how to calculate the special integral related to Chebyshev polynomials?

Question

Chebyshev polynomials of the second kind, $V_n(x)$, can be defined as $$V_n(x)=\frac{\sin(n+1)\theta}{\sin\theta}, \hspace{5mm} x=2\cos\theta,$$ 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: $$\int_0^\pi \left(\frac{\sin nx}{\sin x}\right)^m dx$$ where $n,m\in \mathbb{Z}^+$.

It is easy to see the following result:

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

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

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

Answer 1

Comments.

Trivial cases: For $n=0, m>0$ the result is $0$ . For $m=0$ or $n=1$ the result is $\pi$ . For $n<0, m>0$ it's the result for $n>0, m>0$ multiplicated with $(-1)^m$ . For $m<0, |n|>1$ the integral doesn't work.

Let $\,n,m\in\mathbb{N}\,$ , $\,k\in\mathbb{N}_0\,$ and $\,a_{m,k}\in\mathbb{Q}^+\,$ . Then we set

$$\int\limits_0^\pi \left(\frac{\sin(nx)}{\sin x}\right)^m dx = \pi n^{\frac{1+(-1)^m}{2}} \sum\limits_{k=0}^{ \lfloor \frac{m-1}{2}\rfloor } a_{m,k} n^{2k} $$

where $\,a_{m,k}\,$ can be evaluated with a linear equation system,

if one knows $\enspace\displaystyle \int\limits_0^\pi \left(\frac{\sin(nx)}{\sin x}\right)^m dx\enspace$ for $\enspace\displaystyle n\in\{1,2,…,{ \lfloor \frac{m+1}{2}\rfloor }\}\enspace$ .

Examples:

$\displaystyle \int\limits_0^\pi \left(\frac{\sin(nx)}{\sin x}\right)^1 dx = \pi\enspace$ for odd $\,n\,$ otherwise $\,0\,$

$\displaystyle \int\limits_0^\pi \left(\frac{\sin(nx)}{\sin x}\right)^2 dx = \pi n$

$\displaystyle \int\limits_0^\pi \left(\frac{\sin(nx)}{\sin x}\right)^3 dx =\frac{\pi}{4}(1+3n^2)\enspace$ for odd $\,n\,$ otherwise $\,0\,$

$\displaystyle \int\limits_0^\pi \left(\frac{\sin(nx)}{\sin x}\right)^4 dx = \frac{\pi n}{3}(1+2n^2)$

$\displaystyle \int\limits_0^\pi \left(\frac{\sin(nx)}{\sin x}\right)^5 dx =\frac{\pi}{192}(27+50n^2+115n^4)\enspace$ for odd $\,n\,$ otherwise $\,0\,$

$\displaystyle \int\limits_0^\pi \left(\frac{\sin(nx)}{\sin x}\right)^6 dx = \frac{\pi n}{20}(4+5n^2+11n^4)$

Hint:

For a proof and to get the coefficients it makes sense to use

$$\frac{\sin(nx)}{\sin x} = \sum\limits_{k=0}^{n-1} e^{ix(2k+1-n)}$$

which makes it necessary to use multinomials.

$\frac{\sin(nx)}{\sin x}$ is real therefore the imaginary parts of $\left(\sum\limits_{k=0}^{n-1} e^{ix(2k+1-n)}\right)^m$ can be ignored.

Links:

https://en.wikipedia.org/wiki/Euler%27s_formula

https://en.wikipedia.org/wiki/Multinomial_Theorem

(If the last link doesn’t work please use https://en.wikipedia.org/wiki/Binomial_coefficient#Generalization_and_connection_to_the_binomial_series instead, look for “Generalization to multinomials” .)