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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.

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

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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.

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

In particular, there is this:

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

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

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