Question Number 222812 by MrGaster last updated on 08/Jul/25
Answered by MrGaster last updated on 08/Jul/25
![x=sinh u⇒dx=cosh udu,(√(1+u^2 ))=cosh u u∣_(x=0) =0,u∣_(x=(1/2)) =sinh^(−1) ((1/2)) u=sinh^(−1) ((1/2))ln(((1+(√5))/2)) ∫_0 ^(1/2) ((ln(2x))/( (√(1+x^2 ))))dx=∫_0 ^v ln(2 sinh u)du ln(2 sinh u)=ln(e^u −e^(−u) )=u+ln(1−e^(−2u) ) ln(2 sinh u)du=∫_0 ^v udu+∫_0 ^v ln(1−e^(2u) )du ∫_0 ^v udu=(u^2 /2)∣_0 ^v =(v^2 /0) w=e^(−u) ⇒du=−(dw/w),u∣_0 ^v ⇒w∣_1 ^e^(−u) ∫_0 ^v ln(1−e^(−2u) )du=∫_1 ^v ln(1−w^2 )(−(dw/w))=∫_e^(−v) ^1 ((ln(1−w^2 ))/w)dw ln(1−w^2 )=−Σ_(n=1) ^∞ ∀∣w∣<1 ∫((ln(1−w^2 ))/w)dw=−∫Σ_(n=1) ^∞ (w^(2n+1) /n)dw=−Σ_(n=1) ^∞ (1/n)∫w^(2n−1) dw=−Σ_(n=1) ^∞ (1/n) (w^(2n) /(2n))+C=−(1/2)Σ_(n=1) ^∞ (w^(2n) /n^2 )+C ∫_e^(−v) ^1 ((ln(1−w^2 ))/w)dw=[−(1/2)Σ_(n=0) ^∞ (w^(2n) /n^2 )]_e^(−v) ^1 =−(1/2)Σ_(n=1) ^∞ (1/n^2 )+(1/2)Σ_(n=1) ^∞ (((e^(−2v) )^n )/n^2 )=−(1/2)ζ(2)+(1/2)Li_2 (e^(−2v) ) ζ(2)=(π^2 /6) ∫_0 ^v ln(1−e^(2u) )du=−(π^2 /(12))+(1/2)Li_2 (e^(−2v) ) I=(v^2 /2)+(−(π^2 /(12))+(1/2)Li_2 (e^(−2v) ))=(v^2 /2)−(π^2 /(12))+(1/2)Li_2 (e^(−2v) ) φ=((1+(√5))/2),v=lnφ,e^(−2v) =φ^(−2) Li_2 (φ^(−2) )=Li_2 ((φ^(−1) )^2 )=2[Li_2 (φ^(−1) )+Li_2 (−φ^(−1) )] Li_2 (φ^(−1) )=(π^2 /(10))−(ln φ)^2 Li_2 (−φ^(−1) )+Li_2 (φ^(−2) )=−(1/2)(ln φ)^2 A=Li_2 (φ^(−1) ),B=Li_2 (−φ^(−1) ),C=Li_2 (φ^(−2) ) C=2(A+B) B+C=−(1/2)(ln φ)^2 2A+3B=−(1/2)(ln φ)^2 2((π^2 /(10))−(ln φ)^2 )+3B=−(1/2)(ln φ)^2 (π^2 /5)−2(ln φ)^2 +3B=−(1/2)(ln φ)^2 (π^2 /5)−2(ln φ)^2 +3B=−(1/2)(ln φ)^2 (π^2 /5)−2 (ln φ)^2 +3B=−(1/2)(ln φ)^2 3B=−(π^2 /5)+2(ln φ)^2 −(1/2)(ln φ)^2 =−(π^2 /5)+(3/2)(ln φ)^2 B=−(π^2 /(15))+(1/2)(ln φ)^2 C=2A+2B=2((π^2 /(10))−(lnφ)^2 )+2(−(π^2 /(15))+(1/2)(ln φ)^2 )=(π^2 /(15))−2(ln φ)^2 −((2π^2 )/(15))+(ln φ)^2 =(π^2 /(15))−(ln φ)^2 I=(v^2 /2)−(π^2 /(12))+(1/2)((π^2 /(15))−v^2 )=(v^2 /2)−(π^2 /(12))+(π^2 /(30))−(v^2 /2)=(π^2 /(15))+(π^2 /(30))=π^2 (−(1/(12))+(1/(30)))=π^2 (−(1/(20)))=−(π^2 /(20))](https://www.tinkutara.com/question/Q222813.png)
$${x}=\mathrm{sinh}\:{u}\Rightarrow{dx}=\mathrm{cosh}\:{udu},\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }=\mathrm{cosh}\:{u} \\ $$$${u}\mid_{{x}=\mathrm{0}} =\mathrm{0},{u}\mid_{{x}=\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${u}=\mathrm{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}=\int_{\mathrm{0}} ^{{v}} \mathrm{ln}\left(\mathrm{2}\:\mathrm{sinh}\:{u}\right){du} \\ $$$$\mathrm{ln}\left(\mathrm{2}\:\mathrm{sinh}\:{u}\right)=\mathrm{ln}\left({e}^{{u}} −{e}^{−{u}} \right)={u}+\mathrm{ln}\left(\mathrm{1}−{e}^{−\mathrm{2}{u}} \right) \\ $$$$\mathrm{ln}\left(\mathrm{2}\:\mathrm{sinh}\:{u}\right){du}=\int_{\mathrm{0}} ^{{v}} {udu}+\int_{\mathrm{0}} ^{{v}} \mathrm{ln}\left(\mathrm{1}−{e}^{\mathrm{2}{u}} \right){du} \\ $$$$\int_{\mathrm{0}} ^{{v}} {udu}=\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{0}} ^{{v}} =\frac{{v}^{\mathrm{2}} }{\mathrm{0}} \\ $$$${w}={e}^{−{u}} \Rightarrow{du}=−\frac{{dw}}{{w}},{u}\mid_{\mathrm{0}} ^{{v}} \Rightarrow{w}\mid_{\mathrm{1}} ^{{e}^{−{u}} } \\ $$$$\int_{\mathrm{0}} ^{{v}} \mathrm{ln}\left(\mathrm{1}−{e}^{−\mathrm{2}{u}} \right){du}=\int_{\mathrm{1}} ^{{v}} \mathrm{ln}\left(\mathrm{1}−{w}^{\mathrm{2}} \right)\left(−\frac{{dw}}{{w}}\right)=\int_{{e}^{−{v}} } ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{w}^{\mathrm{2}} \right)}{{w}}{dw} \\ $$$$\mathrm{ln}\left(\mathrm{1}−{w}^{\mathrm{2}} \right)=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\forall\mid{w}\mid<\mathrm{1} \\ $$$$\int\frac{\mathrm{ln}\left(\mathrm{1}−{w}^{\mathrm{2}} \right)}{{w}}{dw}=−\int\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{w}^{\mathrm{2}{n}+\mathrm{1}} }{{n}}{dw}=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int{w}^{\mathrm{2}{n}−\mathrm{1}} {dw}=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\:\frac{{w}^{\mathrm{2}{n}} }{\mathrm{2}{n}}+{C}=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{w}^{\mathrm{2}{n}} }{{n}^{\mathrm{2}} }+{C} \\ $$$$\int_{{e}^{−{v}} } ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{w}^{\mathrm{2}} \right)}{{w}}{dw}=\left[−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{w}^{\mathrm{2}{n}} }{{n}^{\mathrm{2}} }\right]_{{e}^{−{v}} } ^{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({e}^{−\mathrm{2}{v}} \right)^{{n}} }{{n}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2}}\zeta\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left({e}^{−\mathrm{2}{v}} \right) \\ $$$$\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\int_{\mathrm{0}} ^{{v}} \mathrm{ln}\left(\mathrm{1}−{e}^{\mathrm{2}{u}} \right){du}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left({e}^{−\mathrm{2}{v}} \right) \\ $$$${I}=\frac{{v}^{\mathrm{2}} }{\mathrm{2}}+\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left({e}^{−\mathrm{2}{v}} \right)\right)=\frac{{v}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left({e}^{−\mathrm{2}{v}} \right) \\ $$$$\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},{v}=\mathrm{ln}\phi,{e}^{−\mathrm{2}{v}} =\phi^{−\mathrm{2}} \\ $$$$\mathrm{Li}_{\mathrm{2}} \left(\phi^{−\mathrm{2}} \right)=\mathrm{Li}_{\mathrm{2}} \left(\left(\phi^{−\mathrm{1}} \right)^{\mathrm{2}} \right)=\mathrm{2}\left[\mathrm{Li}_{\mathrm{2}} \left(\phi^{−\mathrm{1}} \right)+\mathrm{Li}_{\mathrm{2}} \left(−\phi^{−\mathrm{1}} \right)\right] \\ $$$$\mathrm{Li}_{\mathrm{2}} \left(\phi^{−\mathrm{1}} \right)=\frac{\pi^{\mathrm{2}} }{\mathrm{10}}−\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$$\mathrm{Li}_{\mathrm{2}} \left(−\phi^{−\mathrm{1}} \right)+\mathrm{Li}_{\mathrm{2}} \left(\phi^{−\mathrm{2}} \right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$${A}=\mathrm{Li}_{\mathrm{2}} \left(\phi^{−\mathrm{1}} \right),{B}=\mathrm{Li}_{\mathrm{2}} \left(−\phi^{−\mathrm{1}} \right),{C}=\mathrm{Li}_{\mathrm{2}} \left(\phi^{−\mathrm{2}} \right) \\ $$$${C}=\mathrm{2}\left({A}+{B}\right) \\ $$$${B}+{C}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{A}+\mathrm{3}{B}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$$\mathrm{2}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{10}}−\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \right)+\mathrm{3}{B}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{5}}−\mathrm{2}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} +\mathrm{3}{B}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{5}}−\mathrm{2}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} +\mathrm{3}{B}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{5}}−\mathrm{2}\:\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} +\mathrm{3}{B}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$$\mathrm{3}{B}=−\frac{\pi^{\mathrm{2}} }{\mathrm{5}}+\mathrm{2}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} =−\frac{\pi^{\mathrm{2}} }{\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$${B}=−\frac{\pi^{\mathrm{2}} }{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$${C}=\mathrm{2}{A}+\mathrm{2}{B}=\mathrm{2}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{10}}−\left(\mathrm{ln}\phi\right)^{\mathrm{2}} \right)+\mathrm{2}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \right)=\frac{\pi^{\mathrm{2}} }{\mathrm{15}}−\mathrm{2}\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} −\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{15}}+\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{15}}−\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \\ $$$${I}=\frac{{v}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{15}}−{v}^{\mathrm{2}} \right)=\frac{{v}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{30}}−\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{\pi^{\mathrm{2}} }{\mathrm{15}}+\frac{\pi^{\mathrm{2}} }{\mathrm{30}}=\pi^{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{30}}\right)=\pi^{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{20}}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{20}} \\ $$