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Question Number 142573 by mnjuly1970 last updated on 02/Jun/21

Answered by qaz last updated on 02/Jun/21

$$\csc \mathrm{x}$$

Commented by mnjuly1970 last updated on 02/Jun/21

$$thanksmrqaz..pleaseproof..$$

Commented by qaz last updated on 02/Jun/21

$$\mathrm{i}\mathrm{wish}\mathrm{i}\mathrm{could}...\mathrm{sir}$$

Commented by Dwaipayan Shikari last updated on 03/Jun/21

$$\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{(n\pi +x)}=\frac{1}{sinx}$$

Commented by mnjuly1970 last updated on 03/Jun/21

$$nice...$$

Answered by mindispower last updated on 03/Jun/21

$$letf\left(z\right)=\frac{1}{sin\left(z\right)(z+x)}$$$$=\frac{2i}{e^{iz}-e^{-z}}.\frac{1}{z+x}=f\left(z\right),\underset{R\to \mathrm{\infty }}{\lim }{\int }_{C_R}f\left(z\right)dz=0$$$$C_R={Re}^{i\theta },\theta \in [0,2\pi ]$$$${\int }_{C_R}f\left(z\right)dz=2i\pi Res\left(f\right)$$$$sin\left(z\right)=0\Rightarrow z=n\pi ,n\in \mathbb{Z}$$$$z+x=0\Rightarrow z=-x$$$$Res(f,n\pi )=\underset{z\to n\pi }{\lim }.\frac{z-n\pi }{(n\pi +x)sin\left(z\right)}=\frac{{(-1)}^n}{n\pi +x}$$$$Res(f,-x)=\frac{1}{sin(-x)}$$$$\Rightarrow \sum _{n⩾-\mathrm{\infty }}\frac{{(-1)}^n}{n\pi +x}+\frac{1}{sin(-x)}=\int f\left(z\right)dz=0$$$$\Rightarrow \mathrm{\Sigma }\frac{{(-1)}^n}{n\pi +x}=\frac{1}{sin\left(x\right)}$$

Commented by mnjuly1970 last updated on 03/Jun/21

$$thanksalot...$$

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