There is a series representation as partial fraction expansion where just translated reciprocal functions are summed up, such that the poles of the cotangent function and the reciprocal functions match: $$ \pi \cdot \cot (\pi x) = \lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{x+n}. $$ This identity can be proven with the Herglotz trick.

Sounds funny, especially when you understand german (sorry Gustav ;-), but how does it work?


2 Answers 2


The Herglotz trick is basically to define $$f(x):=\pi\cot\pi x,\quad g(x):=\lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{x+n}$$ and derive enough common properties of these functions to see in the end that they must coincide. Namely, it consists of showing that:

  1. $f$ and $g$ are defined for all non-integral values and are continuous there.
  2. They are periodic of period 1.
  3. They are odd functions.
  4. They satisfy the same functional relation $f(\frac{x}{2})+f(\frac{x+1}{2})=2f(x)$ and $g(\frac{x}{2})+g(\frac{x+1}{2})=2g(x)$.
  5. By defining $h(x)=f(x)-g(x)$, and setting $h(x):=0$ for $x\in\mathbb{Z}$, $h$ becomes a continuous function on all of $\mathbb{R}$ that shares the properties given in (2), (3), and (4).

Now the "trick" is to use all these properties as follows. Since $h$ is a periodic continuous function, it possesses a maximum $m$. Let $x_0$ be a point in $[0,1]$ with $h(x_0)=m$. It follows from (4) that $$h\left(\frac{x_0}{2}\right)+h\left(\frac{x_0+1}{2}\right)=2m,$$ and hence that $h(\frac{x_0}{2})=m$. Iteration gives $h(\frac{x_0}{2^n})=m$ for all $n$, and hence $h(0)=m$ by continuity. But $h(0)=0$, and so $m=0$, that is, $h(x)\leq 0$ for all $x\in\mathbb{R}$. As $h(x)$ is an odd function, $h(x)<0$ is impossible, hence $h(x)=0$ for all $x\in\mathbb{R}$.

  • 4
    $\begingroup$ I took this from "Proofs from THE BOOK" by Aigner and Ziegler, where it is explained in details. $\endgroup$
    – Spenser
    Nov 25, 2013 at 23:12
  • $\begingroup$ This post is very well written and instructive! (+1) $\endgroup$
    – Mark Viola
    May 17, 2018 at 1:15
  • $\begingroup$ It took me a while to figure out why $h_0(\tfrac{x_0}2)$ had to equal $m$, though. $\endgroup$ Aug 25, 2018 at 3:51
  • $\begingroup$ @AntonioDJC it took me a while, too. In case I have to return to this question: if $h(x_0/2)\neq m$ then $h(x_0/2)<m$ because $m$ is the maximum, so $$2m=2h(x_0)=h(x_0/2)+h(\frac{x_0+1}2)<2m$$. $\endgroup$ Apr 7, 2019 at 22:15

I give a pretty detailed proof in this answer, but I am not sure what the Herglotz trick is.

I found a description of the Herglotz trick on the web and it seems to be the method I employed there. That is, look at enough similarities of the functions so that they have to be the same.

  • $\begingroup$ cool, thanks, +1 I'll upvote the linked answer...but it doesn't answer mine. $\endgroup$
    – draks ...
    Nov 25, 2013 at 22:58
  • $\begingroup$ @draks...: I did look up the Herglotz trick just after your comment, and it was apparently what I used. Was there more you needed? $\endgroup$
    – robjohn
    Nov 26, 2013 at 1:17
  • $\begingroup$ no I'm fine now and we both learned something... $\endgroup$
    – draks ...
    Nov 26, 2013 at 6:54

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