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Question Number 133859 by mnjuly1970 last updated on 24/Feb/21

	Answered by Ñï= last updated on 24/Feb/21
	$Σ_(n=−∞) ^(+∞) (1/((x+nπ)^2 )) =Σ_(n=1) ^∞ [(1/((x+nπ)^2 ))+(1/((x−nπ)^2 ))+(1/x^2 )] =−(d/dx)Σ_(n=1) ^∞ ((1/(x+nπ))+(1/(x−nπ))+(1/x)) =−(d/dx)[(1/x)+Σ_(n=1) ^∞ ((2x)/(x^2 −n^2 π^2 ))] =−(d^2 /dx^2 )[lnx+Σ_(n=1) ^∞ ln(x^2 −n^2 π^2 )] =−(d^2 /dx^2 ){ln[Π_(n=1) ^∞ x(x^2 −n^2 π^2 )]} =−(d^2 /dx^2 )(lnsinx) =csc^2 x$
	$\underset{n = - \infty}{\sum^{+ \infty}} \frac{1}{{(x + n π)}^{2}}$ $= \underset{n = 1}{\sum^{\infty}} [\frac{1}{{(x + n π)}^{2}} + \frac{1}{{(x - n π)}^{2}} + \frac{1}{x^{2}}]$ $= - \frac{d}{d x} \underset{n = 1}{\sum^{\infty}} (\frac{1}{x + n π} + \frac{1}{x - n π} + \frac{1}{x})$ $= - \frac{d}{d x} [\frac{1}{x} + \underset{n = 1}{\sum^{\infty}} \frac{2 x}{x^{2} - n^{2} π^{2}}]$ $= - \frac{d^{2}}{{d x}^{2}} [l n x + \underset{n = 1}{\sum^{\infty}} l n (x^{2} - n^{2} π^{2})]$ $= - \frac{d^{2}}{{d x}^{2}} {l n [\underset{n = 1}{\prod^{\infty}} x (x^{2} - n^{2} π^{2})]}$ $= - \frac{d^{2}}{{d x}^{2}} (l n s i n x)$ $= {c s c}^{2} x$
	Commented by mnjuly1970 last updated on 24/Feb/21

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