Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 128806 by LUFFY last updated on 10/Jan/21

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

Answered by Dwaipayan Shikari last updated on 10/Jan/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+k^2}=\frac{1}{2ik}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n-ik}-\frac{1}{n+ik}$$$$=\frac{1}{2ik}(\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{n+ik}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{n-ik})$$$$=\frac{1}{2ik}(\psi (1+ik)-\psi (1-ik))$$$$=\frac{1}{2ik}(\psi \left(ik\right)+\frac{1}{ik}-\psi (1-ik))=-\frac{\psi (1-ik)-\psi \left(ik\right)}{2ik}-\frac{1}{2k^2}$$$$=-\frac{\pi cot\left(\pi ki\right)}{2ki}-\frac{1}{2k^2}=-\frac{\pi }{2k}\left(\frac{e^{\pi k}+e^{-\pi k}}{e^{-\pi k}-e^{\pi k}}\right)-\frac{1}{2k^2}$$$$=\frac{\pi }{2k}coth\left(\pi k\right)-\frac{1}{2k^2}$$$$=\frac{\pi }{2k}+\frac{\pi }{k(e^{2\pi k}-1)}-\frac{1}{2k^2}$$

Answered by Olaf last updated on 10/Jan/21

$$k=0:\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$$$k\ne 0:\frac{\pi }{2}.\frac{\coth \left(k\pi \right)}{k}-\frac{1}{2k^2}$$

Commented by Dwaipayan Shikari last updated on 10/Jan/21

$$Yessir!Notvalidofk=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com