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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.

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    $\begingroup$ For the case $m$ even, see OEIS sequence A163269 $\endgroup$ – Robert Israel Oct 26 '18 at 2:15
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    $\begingroup$ 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 $n$). $\endgroup$ – Carlo Beenakker Oct 26 '18 at 8:01
  • $\begingroup$ @beenakker, Yes, they are good positive examples for my conjecture. $\endgroup$ – Jacob.Z.Lee Oct 26 '18 at 13:31
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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.

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  • $\begingroup$ @beenakker ,Thank you very much! Good reference! What about $P_{k+1}(2n)?$ Is it related to $A^b_k(n) $ or something too? $\endgroup$ – Jacob.Z.Lee Oct 26 '18 at 14:20
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    $\begingroup$ $P_{k+1}(2n)$ is obtained from the polynomial expression for $A_k^b(n)$ by substituting a half-integer value of $n$. $\endgroup$ – Carlo Beenakker Oct 26 '18 at 14:42

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