Exercise 1521Calculus › Single-variable calculus › Integration › Integration techniquesL1Added 2 Nov 2026

Wallis integrals

#Wallis integrals#recurrence#asymptotics#pi

Statement

Let $W_n=\displaystyle\int_0^{\pi/2}\sin^nx\,dx$ .

Show that $nW_n=(n-1)W_{n-2}$ for $n\ge2$ .
Show that $(n+1)W_nW_{n+1}=\dfrac\pi2$ for all $n\ge0$ .
Deduce that $\sqrt{\dfrac\pi{2(n+1)}}\le W_n\le\sqrt{\dfrac\pi{2n}}$ for $n\ge1$ .

Hint

For 1, integrate by parts after writing $\sin^n=\sin^{n-1}\cdot\sin$ . For 2, show that the quantity is constant in $n$ . For 3, use that $(W_n)$ is decreasing.

Show solutionHide solution

Solution

1. With $u=\sin^{n-1}x$ and $v'=\sin x$ (so $v=-\cos x$ ):
$W_n=\bigl[-\cos x\sin^{n-1}x\bigr]_0^{\pi/2}+(n-1)\int_0^{\pi/2}\cos^2x\sin^{n-2}x\,dx=(n-1)(W_{n-2}-W_n),$
using $\cos^2=1-\sin^2$ . Hence $nW_n=(n-1)W_{n-2}$ .

2. Let $u_n=(n+1)W_nW_{n+1}$ . By part 1 (with index $n+1$ ), $(n+1)W_{n+1}=nW_{n-1}$ , so $u_n=nW_{n-1}W_n=u_{n-1}$ . The sequence is constant, equal to $u_0=W_0W_1=\frac\pi2\cdot1$ .

3. Since $0\le\sin x\le1$ , $(W_n)$ is decreasing: $W_{n+1}\le W_n\le W_{n-1}$ . Multiplying by $W_n>0$ :
$W_nW_{n+1}\le W_n^2\le W_nW_{n-1}\ \Longrightarrow\ \frac\pi{2(n+1)}\le W_n^2\le\frac\pi{2n}.$

Consequence. $W_n\sim\sqrt{\frac\pi{2n}}$ . Together with the closed formulas $W_{2p}=\frac{(2p)!}{4^p(p!)^2}\frac\pi2$ , this gives the Wallis product for $\pi$ and the asymptotics of central binomial coefficients.

Wallis integrals

Statement

Solution

More problems in Integration techniques