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# Question-134373

Question Number 134373 by mnjuly1970 last updated on 02/Mar/21
Answered by Dwaipayan Shikari last updated on 02/Mar/21
$${I}\left({a}\right)=\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}^{\mathrm{4}} }+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}{a}}{coth}\left(\pi{a}\right)−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} } \\$$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{−\mathrm{2}{a}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\partial}{\partial{a}}\left(\frac{\pi}{\mathrm{2}{a}}{coth}\left(\pi{a}\right)−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }\right) \\$$$$=−\frac{\pi}{\mathrm{2}{a}^{\mathrm{2}} }{coth}\left(\pi{a}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{2}{a}}{csch}^{\mathrm{2}} \left(\pi{a}\right)+\frac{\mathrm{1}}{{a}^{\mathrm{3}} } \\$$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}{a}^{\mathrm{3}} }{coth}\left(\pi{a}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} }{csch}^{\mathrm{2}} \left(\pi{a}\right)−\frac{\mathrm{1}}{{a}^{\mathrm{4}} } \\$$$${I}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}{coth}\left(\pi\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}{csch}^{\mathrm{2}} \left(\pi\right) \\$$
Commented by mnjuly1970 last updated on 03/Mar/21
$$\:\:\:\:\:\:\:\:\:{thanks}\:{alot}\:{mr}\:{payan}… \\$$