## The Wallis Product

I had seen the Wallis product, but I did not know a derivation of it until a few weeks ago, when I discovered this gem when reading the wonderful book Sources in the Development of Mathematics by Ranjan Roy. I’m storing this example to use next time I teach calculus, as it requires nothing more than integration by parts.

Like in the case of many other such things in mathematics, I don’t know what the Wallis product actually is; I can only discover it by going through a derivation. So, I won’t give away the answer until we get there. But the goal is to write down an infinite product for $\pi$.

In order to do this, we evaluate the integral $I_n = \int_0^\pi \sin^n x\, dx$. The first few are easy:

$I_0 = \int_0^\pi dx=\pi$

$I_1 = \int_0^\pi \sin x\, dx = 2$.

I can keep working out more of these, but now it’s time to tackle the general case, using integration by parts. We assume $n\ge 2$, and we compute $\int_0^\pi u\, dv$, where $u=\sin^{n-1} x$ and $v=\sin x\, dx$, so that

$I_n = \int_0^\pi \sin^n x\, dx = \left[-\sin^{n-1} x\cos x\right]_0^\pi +\int_0^\pi (n-1)\sin^{n-2} x\cos^2 x\, dx$ $= \int_0^\pi (n-1)\sin^{n-2}x(1-\sin^2 x)\, dx = (n-1)(I_{n-2}-I_n)$.

Solving for $I_n$, we get $I_n = \frac{n-1}{n}I_{n-2}$. Now it’s easy to compute more of the $I_n$‘s:

$I_2 = \pi\times\frac{1}{2}$

$I_3 = 2\times\frac{2}{3}$

$I_4 = \pi\times\frac{1}{2}\times\frac{3}{4}$

$I_5 = 2\times\frac{2}{3}\times\frac{4}{5}$,

and in general

$I_{2n} = \pi\times\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{2n-1}{2n}$

$I_{2n+1} = 2\times\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{2n}{2n+1}$.

Now, note that for any $n$, $I_{n-1}>I_n>I_{n+1}$, since $\sin^{n-1} x\ge \sin^n x\ge\sin^{n+1} x$ whenever $0\le x\le\pi$. Furthermore, since $\frac{I_{n+1}}{I_{n-1}}=\frac{n}{n+1}$, this means that eventually $I_{2n-1}$ and $I_{2n}$ get very close together. So

$\pi\times\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{2n-1}{2n}\approx 2\times\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{2n}{2n+1}$,

or

$\frac{\pi}{4}\approx \frac{2}{3}\times\frac{4}{3}\times\frac{4}{5}\times\frac{6}{5}\times\frac{6}{7}\times\cdots\times\frac{2n}{2n+1}$.

(If we wanted, we could be more precise, and give two inequalities for $\frac{\pi}{4}$, but it doesn’t add much value, in my opinion, to do that here.)

Taking the limit, we get

$\frac{\pi}{4}= \frac{2}{3}\times\frac{4}{3}\times\frac{4}{5}\times\frac{6}{5}\times\frac{6}{7}\cdots$.

This is the Wallis product.