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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 x

$$\mathrm{csc}\:\mathrm{x} \\ $$

Commented by mnjuly1970 last updated on 02/Jun/21

  thanks mr qaz ..please proof..

$$\:\:{thanks}\:{mr}\:{qaz}\:..{please}\:{proof}.. \\ $$

Commented by qaz last updated on 02/Jun/21

i wish i could ...sir

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

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

Σ_(n=−∞) ^∞ (((−1)^n )/((nπ+x)))=(1/(sinx))

$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}\pi+{x}\right)}=\frac{\mathrm{1}}{{sinx}} \\ $$

Commented by mnjuly1970 last updated on 03/Jun/21

  nice ...

$$\:\:{nice}\:... \\ $$

Answered by mindispower last updated on 03/Jun/21

let f(z)=(1/(sin(z)(z+x)))  =((2i)/(e^(iz) −e^(−z) )).(1/(z+x))=f(z),lim_(R→∞) ∫_C_R  f(z)dz=0  C_R =Re^(iθ) ,θ∈[0,2π]  ∫_C_R  f(z)dz=2iπRes(f)  sin(z)=0⇒z=nπ,n∈Z  z+x=0⇒z=−x  Res(f,nπ)=lim_(z→nπ) .((z−nπ)/((nπ+x)sin(z)))=(((−1)^n )/(nπ+x))  Res(f,−x)=(1/(sin(−x)))  ⇒Σ_(n≥−∞) (((−1)^n )/(nπ+x))+(1/(sin(−x)))=∫f(z)dz=0  ⇒Σ(((−1)^n )/(nπ+x))=(1/(sin(x)))

$${let}\:{f}\left({z}\right)=\frac{\mathrm{1}}{{sin}\left({z}\right)\left({z}+{x}\right)} \\ $$$$=\frac{\mathrm{2}{i}}{{e}^{{iz}} −{e}^{−{z}} }.\frac{\mathrm{1}}{{z}+{x}}={f}\left({z}\right),\underset{{R}\rightarrow\infty} {\mathrm{lim}}\int_{{C}_{{R}} } {f}\left({z}\right){dz}=\mathrm{0} \\ $$$${C}_{{R}} ={Re}^{{i}\theta} ,\theta\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\int_{{C}_{{R}} } {f}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({f}\right) \\ $$$${sin}\left({z}\right)=\mathrm{0}\Rightarrow{z}={n}\pi,{n}\in\mathbb{Z} \\ $$$${z}+{x}=\mathrm{0}\Rightarrow{z}=−{x} \\ $$$${Res}\left({f},{n}\pi\right)=\underset{{z}\rightarrow{n}\pi} {\mathrm{lim}}.\frac{{z}−{n}\pi}{\left({n}\pi+{x}\right){sin}\left({z}\right)}=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\pi+{x}} \\ $$$${Res}\left({f},−{x}\right)=\frac{\mathrm{1}}{{sin}\left(−{x}\right)} \\ $$$$\Rightarrow\underset{{n}\geqslant−\infty} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\pi+{x}}+\frac{\mathrm{1}}{{sin}\left(−{x}\right)}=\int{f}\left({z}\right){dz}=\mathrm{0} \\ $$$$\Rightarrow\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\pi+{x}}=\frac{\mathrm{1}}{{sin}\left({x}\right)} \\ $$

Commented by mnjuly1970 last updated on 03/Jun/21

   thanks alot...

$$\:\:\:{thanks}\:{alot}... \\ $$

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