Partial fractions for $\pi \cot(\pi z)$

Question

I want to derive $$\pi \cot(\pi z) = \sum_{-\infty}^{\infty}\frac{1}{z-n} + \frac{1}{n}$$ without taking derivatives.

I know through Mittag Leffler that $$\pi \cot(\pi z) = g(z) +\sum_{-\infty}^{\infty}\frac{1}{z-n} + \frac{1}{n}$$ for some entire analytic $g(z)$ but I'm having a hard time showing that $g(z) = 0$. I do not want to take the integral of $\frac{\pi^2}{\sin^2(\pi z)}$.

I have deduced that $g(0) = 0$, $g$ is odd, and $g$ is periodic with period $1$.
If I could deduce that $g$ was bounded even in a strip of length $1$ I wold be finished because of Louiville Theorem. To this end, I can show that $\pi \cot(\pi z)$ is bounded towards infinity but I was not able to show that the sum could be bounded.

Answer 1

I don't know if that is Cauchy's method, but one way to establish $g\equiv 0$ without differentiating is to use a functional equation:

$$2\pi \cot (2\pi z) = \pi \frac{\cos^2(\pi z) - \sin^2(\pi z)}{\sin (\pi z)\cos (\pi z)} = \pi \cot (\pi z) + \pi\cot \left(\pi \left(z+\tfrac{1}{2}\right)\right),$$

and with

$$s(z) = \frac{1}{z} + \sum_{n\neq 0} \left(\frac{1}{z-n} + \frac{1}{n}\right) = \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{z-n}$$

we have

$$\begin{align} s(z) + s\left(z+\tfrac{1}{2}\right) &= \lim_{N\to\infty} \left(\sum_{n=-N}^N \frac{1}{z-n} + \sum_{n=-N}^N\frac{1}{z+\frac{1}{2}-n}\right)\\ &= 2\lim_{N\to\infty} \left(\sum_{n=-N}^N\frac{1}{2z-2n} + \sum_{n=-N}^N \frac{1}{2z+1-n}\right)\\ &= 2\lim_{N\to\infty} \sum_{k=-(2N+1)}^{2N} \frac{1}{2z-k}\\ &= 2s(2z), \end{align}$$

so the entire function $g$ also satisfies the functional equation

$$g(z) + g\left(z+\tfrac{1}{2}\right) = 2g(2z).\tag{$\ast$}$$

But a continuous $1$-periodic function $f$ satisfying the functional equation $(\ast)$ is constant on $\mathbb{R}$. Separately considering the real and imaginary part of $f$, we may assume that $f$ is real-valued (and our $g$ is real-valued on $\mathbb{R}$). By continuity and periodicity, there is an $x_0 \in [0,1]$ with $f(x) \leqslant f(x_0)$ for all $x\in \mathbb{R}$. Let $x_k = 2^{-k}x_0$ for $k\in\mathbb{N}$. Then

$$2 f(x_0) = 2f(x_k) = 2f(2x_{k+1}) = f(x_{k+1}) + f\left(x_{k+1}+\tfrac{1}{2}\right)$$

shows inductively that $f(x_k) = f(x_0)$ for all $k$. Hence, by continuity, $f(0) = \max \{ f(x) : x\in\mathbb{R}\}$. The analogous argument shows $f(0) = \min \{ f(x) : x\in \mathbb{R}\}$, so $f \equiv f(0)$ is constant.

Since $g$ is holomorphic, it follows that $g$ is constant on $\mathbb{C}$, hence $g(z) \equiv g(0) = 0$.

Answer 2

You have to use Cauchy's method (Mittag-Leer Expansion, see the link for detail www.student.math.uwaterloo.ca/~dkotik/Complex.pdf) to verify. Let $$f(z)=\pi\cot(\pi z)-\frac{1}{z}. $$ Since $\lim_{z\to0}f(z)=0$, $z=0$ is a removable singular point of $f(z)$ and we can define $f(0)=0$. Define a sequence of connected simple closed curves by $$ \gamma_n=\{z=x+iy: x=\pm(n+\frac{1}{2}), |y|\le n+\frac{1}{2}\}\cup \{z=x+iy: y=\pm(n+\frac{1}{2}), |x|\le n+\frac{1}{2}\}. $$ Clearly \begin{eqnarray} |\cot(\pi z)|^2&=&\frac{|\cos(\pi z)|^2}{|\sin(\pi z)|^2}=\frac{|e^{i\pi(x+yi)}+e^{-i\pi(x+yi)}|^2}{|e^{i\pi(x+yi)}-e^{-i\pi(x+yi)}|^2}\\ &=&\frac{e^{2\pi y}+e^{-2\pi y}+\cos(2\pi x)}{e^{2\pi y}+e^{-2\pi y}-\cos(2\pi x)} \end{eqnarray} On $\{z=x+iy: x=\pm(n+\frac{1}{2}), |y|\le n+\frac{1}{2}\}$, we have $\cos(2\pi x)=-1$ and hence $$ |\cot(\pi z)|^2=\frac{e^{2\pi y}+e^{-2\pi y}-1}{e^{2\pi y}+e^{-2\pi y}+1}\le1. $$ Similarly, on $\{z=x+iy: y=\pm(n+\frac{1}{2}), |x|\le n+\frac{1}{2}\}$, we have $\cos(2\pi x)=-1$ and hence $$ |\cot(\pi z)|^2=\frac{e^{2\pi y}+e^{-2\pi y}+\cos(2\pi x)}{e^{2\pi y}+e^{-2\pi y}-\cos(2\pi x)}\le\frac{e^{2\pi y}+e^{-2\pi y}+1}{e^{2\pi y}+e^{-2\pi y}-1}\le3. $$ So $f(z)$ is bounded in $\gamma_n$. Clearly the length of $\gamma_n$ is $\mathcal{O}(n)$ and for $\forall z\in\gamma_n$, $|z|\ge n$. Thus the condition of Mittag-Leer Expansion is satisfied. Note $z=k$ $(k=\pm 1,\pm2,\cdots)$ are simple poles and $$ r_k=\text{Res}(f,k\pi)=\frac{\pi\cos(\pi z)}{(\sin(\pi z))'}\big|_{z=k}=1. $$ By Mittag-Leer Expansion, we have $$ f(z)=f(0)+\sum_{k=-\infty}^\infty' r_k\left(\frac{1}{z-z_k}+\frac{1}{z_k}\right)=\sum_{k=-\infty}^\infty'\left(\frac{1}{z-k}+\frac{1}{k}\right) $$ or $$ \pi\cot(\pi z)=\frac{1}{z}+\sum_{k=-\infty}^\infty'\left(\frac{1}{z-k}+\frac{1}{k}\right) $$

Answer 3

I have found an elementary method in Stein and Shakarchi. As mentioned by Daniel Fischer, we are trying to sum $\frac{z}{z^2-n^2}$ over $n$.

Let $z = x + iy$. Then our terms become $\frac{x+iy}{x^2 - y^2 - n^2 + 2 i x y}$. Consider $|y| \gt 1$. The numerator is bounded above by $C|y|$ in the strip $x \in [-1/2, 1/2]$, and the denominator is bounded below by $|y^2 + n^2 - x^2|$ which is again bounded below by $C'|y^2 + n^2|$ so that each term is bounded above $C|\frac{y}{y^2 + n^2}|$. We can dominate the sum by in integral $\int^\infty_0 \frac{y}{y^2 + n^2}dn$, and the integral is finite and independent of $y$ as can be seen by a change of variables $n \rightarrow yn$.

Since our function is periodic, it is bounded for $|y| > 1$ and $x$ anywhere. But so is $\pi \cot(\pi z)$. So the difference is bounded entire and hence constant. But the difference is an odd function so the constant must be 0.