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

prove that coth(x)−(1/x) =Σ_(n=1) ^∞   ((2x)/(x^2  +n^2 π^2 ))  (x≠0)

$${prove}\:{that}\:{coth}\left({x}\right)−\frac{\mathrm{1}}{{x}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \pi^{\mathrm{2}} } \\ $$$$\left({x}\neq\mathrm{0}\right) \\ $$

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

we have proved that  ch(αx)=((sh(πα))/(πα))  +((2α)/π)sh(πα)Σ_(n=1) ^∞  (((−1)^n )/(α^2  +n^2 ))cos(nx)  x=π ⇒ch(πα)=((sh(πα))/(πα)) +((2α)/π)sh(πα)Σ_(n=1) ^∞   (1/(α^2  +n^2 ))  ⇒coth(πα)=(1/(πα)) + ((2α)/π) Σ_(n=1) ^∞   (1/(α^2  +n^2 ))  changement πα=x give  coth(x)=(1/x)  +(2/π) (x/π) Σ_(n=1) ^∞    (1/((x^2 /π^2 ) +n^2 ))  = (1/x)  +Σ_(n=1) ^∞     ((2x)/(x^2  +n^2 π^2 )) ⇒  coth(x)−(1/x) = Σ_(n=1) ^∞     ((2x)/(x^2  +n^2 π^2 )) .

$${we}\:{have}\:{proved}\:{that} \\ $$$${ch}\left(\alpha{x}\right)=\frac{{sh}\left(\pi\alpha\right)}{\pi\alpha}\:\:+\frac{\mathrm{2}\alpha}{\pi}{sh}\left(\pi\alpha\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }{cos}\left({nx}\right) \\ $$$${x}=\pi\:\Rightarrow{ch}\left(\pi\alpha\right)=\frac{{sh}\left(\pi\alpha\right)}{\pi\alpha}\:+\frac{\mathrm{2}\alpha}{\pi}{sh}\left(\pi\alpha\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$$$\Rightarrow{coth}\left(\pi\alpha\right)=\frac{\mathrm{1}}{\pi\alpha}\:+\:\frac{\mathrm{2}\alpha}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} } \\ $$$${changement}\:\pi\alpha={x}\:{give} \\ $$$${coth}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:\:+\frac{\mathrm{2}}{\pi}\:\frac{{x}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\frac{{x}^{\mathrm{2}} }{\pi^{\mathrm{2}} }\:+{n}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{{x}}\:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\:\Rightarrow \\ $$$${coth}\left({x}\right)−\frac{\mathrm{1}}{{x}}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\:. \\ $$

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