Given I n = ∫ − 1 1 ( 1 − x 2 ) n cos ⁡ ( θ x ) d x ,

Question

Given      I    n    =      ∫          −      1        1    (  1  −      x    2        )    n    cos  ⁡  (  θ  x  )  d  x  ,

Tanner Hamilton · Accepted Answer

Looks like it will be easier to prove it using strong induction on n
Show the base case holds. Assume the induction hypothesis for

k \leq n

, ie,

\begin{matrix} (†) & \forall k \leq n : θ^{2 k + 1} I_{k} = k! (P_{k} (θ) \sin (θ) + Q_{k} (θ) \cos (θ)) \end{matrix}

Now, show the inductive step for

n + 1

, ie,

\begin{aligned} θ^{2 (n + 1) + 1} I_{n + 1} & = θ^{2 n + 1} θ^{2} I_{n + 1} \\ \overset{(1)}{=} θ^{2 n + 1} ((2 n + 2) (2 n + 1) I_{n} - (2 n + 2) (2 n) I_{n - 1}) \\ = (2 n + 2) (2 n + 1) [θ^{2 n + 1} I_{n}] - (2 n + 2) (2 n) θ^{2} [θ^{2 (n - 1) + 1} I_{n - 1}] \\ \overset{(†)}{=} (2 n + 2) (2 n + 1) n! (P_{n} (θ) \sin θ + Q_{n} (θ) \cos θ) - θ^{2} (2 n + 2) (2 n) ((n - 1)! P_{n - 1} (θ) \sin θ + Q_{n - 1} \cos θ) \\ = 2 (2 n + 1) (n + 1)! (P_{n} (θ) \sin θ + Q_{n} (θ) \cos θ) - 4 (n + 1)! (θ^{2} P_{n - 1} (θ) \sin θ + θ^{2} Q_{n - 1} \cos θ) \\ = (n + 1)! (P_{n + 1} (θ) \sin θ + Q_{k + 1} (θ) \cos θ) \end{aligned}

where

P_{n + 1} (θ) := 2 (2 n + 1) P_{n} (θ) - 4 θ^{2} P_{n - 1} (θ)

and

Q_{n + 1} (θ) := 2 (2 n + 1) Q_{n} (θ) - 4 θ^{2} Q_{n - 1} (θ)

and we can finish noting that

P_{n + 1}

and

Q_{n + 1}

are both polynomials with integer coefficients and degree at most

2 n + 2 = 2 (n + 1)

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