Recently, I am interested in the following integral related to the Chebychev polynomials

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

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

It is easy to see the following result.

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


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

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

For example, the following results have been proved :

$\displaystyle I_{n1} = \pi\enspace$ for odd $\,n\,$ otherwise $\,0\,$

$\displaystyle I_{n2} = \pi n$

$\displaystyle I_{n3} =\frac{\pi}{4}(1+3n^2)\enspace$ for odd $\,n\,$ otherwise $\,0\,$

$\displaystyle I_{n4}= \frac{\pi n}{3}(1+2n^2)$

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 $I_{nm}$ by the method of interpolation.

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

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 at 14:20
  • 1
    $P_{k+1}(2n)$ is obtained from the polynomial expression for $A_k^b(n)$ by substituting a half-integer value of $n$. – Carlo Beenakker Oct 26 at 14:42

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.