Question Number 220459 by mr W last updated on 12/May/25
$${find}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }=?\:\:\:\:\:\:\:\:\:\:\left({a}\in{R}\right) \\ $$
Answered by MrGaster last updated on 13/May/25
Commented by MrGaster last updated on 13/May/25
Solution 1
Answered by MrGaster last updated on 13/May/25
Commented by MrGaster last updated on 13/May/25
Solution (2)
Commented by mr W last updated on 13/May/25
$${thanks}! \\ $$
Answered by mr W last updated on 13/May/25
![Σ_(n=1) ^∞ (1/(n^2 +a^2 )) =(1/(2ai))Σ_(n=1) ^∞ ((1/(n−ai))−(1/(n+ai))) =(1/(2ai))[Σ_(n=1) ^∞ (1/n)−γ−ψ(1−ai)−Σ_(n=1) ^∞ (1/n)+γ+ψ(1+ai)] =(1/(2ai))[ψ(1+ai)−ψ(1−ai)] =(1/(2ai))[ψ(ai)+(1/(ai))−ψ(ai)−π cot (aπi)] =(1/(2ai))[(1/(ai))−π cot (aπi)] =(1/(2ai))((1/(ai))−((π coth aπ)/i)) =(1/(2a))(π coth aπ−(1/a)) replace a with ai, we get: Σ_(n=1) ^∞ (1/(n^2 −a^2 ))=(1/(2a))((1/a)−π cot aπ) (a∉ Z)](https://www.tinkutara.com/question/Q220479.png)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ai}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−{ai}}−\frac{\mathrm{1}}{{n}+{ai}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ai}}\left[\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\gamma−\psi\left(\mathrm{1}−{ai}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}+\gamma+\psi\left(\mathrm{1}+{ai}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ai}}\left[\psi\left(\mathrm{1}+{ai}\right)−\psi\left(\mathrm{1}−{ai}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ai}}\left[\psi\left({ai}\right)+\frac{\mathrm{1}}{{ai}}−\psi\left({ai}\right)−\pi\:\mathrm{cot}\:\left({a}\pi{i}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ai}}\left[\frac{\mathrm{1}}{{ai}}−\pi\:\mathrm{cot}\:\left({a}\pi{i}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ai}}\left(\frac{\mathrm{1}}{{ai}}−\frac{\pi\:\mathrm{coth}\:{a}\pi}{{i}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}}\left(\pi\:\mathrm{coth}\:{a}\pi−\frac{\mathrm{1}}{{a}}\right) \\ $$$$ \\ $$$${replace}\:{a}\:{with}\:{ai},\:{we}\:{get}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} −{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{a}}\left(\frac{\mathrm{1}}{{a}}−\pi\:\mathrm{cot}\:{a}\pi\right)\:\:\:\left({a}\notin\:{Z}\right) \\ $$