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

  • $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • $\begingroup$ 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

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.