Question Number 147411 by mnjuly1970 last updated on 20/Jul/21
Answered by mnjuly1970 last updated on 20/Jul/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\::\overset{\sqrt{{x}}\::=\:{y}} {=}\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{2}{y}\:{dy}}{\mathrm{1}+{e}^{\:{y}} }\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ydy}}{\mathrm{1}+{e}^{\:{y}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\::=\:\mathrm{2}\:\eta\:\left(\mathrm{2}\right)\:\Gamma\:\left(\mathrm{2}\right)=\mathrm{2}.\frac{\pi^{\:\mathrm{2}} }{\mathrm{12}}\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by tabata last updated on 20/Jul/21
$${sir}\:{whats}\:{the}\:{mean}\:\left(\eta\right) \\ $$
Commented by mnjuly1970 last updated on 20/Jul/21
$$\:\:\eta\:\left({s}\:\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\:{n}−\mathrm{1}} }{{n}^{\:{s}} }\:\:\:\:{eta}\:{function} \\ $$$$\:\:\:\eta\:\left({s}\:\right)=\:\left(\mathrm{1}−\mathrm{2}^{\:\mathrm{1}−{s}} \right)\:\zeta\:\left({s}\right) \\ $$$$\:\:\:\:\:\eta\:\left(\mathrm{2}\right)\:=\:\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−\mathrm{2}} \right)\:\zeta\:\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\zeta\left(\mathrm{2}\right)=\frac{\pi^{\:\mathrm{2}} }{\mathrm{12}}\: \\ $$$$\:\:\:\zeta\:\left({s}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}} }\:\:\:\:{zeta}\:{function}… \\ $$
Commented by Olaf_Thorendsen last updated on 20/Jul/21
$$\mathrm{the}\:\mathrm{Dirichlet}\:\mathrm{eta}\:\mathrm{function} \\ $$
Answered by Olaf_Thorendsen last updated on 20/Jul/21
$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{e}^{\sqrt{{x}}} } \\ $$$$\mathrm{I}\left({x}\right)\:=\:\int\frac{{dx}}{\mathrm{1}+{e}^{\sqrt{{x}}} } \\ $$$$ \\ $$$$\mathrm{I}\left({x}\right)\:=\:\int\left(\mathrm{1}−\frac{{e}^{\sqrt{{x}}} }{\mathrm{1}+{e}^{\sqrt{{x}}} }\right)\:{dx} \\ $$$$\mathrm{I}\left({x}\right)\:=\:\int\left(\mathrm{1}−\mathrm{2}\sqrt{{x}}\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{e}^{\sqrt{{x}}} }{\mathrm{1}+{e}^{\sqrt{{x}}} }\right)\:{dx} \\ $$$$\mathrm{I}\left({x}\right)\:=\:{x}−\mathrm{2}\sqrt{{x}}\mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{\sqrt{{x}}} \right)+\int\frac{\mathrm{ln}\left(\mathrm{1}+{e}^{\sqrt{{x}}} \right)}{\:\sqrt{{x}}}\:{dx} \\ $$$$\mathrm{I}\left({x}\right)\:=\:{x}−\mathrm{2}\sqrt{{x}}\mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{\sqrt{{x}}} \right)−\mathrm{2Li}_{\mathrm{2}} \left(−{e}^{\sqrt{{x}}} \right) \\ $$$$\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)\:=\:\eta\left(\mathrm{2}\right)\:=\:\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−\mathrm{2}} \right)\zeta\left(\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\zeta\left(\mathrm{2}\right) \\ $$$$\mathrm{Li}\left(−{z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left({z}^{\mathrm{2}} \right)−\mathrm{Li}_{\mathrm{2}} \left({z}\right) \\ $$$$\Rightarrow\:\mathrm{Li}\left(−{e}^{\sqrt{{x}}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left({e}^{\mathrm{2}\sqrt{{x}}} \right)−\mathrm{Li}_{\mathrm{2}} \left(\sqrt{{x}}\right) \\ $$$$\mathrm{Li}_{{s}} \left({e}^{\mu} \right)\:\underset{\infty} {\sim}\:−\frac{\mu^{\mathrm{2}} }{\Gamma\left({s}+\mathrm{1}\right)} \\ $$$$\Rightarrow\:\mathrm{Li}_{\mathrm{2}} \left(−{e}^{\sqrt{{x}}} \right)\:\underset{\infty} {\sim}\:−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{4}{x}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}}\:=\:−\frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{I}\left({x}\right)\:\underset{\infty} {\sim}\:{x}−\mathrm{2}\sqrt{{x}}×\sqrt{{x}−}\mathrm{2}\left(−\frac{{x}}{\mathrm{2}}\right)\:\rightarrow\:\mathrm{0} \\ $$$$\mathrm{I}\left({x}\right)\:\underset{\mathrm{0}} {\sim}\:−\mathrm{2Li}_{\mathrm{2}} \left(−\mathrm{1}\right)\:=\:−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\zeta\left(\mathrm{2}\right)\:=\:−\zeta\left(\mathrm{2}\right) \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{e}^{\sqrt{{x}}} }\:=\:\mathrm{I}_{\infty} −\mathrm{I}\left(\mathrm{0}\right)\:=\:\mathrm{0}−\left(−\zeta\left(\mathrm{2}\right)\right) \\ $$$$\mathrm{I}\:=\:\zeta\left(\mathrm{2}\right)\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$
Commented by Tawa11 last updated on 03/Aug/21
$$\mathrm{Great} \\ $$