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Wallis integrals satisfy the two-step recurrence, closed forms, and the adjacent-integral squeeze

Statement

For $n \in N$ , put

$I_{n} : = \int_{0}^{π / 2} \sin^{n} t d t .$

Then $I_{0} = π / 2$ , $I_{1} = 1$ , and for every $n \geq 2$ ,

$I_{n} = \frac{n - 1}{n} I_{n - 2} .$

Consequently, for every $m \geq 0$ ,

$I_{2 m} = \frac{π}{2} \prod_{k = 1}^{m} \frac{2 k - 1}{2 k}, I_{2 m + 1} = \prod_{k = 1}^{m} \frac{2 k}{2 k + 1},$

where an empty product is $1$ . For $m \geq 1$ ,

$I_{2 m + 1} \leq I_{2 m} \leq I_{2 m - 1}, 1 \leq \frac{I_{2 m}}{I_{2 m + 1}} \leq \frac{2 m + 1}{2 m},$

and therefore $I_{2 m} / I_{2 m + 1} \to 1$ .

Facts & Assumptions

Given: The functions $t \mapsto \sin^{n} t$ on $[0, π / 2]$ and the integrals $I_{n}$ .

[L1]

Integration by parts gives $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ when the stated derivatives are integrable (If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ ).

[L2]

$(\sin t)^{'} = \cos t$ , $(\cos t)^{'} = - \sin t$ , $\sin 0 = 0$ , $\cos 0 = 1$ , and $\sin^{2} t + \cos^{2} t = 1$ (The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine).

[L3]

Sine is strictly increasing on $[- π / 2, π / 2]$ , has range $[- 1, 1]$ , and satisfies $\sin (π / 2) = 1$ and $\cos (π / 2) = 0$ (Signs, monotonicity intervals, and ranges of sine and cosine, Quarter-turn values and shifts by pi/2 and pi).

[L4]

The integral is linear; the integral of a constant $c$ on $[a, b]$ is $c (b - a)$ ; and if $f \leq g$ pointwise on $[a, b]$ then $\int_{a}^{b} f \leq \int_{a}^{b} g$ (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ , If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L5]

A finite product in a monoid has empty product equal to the identity and satisfies the recursion that adjoins its last factor (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity).

[L6]

A sequence squeezed between two sequences with the same limit has that limit (The squeeze theorem).

[L7]

The constant $π$ is positive (Pi as twice the smallest positive zero of cosine).

Proof

technique · direct

1.1L2L3L4algebra

By [L3] and [L4], $I_{0} = \int_{0}^{π / 2} 1 d t = π / 2$ , while [L2] gives $I_{1} = [- \cos t]_{0}^{π / 2} = 1$ .

1.2givenL1L2L3algebra

Let $n \geq 2$ . Apply [L1] to $u = \sin^{n - 1} t$ and $v^{'} = \sin t$ . The endpoint term $[- \sin^{n - 1} t \cos t]_{0}^{π / 2}$ is $0$ , including the first legal case $n = 2$ , and [L2] gives $I_{n} = (n - 1) \int_{0}^{π / 2} \sin^{n - 2} t \cos^{2} t d t .$

1.3L2L3L4

For $m \geq 1$ , [L2] and [L3] give $0 \leq \sin^{2 m + 1} t \leq \sin^{2 m} t \leq \sin^{2 m - 1} t$ , so the integral bounds in [L4] yield $I_{2 m + 1} \leq I_{2 m} \leq I_{2 m - 1}$ .

2.1step 1.2L2L4algebra

Substitute $\cos^{2} t = 1 - \sin^{2} t$ in step 1.2 and use [L4]: $I_{n} = (n - 1) (I_{n - 2} - I_{n})$ , hence $I_{n} = ((n - 1) / n) I_{n - 2}$ .

3.1step 1.1step 2.1L5algebra

Iterating step 2.1 separately from the base values of step 1.1 gives the displayed even and odd product formulas; when $m = 0$ , [L5] makes them exactly $I_{0} = π / 2$ and $I_{1} = 1$ .

4.1step 2.1step 1.3step 3.1L7algebra

All $I_{j}$ are positive by the product formulas in step 3.1: their base values are positive by [L7], and every displayed factor is positive. Dividing step 1.3 by $I_{2 m + 1}$ and using step 2.1 at $n = 2 m + 1$ gives $1 \leq \frac{I_{2 m}}{I_{2 m + 1}} \leq \frac{I_{2 m - 1}}{I_{2 m + 1}} = \frac{2 m + 1}{2 m} .$

5.1step 4.1L6∎

Both outer sequences in step 4.1 tend to $1$ , so [L6] gives $I_{2 m} / I_{2 m + 1} \to 1$ .

Wallis integrals satisfy the two-step recurrence, closed forms, and the adjacent-integral squeeze

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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