Show that ∑ n = 0 ∞ 1 n 2 + a 2 = π 2 a coth ⁡ π a + 1 2 a 2 , a &gt; 0

Question

Show that      ∑          n      =      0              ∞            1                  n        2            +              a        2              =      π          2      a        coth  ⁡  π  a  +      1          2              a        2              ,  a  &amp;gt;  0

Camden Stanton · Accepted Answer

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.

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

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