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Question Number 67532 by mathmax by abdo last updated on 28/Aug/19

$$provethat\pi cotan\left(\pi \alpha \right)={lim}_{n\to +\mathrm{\infty }}\sum _{k=-n}^n\frac{1}{\alpha -k}$$

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

$$Usetheformulas\underset{-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}f\left(k\right)=-\sum _{z_k}Res(f\left(z\right)\pi cotan\left(\pi z\right),z_k)and\underset{-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}{(-1)}^{k}f\left(k\right)=-\sum _{z_k}Res(f\left(z\right)\pi csc\left(\pi z\right),z_k)[withcsc\left(\pi x\right)=\frac{1}{sin\left(\pi x\right)}$$$$Soheref\left(k\right)=\frac{1}{\alpha -k}and{z}_k=\alpha$$$$\underset{-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{1}{\alpha -k}=-Res(\frac{\pi cotan\left(\pi z\right)}{\alpha -z},\alpha )=-\underset{z\to \alpha }{\lim }\left(\frac{(z-\alpha )\pi cotan\left(\pi z\right)}{\alpha -z}\right)=\pi cotan\left(\pi \alpha \right)$$

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