Wallis' formula expresses $\pi$ as an infinite product: $$\frac{\pi}{2} = \prod_{n = 1}^{\infty} \frac{4n^{2}}{4n^{2} - 1} = \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5} \cdots$$

It may be derived by taking the limit as $n \to \infty$ of the ratio of the following two integrals. $$\int_{0}^{\frac{\pi}{2}} \sin^{2n}x dx = \frac{1 \cdot 3 \cdots (2n - 1)}{2 \cdot 4 \cdots 2n} \frac{\pi}{2}$$

$$\int_{0}^{\frac{\pi}{2}} \sin^{2n + 1}x dx = \frac{2 \cdot 4 \cdots 2n}{3 \cdot 5 \cdots (2n + 1)}$$