# A Conjecture about the integral related to Chebyshev polynomial

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.

• For the case $m$ even, see OEIS sequence A163269 – Robert Israel Oct 26 '18 at 2:15
• here is a list of some of these generating polynomials of central multinomial coefficients $c_{nm}$: mathoverflow.net/a/313137/11260 ; for $m$ odd there is no polynomial that generates these for all $n$ (only for odd $n$). – Carlo Beenakker Oct 26 '18 at 8:01
• @beenakker, Yes, they are good positive examples for my conjecture. – Jacob.Z.Lee Oct 26 '18 at 13:31

This is proven in arXiv:1002.3844, see top of page 12. The quantity $$A_k^b(n)$$ in that paper is a polynomial in $$n$$ of degree $$\leq k$$ and it is related to the integral $$I_{nm}=\pi P_m(n)$$ in the OP by $$A_{k}^b(n)=P_{k+1}(2n+1)$$.
The polynomial can be expressed as a terminating hypergeometric series, see equation 83. I also note that, according to remark 3 on page 13, the polynomial $$P_m(n)$$ is even for $$m$$ odd and odd for $$m$$ even.
• @beenakker ,Thank you very much! Good reference！ What about $P_{k+1}(2n)?$ Is it related to $A^b_k(n)$ or something too? – Jacob.Z.Lee Oct 26 '18 at 14:20
• $P_{k+1}(2n)$ is obtained from the polynomial expression for $A_k^b(n)$ by substituting a half-integer value of $n$. – Carlo Beenakker Oct 26 '18 at 14:42