Contents
Statement
The infinite product formula for the sine function, due to Euler, is
Taking logarithms and then derivatives, this implies the formula
Conversely, assuming this cotangent identity, then integrating and exponentiating, we deduce that the product formula is correct up to a constant factor , and we further deduce that because . 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
.
Proof
In the right-hand side of the equation
the numerator has a zero of order at , and the denominator has a zero of order at , by MacLaurin expansions.
Lemma
Let and . Then
Proof
The first follows from , , and double-angle formulas. To prove the second identity, put
and note that
Taking the limit as , the result follows.
Lemma
The function , defined by for and for , is a continuous odd function of period .
Proof
The oddness and periodicity assertions are obvious, as is continuity at points . Continuity at points follows from Lemma and periodicity.
Now we prove the cotangent identity, following the notations of the previous two lemmas:
Proof
By continuity and periodicity, attains a maximum value , say at . Since by Lemma , follows (else, implies ). By iterating, for all . By continuity,
Thus for all . From and oddness we get that identically, and so identically. This completes the proof.