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

$$provethatcoth\left(x\right)-\frac{1}{x}=\sum _{n=1}^{\mathrm{\infty }}\frac{2x}{x^2+n^2{\pi }^2}$$$$(x\ne 0)$$

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

$$wehaveprovedthat$$$$ch\left(\alpha x\right)=\frac{sh\left(\pi \alpha \right)}{\pi \alpha }+\frac{2\alpha }{\pi }sh\left(\pi \alpha \right)\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{{\alpha }^2+n^2}cos\left(nx\right)$$$$x=\pi \Rightarrow ch\left(\pi \alpha \right)=\frac{sh\left(\pi \alpha \right)}{\pi \alpha }+\frac{2\alpha }{\pi }sh\left(\pi \alpha \right)\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\alpha }^2+n^2}$$$$\Rightarrow coth\left(\pi \alpha \right)=\frac{1}{\pi \alpha }+\frac{2\alpha }{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\alpha }^2+n^2}$$$$changement\pi \alpha =xgive$$$$coth\left(x\right)=\frac{1}{x}+\frac{2}{\pi }\frac{x}{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\frac{x^2}{{\pi }^2}+n^2}$$$$=\frac{1}{x}+\sum _{n=1}^{\mathrm{\infty }}\frac{2x}{x^2+n^2{\pi }^2}\Rightarrow$$$$coth\left(x\right)-\frac{1}{x}=\sum _{n=1}^{\mathrm{\infty }}\frac{2x}{x^2+n^2{\pi }^2}.$$

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