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Question Number 134373 by mnjuly1970 last updated on 02/Mar/21

	Answered by Dwaipayan Shikari last updated on 02/Mar/21

	$I (a) = \underset{n = - \infty}{\sum^{\infty}} \frac{1}{{(n^{2} + a^{2})}^{2}} = \frac{1}{a^{4}} + 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{{(n^{2} + a^{2})}^{2}}$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + a^{2}} = \frac{π}{2 a} c o t h (π a) - \frac{1}{2 a^{2}}$ $\underset{n = 1}{\sum^{\infty}} \frac{- 2 a}{{(n^{2} + a^{2})}^{2}} = \frac{\partial}{\partial a} (\frac{π}{2 a} c o t h (π a) - \frac{1}{2 a^{2}})$ $= - \frac{π}{2 a^{2}} c o t h (π a) - \frac{π^{2}}{2 a} {c s c h}^{2} (π a) + \frac{1}{a^{3}}$ $\underset{n = 1}{\sum^{\infty}} \frac{2}{{(n^{2} + a^{2})}^{2}} = \frac{π}{2 a^{3}} c o t h (π a) - \frac{π^{2}}{2 a^{2}} {c s c h}^{2} (π a) - \frac{1}{a^{4}}$ $I (1) = \frac{π}{2} c o t h (π) - \frac{π^{2}}{2} {c s c h}^{2} (π)$
	Commented by mnjuly1970 last updated on 03/Mar/21

	$t h a n k s a l o t m r p a y a n . . .$
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