Chebyshev polynomials of the second kind $V_n(x)$ can be defined as $$V_n(x)=\frac{\sin(n+1)\theta}{\sin\theta}, x=2\cos\theta$$ or through the recurrence relation $$V_{n+1}=xV_n-V_{n-1}, V_0=1, V_1=x.$$ First few low-order Chebyshev polynomials of the second kind are as follows: $$V_0=1, V_1=x,V_2=x^2-1,V_3=x^3-2x,$$ $$V_4=x^4-3x^2+1,V_5=x^5-4x^3+3x.$$

I want to know how to calculate the following integral relate to Chebyshev polynomials: $$\int_0^\pi (\frac{\sin nx}{\sin x})^m dx$$ where $n,m\in \mathbb{Z}^+$.

It is easy to see the following result:

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

$$ \int_0^\pi (\frac {\sin nx}{\sin x})^{m} dx=0.$$

I conjectured that the result is a polynomial $P(n)$ of order $m−1$. But I have no idea about the proof.

I prefer to know the other two cases beside the above special case. Thank you.


$$I_{nm}=\frac{1}{\pi}\int_0^\pi \left(\frac{\sin nx}{\sin x}\right)^m dx$$

As stated in the OP, $I_{nm}=0$ for $n$ even and $m$ odd. Otherwise, the nonzero $I_{nm}$ is the central multinomial coefficient $c_{nm}$, the largest coefficient of $(1+x+x^2\cdots x^{n-1})^m$.

A few examples: $$I_{44}=44\;\; \text{and}\;\; (1 + x + x^2 + x^3)^4=1 + \cdots + 40 x^5 + \mathbf{44}\, x^6 + 40 x^7 + \cdots + x^{12}$$ $$I_{34}=19\;\; \text{and}\;\; (1 + x + x^2 )^4=1 + \cdots + 16 x^3 + \mathbf{19}\, x^4 + 16 x^5 + \cdots + x^{8}$$ $$I_{33}=7\;\; \text{and}\;\; (1 + x + x^2 )^3=1 + 3 x + 6 x^2 + \mathbf{7}\, x^3 + 6 x^4 + 3 x^5 + x^6$$

For small $n$ the central multinomial coefficients are listed on OEIS, for example, $I_{6m}$ is given for $m=2,4,6,8,\ldots$ by A063419:

$$ 6, 146, 4332, 135954, 4395456, 144840476, 4836766584, 163112472594, 5542414273884, 189456975899496, 6507792553644256, \ldots$$

An efficient method to calculate these coefficients is described in On a link between Dirichlet kernels and central multinomial coefficients.

Update 1, following the conjecture added in a comment, that $I_{nm}=P_m(n)$ is a polynomial in $n$ of degree $m-1$; I have no proof, but for the record, here is the evidence for small $m$:

$$P_1(n)=1$$ $$P_2(n)=n$$ $$P_{3}(n)=\frac{1}{4} \left(3 n^2+1\right)$$ $$P_4(n)=\frac{1}{3} n\left(2 n^2+1\right)$$ $$P_{5}(n)=\frac{1}{192} \left(115 n^4+50 n^2+27\right)$$ $$P_6(n)=\frac{1}{20} n \left(11 n^4+5 n^2+4\right)$$ $$P_7(n)=\frac{7 \left(841 n^6+385 n^4+259n^2\right)+1125}{11520}$$ $$P_8(n)=\frac{1}{315} n \left(151 n^6+70 n^4+49 n^2+45\right)$$ $$P_9(n)=\frac{259723 n^8+121380 n^6+84882 n^4+64580 n^2+42875}{573440}$$ $$P_{10}(n)=\frac{1}{36288}n \left(15619 n^8+7350 n^6+5187 n^4+4100 n^2+4032\right)$$

For fixed even $m$, these formulas represent closed form expressions for the central multinomial coefficients $c_{nm}$. On OEIS I find that such polynomial expressions are noted by R.H. Hardin (see A071816 and A005900), but without a reference to a general proof.

For fixed odd $m$, only the $c_{nm}$'s with odd $n$ follow. I have checked that there is no degree $m-1$ polynomial expressions for the full set of central multinomial coefficients with odd $m$.

Update 2 I have now located a proof in the literature, see https://mathoverflow.net/a/313854/11260

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  • $\begingroup$ @beenaker how to prove $I_{nm}$ is the largest coefficient of $(1+x+x^2\cdots x^{n-1})^m$. $\endgroup$ – Jacob.Z.Lee Oct 18 '18 at 12:24
  • $\begingroup$ @Beenakker Thanks a lot. Is there any formula on $I_{nm}$? $\endgroup$ – Jacob.Z.Lee Oct 18 '18 at 12:48
  • $\begingroup$ For the proof, just write $\sin(nx)/\sin x=(z^n-z^{-n})/(z-z^{-1)}=z^{n-1}+z^{n-3}+\dots+z^{1-n}=z^{1-n}f(z^2)$ for $z=e^{ix}$ and $f(w)=1+w+\dots+w^{n-1}$, expand the brackets and use the fact that $\int z^{2t}=0$ for $t\ne 0$. $\endgroup$ – Fedor Petrov Oct 18 '18 at 12:56
  • $\begingroup$ @Petrov Nice to see you again. Thanks for the proof. $\endgroup$ – Jacob.Z.Lee Oct 18 '18 at 13:48
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    $\begingroup$ @Beenakker I conjecured that the $I_{nm}$ is a polynomial $P(n)$ of order $m−1$ . $\endgroup$ – Jacob.Z.Lee Oct 20 '18 at 11:49

We use complex analysis, of course, standard way to convert an integral of a periodic function, integrated on a period. In our case: $$ \begin{aligned} J(n,m) &= \int_0^\pi\left(\frac {\sin nt}{\sin t}\right)^m\; dt \\ &= \frac 12 \int_0^{2\pi} \left(\frac {e^{int}-e^{-int}}{e^{it}-e^{-it}}\right)^m \cdot\frac 1{e^{it}}\; ie^{it}\; dt \\ &= \frac 1{2i} \int_\gamma \left(\frac {z^{n}-z^{-n}}{z-z^{-1}}\right)^m\cdot\frac 1z\; \; dz \\ &= 2\pi i\cdot\frac 1{2i} \cdot\text{Residue of } \frac 1z \underbrace{\left(z^{n-1}+\dots z^{n-3}+\dots z^{-(n-3)}+z^{-(n-1)}\right)^m}_{P(n,m)}\ . \end{aligned} $$ Now we have to isolate the residue of the above integral on $\gamma$, the unit circle. We can simplify and need the coefficient of degree zero in $$ P(n,m)=\left(z^{n-1}+\dots z^{n-3}+\dots z^{-(n-3)}+z^{-(n-1)}\right)^m\ . $$ From here, as in the answer of Carlo Beenakker pointing to an OEIS.

Computer check in a special case.

sage: integral( ( sin(7*x) / sin(x) )^9, x, 0, pi )
sage: var('z');
sage: P = (( z^6 + z^4 + z^2 + 1 + z^-2 + z^-4 + z^-6 )^9) / z
sage: P.residue(z)
sage: # also
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  • $\begingroup$ Should it be the following ?$$ \begin{aligned} J(n,m) &= \int_0^\pi\left(\frac {\sin nt}{\sin t}\right)^m\; dt \\ &= \frac 1{2i} \int_0^{2\pi} \left(\frac {e^{int}-e^{-int}}{e^{it}-e^{-it}}\right)^m \cdot\frac 1{e^{it}}\; ie^{it}\; dt \\ &= \frac 1{2i} \int_\gamma \left(\frac {z^{n}-z^{-n}}{z-z^{-1}}\right)^m\cdot\frac 1z\; \; dz \\ &= 2\pi i\cdot\frac 1{2i} \cdot\text{Residue of } \frac 1z \underbrace{\left(z^{n-1}+\dots z^{n-3}+\dots z^{-(n-3)}+z^{-(n-1)}\right)^m}_{P(n,m)}\ . \end{aligned} $$ $\endgroup$ – Jacob.Z.Lee Oct 18 '18 at 13:45
  • $\begingroup$ Yes, this is the value of the integral. For instance in the computed example we need the residue for $$\frac 1z( z^6 + z^4 + z^2 + 1 + z^{-2} + z^{-4} + z^{-6} )^9\ ,$$it is the same as the coefficient in $z^0$ in $$( z^6 + z^4 + z^2 + 1 + z^{-2} + z^{-4} + z^{-6} )^9\ ,$$ i.e. the coefficient in $z^0$ = mid coefficient in $$( z^3 + z^2 + z + 1 + z^{-1} + z^{-2} + z^{-3} )^9\ .$$ Same as taking the mid coefficient in $$( z^6 + z^5 + z^4 + z^3 + z^2 + z+1)^9\ .$$ The $2\pi\; i \ /\ (2i)$ introduces a $\pi$ times this mid coefficient. $\endgroup$ – dan_fulea Oct 18 '18 at 16:41

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