Show that sum_(n=0)^infty 1/(n^2+a^2)=pi/(2a) coth pi a+1/(2a^2), a>0

Hanna Webster 2022-11-21 Answered
Show that
n = 0 1 n 2 + a 2 = π 2 a coth π a + 1 2 a 2 , a > 0
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Answers (1)

Camden Stanton
Answered 2022-11-22 Author has 14 answers
The method of residues applies to sums of the form
n = f ( n ) = k res z = z k π cot π z f ( z )
where z k are poles of f that are not integers. So when f is even in n, you may express as follows:
2 n = 1 f ( n ) + f ( 0 )
For this case, f ( z ) = 1 / ( z 2 + a 2 ) and the poles z ± = ± i a and using the fact that sin i a = i sinh a , we get
n = 1 n 2 + a 2 = π a coth π a
The rest is just a little more algebra.
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