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Question Number 63032 by mathmax by abdo last updated on 28/Jun/19

$$letf\left(z\right)=\frac{1}{sin\left(\pi z\right)}calculateRes(f,n)withnintegr$$

Commented by mathmax by abdo last updated on 28/Jun/19

$$z=nissimplepoleforf\Rightarrow Res(f,n)={lim}_{z\to n}\frac{z-n}{sin\left(\pi z\right)}$$$$changementz-n=tgiveRes(f,n)={lim}_{t\to 0}\frac{t}{sin(\pi (n+t)}$$$$={lim}_{t\to 0}\frac{t}{{(-1)}^{n}sin\left(\pi t\right)}={lim}_{t\to 0}\frac{{(-1)}^n}{\frac{sin\left(\pi t\right)}{\pi t}\pi }=\frac{{(-1)}^n}{\pi }$$

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