nLab product formula for the sine function



The infinite product formula for the sine function, due to Euler, is

sinx=x n=1 (1x 2π 2n 2).\sin x = x \prod_{n=1}^\infty (1 - \frac{x^2}{\pi^2 n^2}).

Taking logarithms and then derivatives, this implies the formula

πcotπx=1x+ n=1 (1x+n+1xn).\pi \cot \pi x = \frac1{x} + \sum_{n=1}^\infty \left(\frac1{x+n} + \frac1{x-n} \right).

Conversely, assuming this cotangent identity, then integrating and exponentiating, we deduce that the product formula is correct up to a constant factor CC, and we further deduce that C=1C = 1 because limx0sinxx=1\underset{x \to 0}{\lim}\; \frac{\sin x}{x} = 1. In this sense, we may say that the product formula is equivalent to this expansion formula for the cotangent.


Various proofs of the product formula are known, some relating the product formula to the theory of the Gamma function, some using complex analysis, among others. Below we prove the corresponding identity for the cotangent stated above, using an ingenious argument due to Herglotz.


limx0πcotπx1x=0\underset{x \to 0}{\lim} \pi \cot \pi x - \frac1{x} = 0.


In the right-hand side of the equation

πcotπx1x=πxcosπxsinπxxsinπx,\pi \cot \pi x - \frac1{x} = \frac{\pi x\cos \pi x - \sin \pi x}{x \sin{\pi x}},

the numerator has a zero of order 33 at x=0x = 0, and the denominator has a zero of order 22 at x=0x = 0, by MacLaurin expansions.


Let f(x)=πcotπxf(x) = \pi \cot \pi x and g(x)=1x+ n=1 (1x+n+1xn)g(x) = \frac1{x} + \sum_{n=1}^\infty \left(\frac1{x+n} + \frac1{x-n} \right). Then

f(x2)+f(x+12)=2f(x);g(x2)+g(x+12)=2g(x).f\left(\frac{x}{2}\right) + f\left(\frac{x+1}{2}\right) = 2f(x); \qquad g\left(\frac{x}{2}\right) + g\left(\frac{x+1}{2}\right) = 2g(x).


The first follows from cosπ(x+1)2=sinπx2\cos\frac{\pi(x+1)}{2} = -\sin\frac{\pi x}{2}, sinπ(x+1)2=cosπx2\sin\frac{\pi(x+1)}{2} = \cos\frac{\pi x}{2}, and double-angle formulas. To prove the second identity, put

g N(x)=1x+ n=1 N(1x+n+1xn)g_N(x) = \frac1{x} + \sum_{n=1}^N \left(\frac1{x+n} + \frac1{x-n} \right)

and note that

g N(x2)+g N(x+12)=2g 2N(x)+2x+2N+1.g_N\left(\frac{x}{2}\right) + g_N\left(\frac{x+1}{2}\right) = 2g_{2N}(x) + \frac{2}{x + 2N + 1}.

Taking the limit as NN \to \infty, the result follows.


The function hh, defined by h(x)=f(x)g(x)h(x) = f(x) - g(x) for xx \in \mathbb{R} \setminus \mathbb{Z} and h(x)=0h(x) = 0 for xx \in \mathbb{Z}, is a continuous odd function of period 11.


The oddness and periodicity assertions are obvious, as is continuity at points xx \notin \mathbb{Z}. Continuity at points xx \in \mathbb{Z} follows from Lemma and periodicity.

Now we prove the cotangent identity, following the notations of the previous two lemmas:


By continuity and periodicity, h(x)h(x) attains a maximum value MM, say at x 0x_0. Since h(x2)+h(x+12)=2h(x)h\left(\frac{x}{2}\right) + h\left(\frac{x+1}{2}\right) = 2h(x) by Lemma , h(x 02)=Mh\left(\frac{x_0}{2}\right) = M follows (else, h(x 02)<Mh\left(\frac{x_0}{2}\right) \lt M implies h(x 0+12)>Mh\left(\frac{x_0+1}{2}\right) \gt M). By iterating, h(x 02 m)=Mh\left(\frac{x_0}{2^m}\right) = M for all mm. By continuity,

M=limmh(x 02 m)=h(0)=0.M = \underset{m \to \infty}{\lim} h\left(\frac{x_0}{2^m}\right) = h(0) = 0.

Thus h(x)M=0h(x) \leq M = 0 for all xx. From h(x)0h(x) \leq 0 and oddness we get that h(x)=0h(x) = 0 identically, and so f(x)=g(x)f(x) = g(x) identically. This completes the proof.

Last revised on June 7, 2023 at 04:21:15. See the history of this page for a list of all contributions to it.