n-1-1-n-2-k-2-k-

Question Number 133075 by LUFFY last updated on 18/Feb/21

\underset{n = 1}{\sum^{\infty}} (\frac{1}{n^{2} + k^{2}}) k \in ? ? ? ? ? ?

Answered by Dwaipayan Shikari last updated on 18/Feb/21

\frac{1}{2 i k} \underset{n = 1}{\sum^{\infty}} \frac{1}{n - i k} - \frac{1}{n + i k}

= \frac{1}{2 i k} ψ (i k) - ψ (- i k) = \frac{1}{2 i k} (ψ (i k) - ψ (1 - i k) + \frac{1}{i k})

= - \frac{π}{2 i k} c o t (π i k) - \frac{1}{2 k^{2}} = - \frac{π i}{2 i k} . \frac{e^{- π k} + e^{π k}}{e^{- π k} - e^{π k}} - \frac{1}{2 k^{2}}

= \frac{π}{2 k} . \frac{e^{π k} + e^{- π k}}{e^{π k} - e^{- π k}} - \frac{1}{2 k^{2}} = \frac{π}{2 k} c o t h (π k) - \frac{1}{2 k^{2}}

Commented by LUFFY last updated on 19/Feb/21

k b e l o n g s t o ? ? ? ? ? ? ?

Commented by Dwaipayan Shikari last updated on 19/Feb/21

k b e l o n g s t o a n y n u m b e r e x c e p t k = 0