# nice-calculus-prove-that-i-0-cos-pix-e-2pi-x-1-dx-2-2-8-ii-compute-n-1-n-4-9n-2-10-

Question Number 134272 by mnjuly1970 last updated on 01/Mar/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{nice}\:\:\:\mathrm{calculus} \\$$$$\:\:\:\:\mathrm{prove}\:\mathrm{that}:\::\: \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{i}\:::\:\: =\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left(\pi{x}\right)}{{e}^{\mathrm{2}\pi\sqrt{{x}}\:} −\mathrm{1}}\:{dx}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}\:}{\mathrm{8}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ii}::\:{compute}:\:\:\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{10}}\:=? \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{m}.{n}… \\$$
Answered by Dwaipayan Shikari last updated on 01/Mar/21
Commented by Dwaipayan Shikari last updated on 01/Mar/21
$${Ramanujan}'{s}\:{letter}\: \\$$
Commented by mnjuly1970 last updated on 02/Mar/21
$$\:\:{thank}\:{you}.. \\$$
Answered by Dwaipayan Shikari last updated on 01/Mar/21
$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{10}}+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{10}} \\$$$$=\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{41}}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +\frac{\mathrm{9}−\sqrt{\mathrm{41}}}{\mathrm{2}}\right)}−\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +\frac{\mathrm{9}+\sqrt{\mathrm{41}}}{\mathrm{2}}\right)} \\$$$$=\frac{\mathrm{1}}{\mathrm{10}}+\sqrt{\frac{\mathrm{2}}{\mathrm{41}}}\left(\frac{\pi}{\:\sqrt{\mathrm{9}−\sqrt{\mathrm{41}}}}{coth}\left(\sqrt{\frac{\mathrm{9}−\sqrt{\mathrm{41}}}{\mathrm{2}}}\pi\right)−\frac{\pi}{\:\sqrt{\mathrm{9}+\sqrt{\mathrm{41}}}}{coth}\left(\sqrt{\frac{\mathrm{9}+\sqrt{\mathrm{41}}}{\mathrm{2}}}\pi\right)\right)+\frac{\mathrm{2}}{\:\sqrt{\mathrm{41}}}\left(\frac{\mathrm{1}}{\mathrm{9}+\sqrt{\mathrm{41}}}−\frac{\mathrm{1}}{\mathrm{9}−\sqrt{\mathrm{41}}}\right) \\$$$$=\sqrt{\frac{\mathrm{2}}{\mathrm{41}}}\left(\frac{\pi}{\:\sqrt{\mathrm{9}−\sqrt{\mathrm{41}}}}{coth}\left(\sqrt{\frac{\mathrm{9}−\sqrt{\mathrm{41}}}{\mathrm{2}}}\pi\right)−\frac{\pi}{\:\sqrt{\mathrm{9}+\sqrt{\mathrm{41}}}}{coth}\left(\sqrt{\frac{\mathrm{9}+\sqrt{\mathrm{41}}}{\mathrm{2}}}\pi\right)\right) \\$$$${Using}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}{x}}{coth}\left(\pi{x}\right)−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} } \\$$