KindiakMath

The Wallis Product

Problem 1. By considering the integral \displaystyle I(n) := \int_0^{\pi} \sin^n(x)\, \mathrm dx, prove the Wallis product

\displaystyle \frac 21 \cdot \frac 23 \cdot \frac 43 \cdot \frac 45 \cdot \cdots := \lim_{n \to \infty} \prod_{k=1}^n \frac{2n \cdot 2n}{(2n-1)(2n+1)} = \frac{\pi}{2}.

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Solution. We leave it as an exercise to check that I(0) = \pi and I(1) = 2. These observations motivate an expression of I(n+2) in terms of I(n). To that end, we integrate by parts to obtain

\begin{aligned} & \int \sin^{n+2}(x)\, \mathrm dx = \int \sin(x) \sin^{n+1}(x)\, \mathrm dx \\ &= \underbrace{-{\cos(x)}}_{\mathrm I} \underbrace{\sin^{n+1}(x)}_{\mathrm S} - \int \underbrace{-{\cos(x)}}_{\mathrm I} \underbrace{(n+1) \sin^n(x) \cos(x) }_{\mathrm D}\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \int \cos^2(x) \sin^n(x)\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \int (1-\sin^2(x)) \sin^n(x)\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \left( \int \sin^n(x)\, \mathrm dx - \int \sin^{n+2}(x)\, \mathrm dx\right). \end{aligned}

By algebruh,

\displaystyle (n+2) \int \sin^{n+2}(x)\, \mathrm dx = -{\cos(x)} \sin^{n+1}(x) + (n+1) \int \sin^n(x)\, \mathrm dx.

Plugging in the limits of [0, \pi],

\displaystyle (n+2) \int_0^{\pi} \sin^{n+2}(x)\, \mathrm dx = (n+1) \int_0^{\pi} \sin^n(x)\, \mathrm dx.

Hence,

\displaystyle \frac{I(n+2)}{I(n)} = \frac{n+1}{n+2}.

In particular,

\displaystyle I(2k+2) = \frac{2k+1}{2k+2} \cdot I(2k) \quad\Rightarrow \quad I(2n+2) = \pi \cdot \prod_{k=1}^n \frac{2k-1}{2k}

Similarly,

\displaystyle I(2n+1) = 2 \cdot \prod_{k=1}^n \frac{2k}{2k+1}.

Dividing the terms appropriately,

\displaystyle \prod_{k=1}^n \frac{2k \cdot 2k}{(2k-1) (2k+1)} = \frac{\pi}{2} \cdot \frac{I(2n+1)}{I(2n+2)}.

It suffices to prove that I(2n+1)/I(2n+2) \to 1. Since 0 \leq \sin(x) \leq 1 for x \in [0, \pi],

0 \leq \sin^{n+1}(x) \leq \sin^n(x) \leq \sin^{n-1}(x) \leq 1.

By the monotonicity of integration, I(n+1) \leq I(n) so that

\displaystyle 1 \leq \frac{I(2n+1)}{I(2n+2)} \leq \frac{I(2n)}{I(2n+2)} = \frac{2n+2}{2n+1} \to 1.

By the squeeze theorem, I(2n+1)/I(2n+2) \to 1, as required.

—Joel Kindiak, 21 Apr 25, 2129H

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