Question Number 142573 by mnjuly1970 last updated on 02/Jun/21
Answered by qaz last updated on 02/Jun/21
$$\mathrm{csc}\:\mathrm{x} \\ $$
Commented by mnjuly1970 last updated on 02/Jun/21
$$\:\:{thanks}\:{mr}\:{qaz}\:..{please}\:{proof}.. \\ $$
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
$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}\pi+{x}\right)}=\frac{\mathrm{1}}{{sinx}} \\ $$
Commented by mnjuly1970 last updated on 03/Jun/21
$$\:\:{nice}\:… \\ $$
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)))](https://www.tinkutara.com/question/Q142628.png)
$${let}\:{f}\left({z}\right)=\frac{\mathrm{1}}{{sin}\left({z}\right)\left({z}+{x}\right)} \\ $$$$=\frac{\mathrm{2}{i}}{{e}^{{iz}} −{e}^{−{z}} }.\frac{\mathrm{1}}{{z}+{x}}={f}\left({z}\right),\underset{{R}\rightarrow\infty} {\mathrm{lim}}\int_{{C}_{{R}} } {f}\left({z}\right){dz}=\mathrm{0} \\ $$$${C}_{{R}} ={Re}^{{i}\theta} ,\theta\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\int_{{C}_{{R}} } {f}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({f}\right) \\ $$$${sin}\left({z}\right)=\mathrm{0}\Rightarrow{z}={n}\pi,{n}\in\mathbb{Z} \\ $$$${z}+{x}=\mathrm{0}\Rightarrow{z}=−{x} \\ $$$${Res}\left({f},{n}\pi\right)=\underset{{z}\rightarrow{n}\pi} {\mathrm{lim}}.\frac{{z}−{n}\pi}{\left({n}\pi+{x}\right){sin}\left({z}\right)}=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\pi+{x}} \\ $$$${Res}\left({f},−{x}\right)=\frac{\mathrm{1}}{{sin}\left(−{x}\right)} \\ $$$$\Rightarrow\underset{{n}\geqslant−\infty} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\pi+{x}}+\frac{\mathrm{1}}{{sin}\left(−{x}\right)}=\int{f}\left({z}\right){dz}=\mathrm{0} \\ $$$$\Rightarrow\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\pi+{x}}=\frac{\mathrm{1}}{{sin}\left({x}\right)} \\ $$
Commented by mnjuly1970 last updated on 03/Jun/21
$$\:\:\:{thanks}\:{alot}… \\ $$