjecture. If this conjecture is true, then for $\forall n,m\in \mathbb{Z}^+ $, we can determine $I_{n,m}$ by the method of interpolation.

If someone can give some suggestion or opinion on the proof of the conjecture, I will appreciate it.