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Question Number 195952 by mnjuly1970 last updated on 13/Aug/23
Ω = Σ_(m=1) ^∞ Σ_(n=1) ^∞ (((−1)^( n+1) )/(m^2 n + mn^( 2) )) = ? −−−−−
$$ \\ $$$$\:\:\:\:\Omega\:=\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\:{n}+\mathrm{1}} }{{m}^{\mathrm{2}} {n}\:+\:{mn}^{\:\mathrm{2}} }\:\:=\:?\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:−−−−− \\ $$
Answered by witcher3 last updated on 14/Aug/23
(1/(mn(n+m)))=(1/m^2 )((1/n)−(1/(n+m))) A=Σ_(m≥1) (1/m^2 )Σ_(n≥1) (((−1)^(n−1) )/n)=((π^2 ln(2))/6) B=Σ_(m≥1) (1/m^2 )Σ_(n≥1) (((−1)^(n−1) )/(n+m)) =Σ_(m≥1) (1/m^2 )Σ_(n≥1) (−1)^(n−1) ∫_0 ^1 x^(n+m−1) dx =∫_0 ^1 Σ(x^m /m^2 )Σ_(n≥1) (−x)^(n−1) dx =∫_0 ^1 Li_2 (x).(dx/(1+x)) =∫_0 ^1 ((Li_2 (x))/(1+x))=ln(2)ζ(2)+∫_0 ^1 ((ln(1−x)ln(1+x))/x)dx =ln(2)ζ(2)+(1/4)∫_0 ^1 (((ln(1−x^2 ))^2 −ln^2 (((1−x)/(1+x))))/x)dx_(=I) I=(1/8)∫_0 ^1 ((ln^2 (1−t))/t)dt−(1/4)∫_0 ^1 ((2ln^2 (t))/((1−t^2 )))+ζ(2)ln(2) =(1/8)∫_0 ^1 ((ln^2 (t))/(1−t))dt−(1/2)∫_0 ^1 ((ln^2 (t))/(1−t^2 ))dt+ζ(2)ln(2) ∫_0 ^1 t^n ln^2 (t)dt=((Γ(3))/((n+1)^3 )) =(2/(8(1+n)^3 ))−Σ_(n≥0) (1/((1+2n)^3 ))+((π^2 ln(2))/6) =(1/4)ζ(3)−((7/8)ζ(3))=−(5/8)ζ(3) Ω=A−B=(π^2 /6)ln(2)−((π^2 /6)ln(2)−(5/8)ζ(3)) Σ_(n,m≥1) (((−1)^(n−1) )/(nm^2 +mn^2 ))=(5/8)ζ(3)
$$\frac{\mathrm{1}}{\mathrm{mn}\left(\mathrm{n}+\mathrm{m}\right)}=\frac{\mathrm{1}}{\mathrm{m}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{m}}\right) \\ $$$$\mathrm{A}=\underset{\mathrm{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{m}^{\mathrm{2}} }\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}=\frac{\pi^{\mathrm{2}} \mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{6}} \\ $$$$\mathrm{B}=\underset{\mathrm{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{m}^{\mathrm{2}} }\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}+\mathrm{m}} \\ $$$$=\underset{\mathrm{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{m}^{\mathrm{2}} }\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{n}+\mathrm{m}−\mathrm{1}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \Sigma\frac{\mathrm{x}^{\mathrm{m}} }{\mathrm{m}^{\mathrm{2}} }\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{x}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right).\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}}=\mathrm{ln}\left(\mathrm{2}\right)\zeta\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\mathrm{ln}\left(\mathrm{2}\right)\zeta\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{ln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\right)^{\mathrm{2}} −\mathrm{ln}^{\mathrm{2}} \left(\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}\right)}{\mathrm{x}}\mathrm{dx}_{=\mathrm{I}} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{t}\right)}{\mathrm{t}}\mathrm{dt}−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2ln}^{\mathrm{2}} \left(\mathrm{t}\right)}{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)}+\zeta\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)}{\mathrm{1}−\mathrm{t}}\mathrm{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\mathrm{dt}+\zeta\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}^{\mathrm{n}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)\mathrm{dt}=\frac{\Gamma\left(\mathrm{3}\right)}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{8}\left(\mathrm{1}+\mathrm{n}\right)^{\mathrm{3}} }−\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2n}\right)^{\mathrm{3}} }+\frac{\pi^{\mathrm{2}} \mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{6}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{3}\right)−\left(\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)\right)=−\frac{\mathrm{5}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$$$\Omega=\mathrm{A}−\mathrm{B}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\mathrm{ln}\left(\mathrm{2}\right)−\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\mathrm{ln}\left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)\right) \\ $$$$\underset{\mathrm{n},\mathrm{m}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{nm}^{\mathrm{2}} +\mathrm{mn}^{\mathrm{2}} }=\frac{\mathrm{5}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$
Answered by qaz last updated on 14/Aug/23
Ω=Σ_(m,n=1) ^∞ (((−1)^(n+1) )/(nm(n+m)))=Σ_(m,n=1) ^∞ (((−1)^(n+1) )/(nm))∫_0 ^1 x^(n+m−1) dx =−∫_0 ^1 ((ln(1+x)ln(1−x))/x)dx=(5/8)ζ(3)
$$\Omega=\underset{{m},{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{nm}\left({n}+{m}\right)}=\underset{{m},{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{nm}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+{m}−\mathrm{1}} {dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}=\frac{\mathrm{5}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$

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