# Contents

## Statement

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

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

Taking logarithms and then derivatives, this implies the formula

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

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

## Proof

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.

###### Lemma

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

###### Proof

In the right-hand side of the equation

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

the numerator has a zero of order $3$ at $x = 0$, and the denominator has a zero of order $2$ at $x = 0$, by MacLaurin expansions.

###### Lemma

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

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

###### Proof

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

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

and note that

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

Taking the limit as $N \to \infty$, the result follows.

###### Lemma

The function $h$, defined by $h(x) = f(x) - g(x)$ for $x \in \mathbb{R} \setminus \mathbb{Z}$ and $h(x) = 0$ for $x \in \mathbb{Z}$, is a continuous odd function of period $1$.

###### Proof

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

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

###### Proof

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

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

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