Question Number 134373 by mnjuly1970 last updated on 02/Mar/21
![](https://www.tinkutara.com/question/19195.png)
Answered by Dwaipayan Shikari last updated on 02/Mar/21
![I(a)=Σ_(n=−∞) ^∞ (1/((n^2 +a^2 )^2 ))=(1/a^4 )+2Σ_(n=1) ^∞ (1/((n^2 +a^2 )^2 )) Σ_(n=1) ^∞ (1/(n^2 +a^2 ))=(π/(2a))coth(πa)−(1/(2a^2 )) Σ_(n=1) ^∞ ((−2a)/((n^2 +a^2 )^2 ))=(∂/∂a)((π/(2a))coth(πa)−(1/(2a^2 ))) =−(π/(2a^2 ))coth(πa)−(π^2 /(2a))csch^2 (πa)+(1/a^3 ) Σ_(n=1) ^∞ (2/((n^2 +a^2 )^2 ))=(π/(2a^3 ))coth(πa)−(π^2 /(2a^2 ))csch^2 (πa)−(1/a^4 ) I(1)=(π/2)coth(π)−(π^2 /2)csch^2 (π)](https://www.tinkutara.com/question/Q134374.png)
$${I}\left({a}\right)=\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}^{\mathrm{4}} }+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}{a}}{coth}\left(\pi{a}\right)−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{−\mathrm{2}{a}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\partial}{\partial{a}}\left(\frac{\pi}{\mathrm{2}{a}}{coth}\left(\pi{a}\right)−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }\right) \\ $$$$=−\frac{\pi}{\mathrm{2}{a}^{\mathrm{2}} }{coth}\left(\pi{a}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{2}{a}}{csch}^{\mathrm{2}} \left(\pi{a}\right)+\frac{\mathrm{1}}{{a}^{\mathrm{3}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}{a}^{\mathrm{3}} }{coth}\left(\pi{a}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} }{csch}^{\mathrm{2}} \left(\pi{a}\right)−\frac{\mathrm{1}}{{a}^{\mathrm{4}} } \\ $$$${I}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}{coth}\left(\pi\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}{csch}^{\mathrm{2}} \left(\pi\right) \\ $$
Commented by mnjuly1970 last updated on 03/Mar/21
![thanks alot mr payan...](https://www.tinkutara.com/question/Q134398.png)
$$\:\:\:\:\:\:\:\:\:{thanks}\:{alot}\:{mr}\:{payan}… \\ $$