7

As far as we know, Euler was the first to prove $$ \pi \cot(\pi z) = \frac{1}{z} + \sum_{k=1}^\infty \left( \frac{1}{z-k} + \frac{1}{z+k} \right).$$ I've seen several modern proofs of it and they all seem to rely either on the Herglotz trick or on the residue theorem. I recon Euler had neither nor at his disposal, so how did he prove it?

Added: Did Euler prove it for complex $z$ or just reals?

Community
  • 1
Leo
  • 7,670
  • 5
  • 31
  • 63
  • You may wish to read Euler's original work "Intoduction in Analysin Infinitorum", §178. In Latin: https://scholarlycommons.pacific.edu/cgi/viewcontent.cgi?article=1100&context=euler-works. and in an english translation http://www.17centurymaths.com/contents/introductiontoanalysisvol1.htm. Look for chapter 9 (product formula for $\sin$) and 10 (proof of series for $\cot$). – Dr. Wolfgang Hintze Mar 28 '21 at 16:19

2 Answers2

4

In my opinion, Euler just considered the logarithmic derivative ($\frac{d}{dz}\log(\cdot)$) of both sides of $$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) \tag{1}$$ to derive the above identity, disregarding possible convergence issues. $(1)$ was well-known to him, and the key for his solution of the Basel problem. The possibility to apply the logarithmic derivative to both sides of $(1)$ follows from the whole Mittag-Leffler/Weierstrass products machinery.

It is possible to prove $(1)$ for $z\in\mathbb{R}$ (together with its uniform convergence over any compact subset of $\mathbb{R}$) by avoiding complex analysis, just exploiting the properties of Chebyshev polynomials of the second kind, but I am not so sure that Euler was aware of that (also because Chebyshev came about 100 years later than Euler).

Community
  • 1
Jack D'Aurizio
  • 353,855
4

Euler did have access to the infinite product representation of the sine function

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

Then, we have

$$\begin{align} \frac{d \log(\sin(\pi z))}{dz}&=\pi \cot(\pi z)\\\\ &=\frac1z+\sum_{n=1}^\infty \frac{2z}{z^2-n^2}\\\\ &=\frac1z+\sum_{n=1}^\infty \left(\frac{1}{z-n}+\frac{1}{z+n}\right)\\\\ \end{align}$$

as was to be shown!

Mark Viola
  • 179,405
  • Yes, but how do we justify that logarithm and infinite product can be interchanged? – aiyy Sep 23 '21 at 16:29
  • @aiyy The logarithm is a continuous function on the positive reals. Can you show that if $a_k\to a$ that $\log\left(a_k\right)\to\log(a)$? – Mark Viola Sep 23 '21 at 20:17