prove-that-coth-x-1-x-n-1-2x-x-2-n-2-pi-2-x-0-

Question Number 38207 by prof Abdo imad last updated on 22/Jun/18

p r o v e t h a t c o t h (x) - \frac{1}{x} = \sum_{n = 1}^{\infty} \frac{2 x}{x^{2} + n^{2} π^{2}}

(x \neq 0)

Commented by math khazana by abdo last updated on 25/Jun/18

w e h a v e p r o v e d t h a t

c h (α x) = \frac{s h (π α)}{π α} + \frac{2 α}{π} s h (π α) \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{α^{2} + n^{2}} c o s (n x)

x = π \Rightarrow c h (π α) = \frac{s h (π α)}{π α} + \frac{2 α}{π} s h (π α) \sum_{n = 1}^{\infty} \frac{1}{α^{2} + n^{2}}

\Rightarrow c o t h (π α) = \frac{1}{π α} + \frac{2 α}{π} \sum_{n = 1}^{\infty} \frac{1}{α^{2} + n^{2}}

c h a n g e m e n t π α = x g i v e

c o t h (x) = \frac{1}{x} + \frac{2}{π} \frac{x}{π} \sum_{n = 1}^{\infty} \frac{1}{\frac{x^{2}}{π^{2}} + n^{2}}

= \frac{1}{x} + \sum_{n = 1}^{\infty} \frac{2 x}{x^{2} + n^{2} π^{2}} \Rightarrow

c o t h (x) - \frac{1}{x} = \sum_{n = 1}^{\infty} \frac{2 x}{x^{2} + n^{2} π^{2}} .