Series expansion of $\coth x$ using the Fourier transform

Question

Hi I have research about the series of coth but all of the solutions emerges from integral on a contour, Could you calculate the fourier transform of coth? Is that possible at all?My goal is to reach : $$\coth x = \frac{1}{x} + 2 \sum_{n=1}^\infty \frac{x}{x^2+\pi^2n^2}$$ through fourier.

Answer 1

The expansion can be derived from the complex Fourier series of the $2 \pi$-periodic function $f(x) = e^{ax}$ for $- \pi < x < \pi$.

By definition of the complex Fourier series, $$e^{ax} = \lim_{N \to \infty}\sum_{n=-N}^{N} c_{n} e^{inx} $$

where $$ \begin{align} c_{n} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{ax} e^{-inx} \ dx &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{(a-in)x} \ dx \\ &= \frac{1}{2 \pi} \frac{e^{(a-in)x}}{a-in} \Big|^{\pi}_{-\pi} \\ &= \frac{1}{2 \pi} \frac{1}{a-in} \Big( e^{a \pi}e^{-in \pi} - e^{- a \pi}e^{i n \pi} \Big) \\ &= \frac{1}{\pi} \frac{(-1)^{n}}{a-in}\frac{e^{a \pi}-e^{- a \pi}}{2} \\ &= \frac{(-1)^{n}}{\pi} \frac{a+in}{a^{2}+n^{2}} \sinh a \pi \end{align}$$

At $x= \pi$ (a point of discontinuity), $$\frac{e^{a \pi}+e^{- a \pi}}{2} = \cosh a \pi = \lim_{N \to \infty}\sum_{n=-N}^{N} c_{n} (-1)^{n} = \frac{\sinh a \pi}{\pi} \lim_{N \to \infty} \sum_{n=-N}^{N} \frac{a+in}{a^{2}+n^{2}}$$

which implies

$$ \begin{align} \coth a \pi &= \frac{1}{\pi} \lim_{N \to \infty}\sum_{n=-N}^{N} \frac{a+in}{a^{2}+n^{2}} \\ &= \frac{1}{\pi} \lim_{N \to \infty}\sum_{n=-N}^{N} \frac{a}{a^{2}+n^{2}} + \frac{i}{\pi} \lim_{N \to \infty} \sum_{n=-N}^{N} \frac{n}{a^{2}+n^{2}} \\ &= \frac{1}{\pi} \lim_{N \to \infty}\sum_{n=-N}^{N} \frac{a}{a^{2}+n^{2}}+ 0 \\ &= \frac{1}{\pi}\sum_{n=-\infty}^{\infty} \frac{a}{a^{2}+n^{2}} \end{align}$$

And replacing $a$ with $ \displaystyle \frac{x}{\pi}$, we get

$$ \begin{align} \coth x &= \sum_{n=-\infty}^{\infty} \frac{x}{x^{2}+\pi^{2}n^{2}} \\ &= \frac{1}{x} + 2 \sum_{n=1}^{\infty} \frac{x}{x^{2}+\pi^{2}n^{2}} \end{align}$$