Integrating Chebyshev polynomial of the first kind

Question

I'm trying to evaluate the integral of the Chebyshev polynomials of the first kind on the interval $-1 \leq x \leq 1 $ . My idea is to use the closed form $$T_n(x) = \frac{z_1^n + z_2^{-n} }{2}$$ where $z_1 = (x + \sqrt{x^2 - 1})$ and $z_2 = (x - \sqrt{x^2 - 1})$, giving the following integral: $$ \int _{-1}^{1}\!1/2\, \left( x+\sqrt {-1+{x}^{2}} \right) ^{n}+1/2\, \left( x-\sqrt {-1+{x}^{2}} \right) ^{n}{}{dx} $$ I'm stuck at integrating $z_1$ and $z_2$. I tried integrating by parts $n$ times, but i'm looking for a general formula. My calculus is pretty rusty so i'm not sure if this is the way to go. Any tips? Thanks a lot.

Answer 1

Wikipedia has a nice article on the Chebychev polynomials: http://en.wikipedia.org/wiki/Chebyshev_polynomials.

In particular, there is this:

$\int T_n(x) dx = \frac{n T_{n+1}(x)}{n^2-1}-\frac{x T_n(x)}{n-1}$.

Answer 2

It is actually not that hard. You can derive a lot of relations on the wiki page yourself by substituting $\cos\theta$ for $x$ and use the defining relation of Chebyshev polynomials:

$$T_n(\cos\theta) = \cos( n\theta)$$

For example, one have: $$\begin{align}\int T_n(x) dx = & \int T_n(\cos\theta) d\cos \theta\\ = & -\int \cos(n\theta)\sin\theta d\theta\\ = & -\frac12 \int \left(\sin((n+1)\theta) - \sin((n-1)\theta)\right)d\theta\\ = & \frac12 \left(\frac{\cos((n+1)\theta)}{n+1} - \frac{\cos((n-1)\theta)}{n-1}\right) + \text{const.}\\ = & \frac12 \left(\frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)}{n-1}\right) + \text{const.} \end{align} $$