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

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

Answer 2

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.

Answer 3

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 4

More systematic way of doing this by some help from Wikipedia's article:

$$T_n^\prime(x)=nU_{n-1}\tag1$$ $$T_n(x)=\frac12(U_{n}(x)-U_{n-2}(x))\tag2$$ where $T_n(x)$ and $U_n(x)$ are Chebyshev polynomials of the first and the second kind respectively. Then, $$\int T_n(x)dx=\frac12\int(U_{n}(x)-U_{n-2}(x))dx\\ =\frac12\int\left(\frac{T^\prime_{n+1}(x)}{n+1}-\frac{T^\prime_{n-1}(x)}{n-1}\right)dx =\frac12\left(\frac{T_{n+1}(x)}{n+1}-\frac{T_{n-1}(x)}{n-1}\right)+C$$

Answer 5

I read your question as asking for a brute force approach to the integral.

Defining $$ I_n^{\pm} := \int_{-1}^1 dx ( x \pm \sqrt{x^2-1})^n $$ it is obvious that $I_0^{\pm}=2$.
Using trigonometric substitutions ($x=\cos(u)$) and $\sin(2u)$ and $\cos(2u)$ sum formulas, you can calculate $$ I_1^{+/-} := \pm i \pi/2 $$

For higher orders, it is easier to calculate derivatives of $x^k\sqrt{x^2-1}$ but you will have quite a bit of housekeeping to do.

If you stay on course or use computer algebra for that, you will find $$ I_{2k+1, k>1}^{\pm} = 0 $$ and $$ I_{2k, k\geq 1}^{\pm} = -\frac{2}{ (k-1)(k+1)} = -\frac{2}{k^2-1} $$

... or alternatively the known result for the Chebyshev polynomials $$ \int_{-1}^1 dx\, T_k(x) = \frac{(-1)^k+1}{k^2-1}, k \neq 1 $$ and $0$ for $k=1$.