Question Number 38207 by prof Abdo imad last updated on 22/Jun/18
$${prove}\:{that}\:{coth}\left({x}\right)−\frac{\mathrm{1}}{{x}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \pi^{\mathrm{2}} } \\ $$$$\left({x}\neq\mathrm{0}\right) \\ $$
Commented by math khazana by abdo last updated on 25/Jun/18
$${we}\:{have}\:{proved}\:{that} \\ $$$${ch}\left(\alpha{x}\right)=\frac{{sh}\left(\pi\alpha\right)}{\pi\alpha}\:\:+\frac{\mathrm{2}\alpha}{\pi}{sh}\left(\pi\alpha\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }{cos}\left({nx}\right) \\ $$$${x}=\pi\:\Rightarrow{ch}\left(\pi\alpha\right)=\frac{{sh}\left(\pi\alpha\right)}{\pi\alpha}\:+\frac{\mathrm{2}\alpha}{\pi}{sh}\left(\pi\alpha\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$$$\Rightarrow{coth}\left(\pi\alpha\right)=\frac{\mathrm{1}}{\pi\alpha}\:+\:\frac{\mathrm{2}\alpha}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$$${changement}\:\pi\alpha={x}\:{give} \\ $$$${coth}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:\:+\frac{\mathrm{2}}{\pi}\:\frac{{x}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\frac{{x}^{\mathrm{2}} }{\pi^{\mathrm{2}} }\:+{n}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{{x}}\:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\:\Rightarrow \\ $$$${coth}\left({x}\right)−\frac{\mathrm{1}}{{x}}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\:. \\ $$