Consider the function defined by the integral:
$$ f_n(x) = -\int_0^\pi \cos(n\theta)\log(|x-\cos\theta|)\,d\theta $$
I want to show that $f_n$ is a multiple of the nth Chebyshev polynomial on the interval $[-1, 1]$, where the Chebyshev polynomials are defined by
$$ T_n(x) = \cos\left(n\cos^{-1}(x)\right) $$
In other words, for each $n$, there is some $C$ such that $f_n(x) = C\cdot T_n(x)$ for $-1\leq x\leq 1$. Unfortunately, the integral looks pretty complicated and I'm not sure how to solve it. On the other hand, I'm reasonably sure that the claim is true from numerical evidence. Also, can we get an explicit formula for $C$ in terms of $n$?
Best Answer
Although your problem is classic, in order to prove the identity, we need to introduce Chebyshev's polynomials of the second kind: $$\begin{cases} U_0 (x) = 1\\ U_n (x) = \frac{\sin \left[(n+1)\arccos x\right]}{\sin (\arccos x)}, n\geq 1. \end{cases}$$ Now the below integral must hold for all $n$: $$I_n:=\int_{-1}^{1}\frac{\sqrt{1-t^2}\cdot U_{n-1}(t)}{x-t}~\mathrm{d}t = \pi \cdot T_n (x)$$ To prove the integral stated above by using the recursion formula of $T_n (x)$ and $U_n (x)$ for induction. Hence it is vital to prove it's true for $n=1,2$. Initially, we have: $$\int_{0}^{\pi} \frac{1}{\cos x - \cos v}dx = \frac{1}{2}\int_{-\pi}^{\pi}\frac{1}{\cos x - \cos v}dx = \frac{1}{2}\int_{-\pi}^{\pi} \frac{e^{ix}}{(e^{ix} - e^{iv})(e^{ix}-e^{-iv})}~dx$$$$=\int_{\vert z \vert= 1}\frac{-i}{(z - e^{iv})(z-e^{-iv})}~dz = -\frac{i}{e^{iv}-e^{-iv}}\left(\int_{\vert z\vert=1}\frac{1}{z-e^{iv}}dz-\int_{\vert z\vert=1}\frac{1}{z-e^{-iv}}dz\right)=0$$ For $n=0$: $$I_1= \int_{-1}^{1}\frac{\sqrt{1-t^2}}{\cos \theta-t}~\mathrm{d}t=\int_{-\pi}^{\pi}\frac{\sin^2 k}{2(\cos \theta-\cos k)}~\mathrm{d}k=\int_{-\pi}^{\pi}\frac{1-\cos 2k}{4(\cos \theta-\cos k)}~\mathrm{d}k$$$$=\int_{-\pi}^{\pi}\frac{\cos 2k}{4(\cos k-\cos \theta )}~\mathrm{d}k$$ The last expression of $I_1$ is worth trying to train complex analysis method. Then we obtain: $I_1 = \pi x = \pi\cdot T_1 (x)$. The second integral follows easily since: $$I_2 = \int_{-1}^{1} \frac{2t\cdot\sqrt{1-t^2}}{x-t}~\mathrm{d}t= \int_{-1}^{1} \frac{2x\cdot\sqrt{1-t^2}}{x-t}~\mathrm{d}t -\int_{-1}^{1} 2\cdot\sqrt{1-t^2}~\mathrm{d}t$$ $$=2x\cdot I_1 - \pi = \pi\cdot(2x^2-1)=\pi T_1(x)$$
At this time, we will assume $I_k = \pi \cdot T_k(x)$ from $k=1,...,n$. The work to do now is to show the equality holds for $k=n+1$. $$I_{n+1} = \int_{-1}^{1}\frac{\sqrt{1-t^2}\cdot U_{n}(t)}{x-t}~\mathrm{d}t= \int_{-1}^{1}\frac{2t\sqrt{1-t^2}\cdot U_{n-1}(t)}{x-t}~\mathrm{d}t- \int_{-1}^{1}\frac{\sqrt{1-t^2}\cdot U_{n-2}(t)}{x-t}~\mathrm{d}t$$
$$= \int_{-1}^{1}\frac{2x\sqrt{1-t^2}\cdot U_{n-1}(t)}{x-t}~\mathrm{d}t - 2\int_{-1}^{1} \sqrt{1-t^2}U_{n-1}(t)dt - \pi \cdot T_{n-1} (x)= \pi \cdot 2x\cdot T_n (x) - \pi \cdot T_{n-1} (x)$$$$\Rightarrow I_{n+1}= \pi \cdot T_{n+1} (x)$$
In the calculation above, we have: $$\int_{-1}^{1} U_{n-1} (x)\sqrt{1-x^2}\mathrm{d}x = \int_{0}^{\pi} \sin t\cdot \sin nt dt = \frac{1}{2}\left(\int_{0}^{\pi} \cos (n-1)t \mathrm{d}t - \int_{0}^{\pi} \cos (n+1)t\mathrm{d}t \right)=0 $$
For the integral $I_n$, substitute $t = \cos u$ and integrate by parts: $$I_n = \int_{0}^{\pi} \frac{\sin^2 u\cdot U_{n-1} (\cos u)}{x-\cos u} ~ \mathrm{d}u= \int_{0}^{\pi} \frac{\sin t\cdot \sin nt}{x-\cos t}~\mathrm{d}t$$$$=\ln \vert x - \cos u\vert. \sin nt\Bigg|_0^{\pi}-n\cdot\int_{0}^{\pi}\cos nu.\ln \vert x - \cos u\vert\mathrm{d}u=-n\cdot f_n(x) $$ $$\Rightarrow f_n (x) = \frac{\pi}{n}\cdot T_n(x)$$