# 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 a lot.

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 '19 at 9:35
• Yes. My typo. I'll fix it. – marty cohen May 2 '19 at 11:14