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Question Number 134272 by mnjuly1970 last updated on 01/Mar/21

nice calculus prove that: : i :: =∫_0 ^( ∞) ((cos(πx))/(e^(2π(√x) ) −1)) dx=((2−(√2) )/8) ii:: compute: Σ_(n=−∞) ^∞ (1/(n^4 +9n^2 +10)) =? ...m.n...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{nice}\:\:\:\mathrm{calculus} \\ $$$$\:\:\:\:\mathrm{prove}\:\mathrm{that}:\::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{i}\:::\:\: =\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left(\pi{x}\right)}{{e}^{\mathrm{2}\pi\sqrt{{x}}\:} −\mathrm{1}}\:{dx}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}\:}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ii}::\:{compute}:\:\:\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{10}}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}... \\ $$

Answered by Dwaipayan Shikari last updated on 01/Mar/21

Commented by Dwaipayan Shikari last updated on 01/Mar/21

Ramanujan′s letter

$${Ramanujan}'{s}\:{letter}\: \\ $$

Commented by mnjuly1970 last updated on 02/Mar/21

thank you..

$$\:\:{thank}\:{you}.. \\ $$

Answered by Dwaipayan Shikari last updated on 01/Mar/21

Σ_(n=−∞) ^∞ (1/(n^4 +9n^2 +10))=(1/(10))+2Σ_(n=1) ^∞ (1/(n^4 +9n^2 +10)) =(1/(10))+(2/( (√(41))))Σ_(n=1) ^∞ (1/((n^2 +((9−(√(41)))/2))))−(1/((n^2 +((9+(√(41)))/2)))) =(1/(10))+(√(2/(41)))((π/( (√(9−(√(41))))))coth((√((9−(√(41)))/2))π)−(π/( (√(9+(√(41))))))coth((√((9+(√(41)))/2))π))+(2/( (√(41))))((1/(9+(√(41))))−(1/(9−(√(41))))) =(√(2/(41)))((π/( (√(9−(√(41))))))coth((√((9−(√(41)))/2))π)−(π/( (√(9+(√(41))))))coth((√((9+(√(41)))/2))π)) Using Σ_(n=1) ^∞ (1/(n^2 +x^2 ))=(π/(2x))coth(πx)−(1/(2x^2 ))

$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{10}}+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{10}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{41}}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +\frac{\mathrm{9}−\sqrt{\mathrm{41}}}{\mathrm{2}}\right)}−\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +\frac{\mathrm{9}+\sqrt{\mathrm{41}}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}+\sqrt{\frac{\mathrm{2}}{\mathrm{41}}}\left(\frac{\pi}{\:\sqrt{\mathrm{9}−\sqrt{\mathrm{41}}}}{coth}\left(\sqrt{\frac{\mathrm{9}−\sqrt{\mathrm{41}}}{\mathrm{2}}}\pi\right)−\frac{\pi}{\:\sqrt{\mathrm{9}+\sqrt{\mathrm{41}}}}{coth}\left(\sqrt{\frac{\mathrm{9}+\sqrt{\mathrm{41}}}{\mathrm{2}}}\pi\right)\right)+\frac{\mathrm{2}}{\:\sqrt{\mathrm{41}}}\left(\frac{\mathrm{1}}{\mathrm{9}+\sqrt{\mathrm{41}}}−\frac{\mathrm{1}}{\mathrm{9}−\sqrt{\mathrm{41}}}\right) \\ $$$$=\sqrt{\frac{\mathrm{2}}{\mathrm{41}}}\left(\frac{\pi}{\:\sqrt{\mathrm{9}−\sqrt{\mathrm{41}}}}{coth}\left(\sqrt{\frac{\mathrm{9}−\sqrt{\mathrm{41}}}{\mathrm{2}}}\pi\right)−\frac{\pi}{\:\sqrt{\mathrm{9}+\sqrt{\mathrm{41}}}}{coth}\left(\sqrt{\frac{\mathrm{9}+\sqrt{\mathrm{41}}}{\mathrm{2}}}\pi\right)\right) \\ $$$${Using}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}{x}}{coth}\left(\pi{x}\right)−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$

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