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Exact integrals of Chebyshev polynomials

Statement

For every integer $k \geq 0$ , let $μ_{k} : = \int_{- 1}^{1} T_{k} (x) d x$ , where $T_{k}$ is the first-kind Chebyshev polynomial. Then $μ_{k} = {\begin{cases} 2, & k = 0, \\ 0, & k odd, \\ \frac{2}{1 - k^{2}}, & k \geq 2 even . \end{cases}$ Equivalently, $μ_{2 r} = 2 / (1 - 4 r^{2})$ for every integer $r \geq 0$ .

Facts & Assumptions

Given: An integer $k \geq 0$ and the polynomial $T_{k}$ of Chebyshev polynomials of the first and second kinds by their three-term recurrences.

[A1]

$k$ is a nonnegative integer.

[F1]

The Chebyshev recurrence agrees with cosine multiple angles: $T_{k} (\cos θ) = \cos (k θ)$ for every real $θ$ .

[F2]

Monotone change of variable for Riemann-integrable functions: a monotone differentiable bijection with integrable derivative changes a Riemann integral by the absolute derivative.

[F3]

Signs, monotonicity intervals, and ranges of sine and cosine and The derivatives of sine and cosine are cosine and minus sine: $\cos$ decreases on $[0, π]$ , $\sin θ \geq 0$ there, and $(\cos θ)^{'} = - \sin θ$ .

[F4]

Chebyshev polynomials of the first and second kinds by their three-term recurrences and A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: $T_{k}$ is a continuous polynomial and is Riemann-integrable on $[- 1, 1]$ .

[F5]

The addition formulas for sine and cosine: $\sin u \cos v = \frac{1}{2} (\sin (u + v) + \sin (u - v))$ .

[F6]

The derivatives of sine and cosine are cosine and minus sine and The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : an antiderivative of $\sin (m θ)$ for nonzero integer $m$ is $- \cos (m θ) / m$ .

[F7]

Quarter-turn values and shifts by pi/2 and pi and The principle of mathematical induction: induction using the shift by $π$ gives $\cos (m π) = (- 1)^{m}$ for natural $m \geq 0$ .

Proof

1.1givenA1F1F2F3F4algebra

Fix the integer $k$ in [A1]. The map $θ \mapsto \cos θ$ is a decreasing differentiable bijection from $[0, π]$ to $[- 1, 1]$ with derivative $- \sin θ$ ; [F2]–[F4] verify the change-of-variable hypotheses. Therefore [F1]–[F4] give $μ_{k} = \int_{0}^{π} T_{k} (\cos θ) \sin θ d θ = \int_{0}^{π} \cos (k θ) \sin θ d θ .$

2.1givenA1F5F6F7step 1.1algebra

For the nonnegative integer $k$ in [A1], the case $k = 0$ gives $μ_{0} = \int_{0}^{π} \sin θ d θ = 2$ by step 1.1. For $k \geq 1$ , [F5] gives $\sin θ \cos (k θ) = \frac{1}{2} (\sin ((k + 1) θ) - \sin ((k - 1) θ))$ . [F6]–[F7] evaluate each nonzero-frequency sine integral; the $m = 0$ term is identically zero.

3.1givenF6F7step 2.1algebra∎

If $k$ is odd, both positive integers among $k - 1$ and $k + 1$ are even, so [F6]–[F7] and step 2.1 give $μ_{k} = 0$ . If $k \geq 2$ is even, both $k - 1$ and $k + 1$ are odd, so they give $μ_{k} = 1 / (k + 1) - 1 / (k - 1) = 2 / (1 - k^{2})$ . Together with the $k = 0$ case, this proves the formula.

Depends on

Used by

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Sources

L. Ridgway Scott, Numerical Analysis, 2nd ed., §13.4.1 (standard reference, not scraped)
Lloyd N. Trefethen, Is Gauss Quadrature Better than Clenshaw–Curtis?, §2 (standard reference, not scraped)