Wallis derived this infinite product using interpolation, though his method is not regarded as rigorous. A modern derivation can be found by examining ${\displaystyle \int _{0}^{\pi }\sin ^{n}x\,dx}$  for even and odd values of ${\displaystyle n}$ , and noting that for large ${\displaystyle n}$ , increasing ${\displaystyle n}$  by 1 results in a change that becomes ever smaller as ${\displaystyle n}$  increases. Let^[2]

${\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}x\,dx.}$ 

(This is a form of Wallis' integrals.) Integrate by parts:

${\displaystyle {\begin{aligned}u&=\sin ^{n-1}x\\\Rightarrow du&=(n-1)\sin ^{n-2}x\cos x\,dx\\dv&=\sin x\,dx\\\Rightarrow v&=-\cos x\end{aligned}}}$ 

${\displaystyle {\begin{aligned}\Rightarrow I(n)&=\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=-\sin ^{n-1}x\cos x{\Biggl |}_{0}^{\pi }-\int _{0}^{\pi }(-\cos x)(n-1)\sin ^{n-2}x\cos x\,dx\\[6pt]{}&=0+(n-1)\int _{0}^{\pi }\cos ^{2}x\sin ^{n-2}x\,dx,\qquad n>1\\[6pt]{}&=(n-1)\int _{0}^{\pi }(1-\sin ^{2}x)\sin ^{n-2}x\,dx\\[6pt]{}&=(n-1)\int _{0}^{\pi }\sin ^{n-2}x\,dx-(n-1)\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=(n-1)I(n-2)-(n-1)I(n)\\[6pt]{}&={\frac {n-1}{n}}I(n-2)\\[6pt]\Rightarrow {\frac {I(n)}{I(n-2)}}&={\frac {n-1}{n}}\\[6pt]\end{aligned}}}$ 

Now, we make two variable substitutions for convenience to obtain:

${\displaystyle I(2n)={\frac {2n-1}{2n}}I(2n-2)}$ 

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

We obtain values for ${\displaystyle I(0)}$  and ${\displaystyle I(1)}$  for later use.

${\displaystyle {\begin{aligned}I(0)&=\int _{0}^{\pi }dx=x{\Biggl |}_{0}^{\pi }=\pi \\[6pt]I(1)&=\int _{0}^{\pi }\sin x\,dx=-\cos x{\Biggl |}_{0}^{\pi }=(-\cos \pi )-(-\cos 0)=-(-1)-(-1)=2\\[6pt]\end{aligned}}}$ 

Now, we calculate for even values ${\displaystyle I(2n)}$  by repeatedly applying the recurrence relation result from the integration by parts. Eventually, we end get down to ${\displaystyle I(0)}$ , which we have calculated.

${\displaystyle I(2n)=\int _{0}^{\pi }\sin ^{2n}x\,dx={\frac {2n-1}{2n}}I(2n-2)={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}I(2n-4)}$ 

${\displaystyle ={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}\cdot {\frac {2n-5}{2n-4}}\cdot \cdots \cdot {\frac {5}{6}}\cdot {\frac {3}{4}}\cdot {\frac {1}{2}}I(0)=\pi \prod _{k=1}^{n}{\frac {2k-1}{2k}}}$ 

Repeating the process for odd values ${\displaystyle I(2n+1)}$ ,

${\displaystyle I(2n+1)=\int _{0}^{\pi }\sin ^{2n+1}x\,dx={\frac {2n}{2n+1}}I(2n-1)={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}I(2n-3)}$ 

${\displaystyle ={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}\cdot {\frac {2n-4}{2n-3}}\cdot \cdots \cdot {\frac {6}{7}}\cdot {\frac {4}{5}}\cdot {\frac {2}{3}}I(1)=2\prod _{k=1}^{n}{\frac {2k}{2k+1}}}$ 

We make the following observation, based on the fact that ${\displaystyle \sin {x}\leq 1}$ 

${\displaystyle \sin ^{2n+1}x\leq \sin ^{2n}x\leq \sin ^{2n-1}x,0\leq x\leq \pi }$ 

${\displaystyle \Rightarrow I(2n+1)\leq I(2n)\leq I(2n-1)}$ 

Dividing by ${\displaystyle I(2n+1)}$ :

${\displaystyle \Rightarrow 1\leq {\frac {I(2n)}{I(2n+1)}}\leq {\frac {I(2n-1)}{I(2n+1)}}={\frac {2n+1}{2n}}}$ , where the equality comes from our recurrence relation.

By the squeeze theorem,

${\displaystyle \Rightarrow \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}}=1}$ 

${\displaystyle \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}}={\frac {\pi }{2}}\lim _{n\rightarrow \infty }\prod _{k=1}^{n}\left({\frac {2k-1}{2k}}\cdot {\frac {2k+1}{2k}}\right)=1}$ 

${\displaystyle \Rightarrow {\frac {\pi }{2}}=\prod _{k=1}^{\infty }\left({\frac {2k}{2k-1}}\cdot {\frac {2k}{2k+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot \cdots }$ 

Proof using Laplace's methodEdit

See the main page on Gaussian integral.

While the proof above is typically featured in modern calculus textbooks, the Wallis product is, in retrospect, an easy corollary of the later Euler infinite product for the sine function.

${\displaystyle {\frac {\sin x}{x}}=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{n^{2}\pi ^{2}}}\right)}$ 

Let ${\displaystyle x={\frac {\pi }{2}}}$ :

${\displaystyle {\begin{aligned}\Rightarrow {\frac {2}{\pi }}&=\prod _{n=1}^{\infty }\left(1-{\frac {1}{4n^{2}}}\right)\\[6pt]\Rightarrow {\frac {\pi }{2}}&=\prod _{n=1}^{\infty }\left({\frac {4n^{2}}{4n^{2}-1}}\right)\\[6pt]&=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdots \end{aligned}}}$    ^[1]

Stirling's approximation for the factorial function ${\displaystyle n!}$  asserts that

${\displaystyle n!={\sqrt {2\pi n}}{\left({\frac {n}{e}}\right)}^{n}\left[1+O\left({\frac {1}{n}}\right)\right].}$ 

Consider now the finite approximations to the Wallis product, obtained by taking the first ${\displaystyle k}$  terms in the product

${\displaystyle p_{k}=\prod _{n=1}^{k}{\frac {2n}{2n-1}}{\frac {2n}{2n+1}},}$ 

where ${\displaystyle p_{k}}$  can be written as

${\displaystyle {\begin{aligned}p_{k}&={1 \over {2k+1}}\prod _{n=1}^{k}{\frac {(2n)^{4}}{[(2n)(2n-1)]^{2}}}\\[6pt]&={1 \over {2k+1}}\cdot {{2^{4k}\,(k!)^{4}} \over {[(2k)!]^{2}}}.\end{aligned}}}$ 

Substituting Stirling's approximation in this expression (both for ${\displaystyle k!}$  and ${\displaystyle (2k)!}$ ) one can deduce (after a short calculation) that ${\displaystyle p_{k}}$  converges to ${\displaystyle {\frac {\pi }{2}}}$  as ${\displaystyle k\rightarrow \infty }$ .

The Riemann zeta function and the Dirichlet eta function can be defined:^[1]

${\displaystyle {\begin{aligned}\zeta (s)&=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},\Re (s)>1\\[6pt]\eta (s)&=(1-2^{1-s})\zeta (s)\\[6pt]&=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}},\Re (s)>0\end{aligned}}}$ 

Applying an Euler transform to the latter series, the following is obtained:

${\displaystyle {\begin{aligned}\eta (s)&={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {1}{n^{s}}}-{\frac {1}{(n+1)^{s}}}\right],\Re (s)>-1\\[6pt]\Rightarrow \eta '(s)&=(1-2^{1-s})\zeta '(s)+2^{1-s}(\ln 2)\zeta (s)\\[6pt]&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {\ln n}{n^{s}}}-{\frac {\ln(n+1)}{(n+1)^{s}}}\right],\Re (s)>-1\end{aligned}}}$ 

${\displaystyle {\begin{aligned}\Rightarrow \eta '(0)&=-\zeta '(0)-\ln 2=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[\ln n-\ln(n+1)\right]\\[6pt]&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}\\[6pt]&=-{\frac {1}{2}}\left(\ln {\frac {1}{2}}-\ln {\frac {2}{3}}+\ln {\frac {3}{4}}-\ln {\frac {4}{5}}+\ln {\frac {5}{6}}-\cdots \right)\\[6pt]&={\frac {1}{2}}\left(\ln {\frac {2}{1}}+\ln {\frac {2}{3}}+\ln {\frac {4}{3}}+\ln {\frac {4}{5}}+\ln {\frac {6}{5}}+\cdots \right)\\[6pt]&={\frac {1}{2}}\ln \left({\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot \cdots \right)={\frac {1}{2}}\ln {\frac {\pi }{2}}\\\Rightarrow \zeta '(0)&=-{\frac {1}{2}}\ln \left(2\pi \right)\end{aligned}}}$ 

• John Wallis, English mathematician who is given partial credit for the development of infinitesimal calculus and pi.
• Viète's formula, a different infinite product formula for ${\displaystyle \pi }$ .
• Leibniz formula for π, an infinite sum that can be converted into an infinite Euler product for ${\displaystyle \pi }$ .
• Wallis sieve

• "Wallis formula", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• "Why does this product equal π/2? A new proof of the Wallis formula for π." 3Blue1Brown. April 20, 2018. Archived from the original on 2021-12-12 – via YouTube.