Wallis product

Proof using integration

Wallis derived this infinite product using interpolation, though his method is not regarded as rigorous. A modern derivation can be found by examining \int_0^\pi \sin^n x,dx for even and odd values of n, and noting that for large n, increasing n by 1 results in a change that becomes ever smaller as n increases. Let

:I(n) = \int_0^\pi \sin^n x,dx.

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

:\begin{align} 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{align}

:\begin{align} \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^2x \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{align} Now, we make two variable substitutions for convenience to obtain: :I(2n) = \frac{2n-1}{2n}I(2n-2) :I(2n+1) = \frac{2n}{2n+1}I(2n-1)

We obtain values for I(0) and I(1) for later use.

:\begin{align} 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{align}

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

: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)

:=\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 I(2n+1),

: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)

:=\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 \sin{x} \leq 1

:\sin^{2n+1}x \le \sin^{2n}x \le \sin^{2n-1}x, 0 \le x \le \pi

:\Rightarrow I(2n+1) \le I(2n) \le I(2n-1)

Dividing by I(2n+1):

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

By the squeeze theorem,

:\Rightarrow \lim_{n\rightarrow\infty} \frac{I(2n)}{I(2n+1)}=1

:\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

:\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 method]]

See the main page on Gaussian integral.

Proof using Euler's infinite product for the sine function

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.

:\frac{\sin x}{x} = \prod_{n=1}^\infty\left(1 - \frac{x^2}{n^2\pi^2}\right)

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

:\begin{align} \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{align}

Relation to Stirling's approximation

Stirling's approximation for the factorial function n! asserts that :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 k terms in the product :p_k = \prod_{n=1}^{k} \frac{2n}{2n - 1}\frac{2n}{2n + 1},

where p_k can be written as :\begin{align} p_k &= {1 \over {2k + 1}} \prod_{n=1}^{k} \frac{(2n)^4}{[(2n)(2n - 1)]^2} \[6pt] &= {1 \over {2k + 1}} \cdot . \end{align}

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

Derivative of the Riemann zeta function at zero

The Riemann zeta function and the Dirichlet eta function can be defined: :\begin{align} \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{align}

Applying an Euler transform to the latter series, the following is obtained: :\begin{align}

\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{align}

:\begin{align} \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{align}

Notes

References

1. "Wallis Formula".
2. "Integrating Powers and Product of Sines and Cosines: Challenging Problems".