# Integrating Chebyshev polynomial of the first kind

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 alot.

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}

An alternative easy way to do this would be to convert everything to the complex exponential. Given

$$T_n(\cos\theta) = \cos(n\theta) = \frac{1}{2}\left( e^{in\theta} + e^{-in\theta} \right)$$

and since $\frac{d\cos\theta}{d\theta} = \frac{i}{2}\left( e^{i\theta} - e^{-i\theta} \right)$ we have

\begin{align}\int T_n(x)dx & = \int T_n(\cos\theta)d\cos\theta \\ & = \int \cos(n\theta)d\cos\theta \\ &= \frac{i}{4}\int \left( e^{in\theta} + e^{-in\theta} \right)\left( e^{i\theta} - e^{-i\theta} \right)d\theta \end{align}

which is straightforward to integrate. At the end of the day you will get

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

after converting the complex exponential back to the trigonometric form and using the definition of the $n$-th Chebyshev polynomial as given above.

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

In particular, there is this:

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

• Best explanation in this video youtube.com/watch?v=8EYEgxhsGFA in french – Zbigniew Mar 1 '15 at 19:36
• Isn't there a minus instead of a plus between the two polynomial terms? – Mayou36 May 2 at 9:35
• Yes. My typo. I'll fix it. – marty cohen May 2 at 11:14