# Proving the Partial Fraction Decomposition of the Hyperbolic Cotangent Function by using Poisson Summation

While skimming through the wonderful post What are some examples of colorful language in serious mathematics papers? on MathOverflow an example given by Ben Green made me curious. He referred to the Poisson Summation Formula, which I heard from for the first time. According to WolframMathWorld the Poisson Summation Formula is given by

$$\sum_{n=-\infty}^\infty f(x+n)~=~\sum_{k=-\infty}^\infty e^{2\pi i kx}\int_{-\infty}^\infty f(x')e^{-2\pi ikx'}\mathrm dx'\tag1$$

Denote the Fourier Transform of $$f(x)$$ by $$\hat f(\xi)$$. Evaluating at $$x=0$$ reduces $$(1)$$ to the more common formulation, as it is also stated on Wikipedia

$$\sum_{n=-\infty}^\infty f(n)~=~\sum_{k=-\infty}^\infty \hat f(k)\tag2$$

Anyway, within the Wikipedia entry there is an Example section. One of the examples is the partial fraction decomposition of the hyperbolic cotagent function. To be precise it is given that

$$f(x)=e^{-ax}$$ for $$0\leq x$$ and $$f(x)=0$$ for $$x<0$$ to get $$\coth(x)=\frac1x+2x\sum\limits_{n\in\mathbb Z^+}\frac1{x^2+\pi^2n^2}$$, the partial fraction decomposition of $$\coth(x)$$

As far as I can tell the here used function $$f(x)$$ is wrong. I think it should be a bilateral exponential which vanishs for $$\pm\infty$$ but has an actual value for $$x<0$$. In other words I would say that the function $$f(x)$$ should be defined as

$$f(x)=\begin{cases}e^{-ax}&,x\ge 0\\e^{ax}&,x<0\end{cases}$$

Here $$a$$ denotes a positive real constant. Clearly $$f(x)$$ is an even function from which it follows directly that $$f(x)=f(-x)$$. Utlizing this definition we can explicitly write the LHS of $$(2)$$ as

$$\sum_{n=-\infty}^\infty f(n)=\hat f(0)+2\sum_{n=1}^\infty f(n)$$

Now note that $$f(n)=e^{-an}$$ for all $$n\ge1$$. Thus, we obtain that

$$f(0)+2\sum_{n=1}^\infty f(n)=1+2\sum_{n=1}^\infty e^{-an}=1+\frac{2e^{-a}}{1-e^{-a}}=\frac{e^a+1}{e^a-1}=\coth\left(\frac a2\right)$$

The Fourier Transform $$\hat f(\xi)$$ of $$f(x)$$ can be computed by splitting the region of integration and recombining the results afterwards. Overall this sums up to

$$\hat f(\xi)=\frac1{a+2\pi i\xi}+\frac1{a-2\pi i\xi}=\frac{2a}{a^2+4\pi^2\xi^2}$$

Again $$\hat f(\xi)$$ is clearly even from which we can conclude that $$\hat f(\xi)=\hat f(-\xi)$$ and therefore the RHS of $$(2)$$ becomes

$$\sum_{k=-\infty}^\infty \hat f(k)=\hat f(0)+2\sum_{n=1}^\infty \hat f(k)$$

Finally we obtain

$$\hat f(0)+2\sum_{n=1}^\infty \hat f(k)=\frac2a f(0)+2\sum_{n=1}^\infty \frac{2a}{a^2+4\pi^2k^2}=\frac2a+2\left(\frac a2\right)\sum_{n=1}^\infty \frac1{\left(\frac a2\right)^2+\pi^2k^2}$$

Utilizing the whole equation $$(2)$$ again and by chosing $$a=2a$$ we obtain the desired forumla.

$$\therefore~\coth(a)~=~\frac1a+2a\sum_{n=1}^\infty \frac1{a^2+\pi^2k^2}\tag2$$

However, I claim that the function Wikipedia gives does not produce the here proved formula $$(3)$$. According to my own calculation the RHS indeed equals the partial fraction decomposition of the hyperbolic cotangent function whereas the LHS is missing a factor $$2$$ infront of the infinite sum from where it is not deducable that the LHS is in fact the hyperbolic cotagent.

Two questions, closely related: First of all is the identity denoted as $$(3)$$ derivable from the applying Poisson Summation on the function given on Wikipedia? Is my own solution one the on hand correct but perhaps to complicated since my choice of $$f(x)$$ for $$x<0$$ might be redundant if the function given by Wikipedia is sufficient for this purpose? I appreciate every suggestion of either improving my own proof or explaining me what I am missing about the function stated on Wikipedia.

Let $$a \not \in 2i \pi \mathbb{Z}$$ $$G(x) = e^{-ax} 1_{x \in [0,1]}, \qquad g(x) = \sum_n G(x+n)$$ piecewise $$C^1$$ and $$1$$-periodic.
Then $$\hat{g}(k) = \int_0^1 g(x)e^{-2i \pi kx}dx = \frac{1-e^{-a}}{a+2i \pi k}$$ And by the Fourier series theorem for piecewise $$C^1$$ periodic functions $$\frac{1+e^{-a}}{2} = \lim_{x \to 0} \frac{G(x)+G(-x)}{2} = \lim_{K \to \infty} \sum_{k=-K}^K \hat{g}(k) = (1-e^{-a}) \lim_{K \to \infty} \sum_{k=-K}^K \frac{1}{a+2i \pi k}$$