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Question Number 110910 by mathdave last updated on 31/Aug/20
mr M.N july 1970 the question you posted earlier here goes the solution ∫_0 ^(1/2) ((ln^2 (1−x))/x)dx
$${mr}\:{M}.{N}\:{july}\:\mathrm{1970}\:{the}\:{question}\:{you}\:{posted}\:{earlier}\:{here}\:{goes}\:{the}\:{solution} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$ \\ $$
Answered by mathdave last updated on 31/Aug/20
solution let I=∫_0 ^(1/2) ((ln(1−x)ln(1−x))/x)dx let u=ln(1−x) ,du=−(1/(1−x))dx,∫dv=∫((ln(1−x))/x)dx,v=−Li_2 (x) using IBP I=−ln(1−x)Li_2 (x)∣_0 ^(1/2) −∫_0 ^(1/2) ((Li_2 (x))/(1−x))dx I=ln((1/2))Li_2 ((1/2))−∫_0 ^(1/2) ((Li_2 (x))/(1−x))dx now solve ∫_0 ^(1/2) ((Li_2 (x))/(1−x))dx (let t=1−x,x=1−t,dx=−dt) ∵∫_0 ^(1/2) ((Li_2 (x))/(1−x))dx=−∫_0 ^(1/2) ((Li_2 (1−t))/t)dt=∫_(1/2) ^1 ((Li_2 (1−t))/t)dt recall that Li_2 (z)+Li_2 (1−z)=(π^2 /6)−ln(z)ln(1−z) ∵∫_(1/2) ^1 ((Li_2 (1−t))/t)dt=(π^2 /6)∫_(1/2) ^1 (1/t)dt−∫_(1/2) ^1 ((Li_2 (t))/t)dt−∫_(1/2) ^1 ((ln(1−t)ln(t))/t)dt let A=∫_(1/2) ^1 ((Li_2 (t))/t)dt=∫_(1/2) ^0 ((Li_2 (t))/t)dt+∫_0 ^1 ((Li_2 (t))/t)dt A=∫_0 ^1 ((Li_2 (t))/t)dt−∫_0 ^(1/2) ((Li_2 (t))/t)dt note that Li_((n+1)) (z)=∫_b ^a ((Li_n (z))/z)dz A=Li_3 (1)−Li_3 ((1/2))........(x) recall that Li_n (z)=Σ_(k=1) ^∞ (z^k /k^n ) when n=3,z=1 Li_3 (1)=Σ_(k=1) ^∞ (1/k^3 )=ζ(3) and when n=3,z=(1/(2 )) then Li_3 ((1/2))=Σ_(k=1) ^∞ (1/(k^3 2^k ))=(1/(24))(21ζ(3)+4ln^3 (2)−2π^2 ln(2)) A=Li_3 (1)−Li_3 ((1/2))=((ζ(3))/8)−((ln^3 (2))/6)+(1/(12))π^2 ln2.........(1) and again let B=∫_(1/2) ^1 ((ln(1−t)ln(t))/t)dt let u=lnt,du=(1/t),∫dv=∫((ln(1−t))/t)dt,v=−Li_2 (t) using IBP B=−Li_2 (t)ln(t)∣_(1/2) ^1 +∫_(1/2) ^1 ((Li_2 (t))/t)dt B=Li_2 ((1/2))ln((1/2))+∫_(1/2) ^1 ((Li_2 (t))/t)dt recall that ∫_(1/2) ^1 ((Li_2 (t))/t)dt=((ζ(3))/8)−((ln^3 (2))/6)+(π^2 /(12))ln2 ∵∫_(1/2) ^1 ((ln(1−t)ln(t))/t)dt=Li_2 ((1/2))ln((1/2))+((ζ(3))/8)−((ln^3 (2))/6)+(π^2 /(12))ln2 B=−(π^2 /(12))ln2+((ln^3 (2))/2)+((ζ(3))/8)−((ln^3 (2))/6)+(π^2 /(12))ln2 B=∫_(1/2) ^1 ((ln(1−t)ln(t))/t)dt=((ζ(3))/8)+((ln^3 (2))/3)......(2) but I_1 =∫_(1/2) ^1 ((Li_2 (1−t))/t)dt=((.π^2 )/6)∫_(1/2) ^1 (1/t)dt−A−B and I=−ln((1/2))Li_2 ((1/2))−I_1 so I_1 =−(π^2 /6)ln((1/2))−((ζ(3))/8)−((ln^3 (2))/3)−(π^2 /(12))ln2−((ζ(3))/8)−((ln^3 (2))/3) I_1 =(π^2 /(12))ln2−((ζ(3))/4)−((ln^3 (2))/6) but I=−ln((1/2))Li_2 ((1/2))−I_1 so I=ln2((π^2 /(12))−((ln^2 (2))/2))−((π^2 /(12))ln2−((ζ(3))/4)−((ln^3 (2))/6)) I=(π^2 /(12))ln2−((ln^3 (2))/2)−(π^2 /(12))ln2+((ζ(3))/4)+((ln^3 (2))/6) I=((ζ(3))/4)+((ln^3 (2))/6)−((ln^3 (2))/2)=((ζ(3))/4)+((ln^3 (2)−3ln^3 (2))/6) ∵I=((ζ(3))/4)−((ln^3 (2))/3) ∵∫_0 ^(1/2) ((ln^2 (1−x))/x)dx=(1/4)ζ(3)−(1/3)ln^3 (2) by mathdave(31/08/2020)
$${solution}\: \\ $$$${let}\:{I}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)\mathrm{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$${let}\:{u}=\mathrm{ln}\left(\mathrm{1}−{x}\right) \\ $$$$,{du}=−\frac{\mathrm{1}}{\mathrm{1}−{x}}{dx},\int{dv}=\int\frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx},{v}=−{Li}_{\mathrm{2}} \left({x}\right) \\ $$$${using}\:{IBP} \\ $$$${I}=−\mathrm{ln}\left(\mathrm{1}−{x}\right){Li}_{\mathrm{2}} \left({x}\right)\mid_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{Li}_{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$${I}=\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{Li}_{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$${now}\:{solve}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{Li}_{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}}{dx}\:\:\:\:\left({let}\:\right. \\ $$$$\left.{t}=\mathrm{1}−{x},{x}=\mathrm{1}−{t},{dx}=−{dt}\right) \\ $$$$\because\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{Li}_{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}}{dx}=−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{Li}_{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}{dt}=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}{dt} \\ $$$${recall}\:{that} \\ $$$${Li}_{\mathrm{2}} \left({z}\right)+{Li}_{\mathrm{2}} \left(\mathrm{1}−{z}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{ln}\left({z}\right)\mathrm{ln}\left(\mathrm{1}−{z}\right) \\ $$$$\because\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}{dt}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}}{dt}−\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt}−\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)\mathrm{ln}\left({t}\right)}{{t}}{dt} \\ $$$${let}\: \\ $$$${A}=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt}=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{0}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt} \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt}−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt} \\ $$$${note}\:{that}\:{Li}_{\left({n}+\mathrm{1}\right)} \left({z}\right)=\int_{{b}} ^{{a}} \frac{{Li}_{{n}} \left({z}\right)}{{z}}{dz} \\ $$$${A}={Li}_{\mathrm{3}} \left(\mathrm{1}\right)−{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)……..\left({x}\right) \\ $$$${recall}\:{that}\:{Li}_{{n}} \left({z}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{z}^{{k}} }{{k}^{{n}} }\:\:{when}\:\:{n}=\mathrm{3},{z}=\mathrm{1} \\ $$$${Li}_{\mathrm{3}} \left(\mathrm{1}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{3}} }=\zeta\left(\mathrm{3}\right) \\ $$$${and}\:{when}\:{n}=\mathrm{3},{z}=\frac{\mathrm{1}}{\mathrm{2}\:}\:\:\:{then} \\ $$$${Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{3}} \mathrm{2}^{{k}} }=\frac{\mathrm{1}}{\mathrm{24}}\left(\mathrm{21}\zeta\left(\mathrm{3}\right)+\mathrm{4ln}^{\mathrm{3}} \left(\mathrm{2}\right)−\mathrm{2}\pi^{\mathrm{2}} \mathrm{ln}\left(\mathrm{2}\right)\right) \\ $$$${A}={Li}_{\mathrm{3}} \left(\mathrm{1}\right)−{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{8}}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{12}}\pi^{\mathrm{2}} \mathrm{ln2}………\left(\mathrm{1}\right)\:\: \\ $$$${and}\:{again}\:{let}\:\:{B}=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)\mathrm{ln}\left({t}\right)}{{t}}{dt} \\ $$$${let} \\ $$$${u}=\mathrm{ln}{t},{du}=\frac{\mathrm{1}}{{t}},\int{dv}=\int\frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt},{v}=−{Li}_{\mathrm{2}} \left({t}\right) \\ $$$${using}\:{IBP} \\ $$$${B}=−{Li}_{\mathrm{2}} \left({t}\right)\mathrm{ln}\left({t}\right)\mid_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} +\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt} \\ $$$${B}={Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt}\:\:\:{recall}\:{that} \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt}=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{8}}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{6}}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{ln2} \\ $$$$\because\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)\mathrm{ln}\left({t}\right)}{{t}}{dt}={Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{8}}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{6}}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{ln2} \\ $$$${B}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{ln2}+\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{8}}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{6}}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{ln2} \\ $$$${B}=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{t}\right)\mathrm{ln}\left({t}\right)}{{t}}{dt}=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{8}}+\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{3}}……\left(\mathrm{2}\right) \\ $$$${but} \\ $$$${I}_{\mathrm{1}} =\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}{dt}=\frac{.\pi^{\mathrm{2}} }{\mathrm{6}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}}{dt}−{A}−{B} \\ $$$${and}\:\:{I}=−\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{I}_{\mathrm{1}} \:\:\:\:\:{so} \\ $$$${I}_{\mathrm{1}} =−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{8}}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{3}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{ln2}−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{8}}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$${I}_{\mathrm{1}} =\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{ln2}−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{6}} \\ $$$${but}\:\:{I}=−\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{I}_{\mathrm{1}} \:\:{so} \\ $$$${I}=\mathrm{ln2}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}\right)−\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{ln2}−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{6}}\right) \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{ln2}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\mathrm{ln2}+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}}+\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{6}} \\ $$$${I}=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}}+\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{6}}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{2}}=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}}+\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)−\mathrm{3ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{6}} \\ $$$$\because{I}=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}}−\frac{\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$$\because\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right) \\ $$$${by}\:{mathdave}\left(\mathrm{31}/\mathrm{08}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 31/Aug/20
taieballah anfasakom.peace beupon you .thank you sir...
$${taieballah}\:{anfasakom}.{peace} \\ $$$${beupon}\:{you}\:.{thank}\:{you}\:{sir}… \\ $$
Commented by I want to learn more last updated on 31/Aug/20
Weldone sir
$$\mathrm{Weldone}\:\mathrm{sir} \\ $$
Answered by mathmax by abdo last updated on 01/Sep/20
A =∫_0 ^(1/2) ((ln^2 (1−x))/x)dx ⇒A =_(1−x=u) ∫_1 ^(1/2) ((ln^2 u)/(1−u)) (−du) =∫_(1/2) ^1 ln^2 u(Σ_(n=0) ^∞ u^n )du =Σ_(n=0) ^∞ ∫_(1/2) ^1 u^n ln^2 u du U_n =∫_(1/2) ^1 u^n ln^2 (u)du =_(byparts) [(u^(n+1) /(n+1))ln^2 u]_(1/2) ^1 −∫_(1/2) ^1 (u^(n+1) /(n+1))×((2lnu)/u)du =−((ln^2 (2))/((n+1)2^(n+1) )) −(2/(n+1)) ∫_(1/2) ^1 u^n lnu du but ∫_(1/2) ^1 u^n lnu du =[(u^(n+1) /(n+1))lnu]_(1/2) ^1 −∫_(1/2) ^1 (u^(n+1) /(n+1))(du/u) =((ln2)/((n+1)2^(n+1) )) −(1/(n+1)) ∫_(1/2) ^(1 ) u^(n ) du =((ln(2))/((n+1)2^(n+1) ))−(1/((n+1)^2 )){1−(1/2^(n+1) )} ⇒ U_n =−((ln^2 2)/((n+1)2^(n+1) ))−((2ln2)/((n+1)^2 2^(n+1) )) +(2/((n+1)^3 ))(1−(1/2^(n+1) )) ⇒ A =−ln^2 2 Σ_(n=0) ^∞ (1/((n+1)2^(n+1) ))−2ln2 Σ_(n=0) ^∞ (1/((n+1)^2 2^(n+1) )) +2Σ_(n=0) ^∞ (1/((n+1)^3 )) −2 Σ_(n=0) ^∞ (1/((n+1)^3 2^(n+1) )) =−ln^2 2 Σ_(n=1) ^∞ (1/(n2^n ))−2ln2 Σ_(n=1) ^∞ (1/(n^2 2^n )) +2ξ(3)−2Σ_(n=1) ^∞ (1/(n^3 2^n )) =−ln^3 (2)+2ξ(3) −2ln(2) Σ_(n=1) ^∞ (1/(n^2 2^n )) −2 Σ_(n=1) ^∞ (1/(n^3 2^n )) we have Σ_(n=1) ^∞ (x^(n−1) /n) =−((ln(1−x))/x) ⇒ ∫Σ (x^(n−1) /n)dx =−∫((ln(1−x))/x)dx +c ⇒Σ_(n=1) ^∞ (x^n /(n^2 )) =−∫_0 ^x ((ln(1−t))/t) dt (c=0) ...be continued...
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:\Rightarrow\mathrm{A}\:=_{\mathrm{1}−\mathrm{x}=\mathrm{u}} \:\:\:\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{u}}{\mathrm{1}−\mathrm{u}}\:\left(−\mathrm{du}\right) \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\mathrm{ln}^{\mathrm{2}} \mathrm{u}\left(\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{u}^{\mathrm{n}} \:\right)\mathrm{du}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\mathrm{u}^{\mathrm{n}} \:\mathrm{ln}^{\mathrm{2}} \mathrm{u}\:\mathrm{du} \\ $$$$\mathrm{U}_{\mathrm{n}} =\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\mathrm{u}^{\mathrm{n}} \:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{u}\right)\mathrm{du}\:=_{\mathrm{byparts}} \:\:\:\left[\frac{\mathrm{u}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\mathrm{ln}^{\mathrm{2}} \mathrm{u}\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} −\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{\mathrm{u}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}×\frac{\mathrm{2lnu}}{\mathrm{u}}\mathrm{du} \\ $$$$=−\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\:−\frac{\mathrm{2}}{\mathrm{n}+\mathrm{1}}\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\mathrm{u}^{\mathrm{n}} \:\mathrm{lnu}\:\mathrm{du}\:\:\mathrm{but} \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\mathrm{u}^{\mathrm{n}} \:\mathrm{lnu}\:\mathrm{du}\:=\left[\frac{\mathrm{u}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\mathrm{lnu}\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} −\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{\mathrm{u}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\frac{\mathrm{du}}{\mathrm{u}}\:=\frac{\mathrm{ln2}}{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{n}+\mathrm{1}} } \\ $$$$−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}\:} \:\mathrm{u}^{\mathrm{n}\:} \mathrm{du}\:=\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\right\}\:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} =−\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{2}}{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{n}+\mathrm{1}} }−\frac{\mathrm{2ln2}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\:+\frac{\mathrm{2}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }\right)\:\Rightarrow \\ $$$$\mathrm{A}\:=−\mathrm{ln}^{\mathrm{2}} \mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{n}+\mathrm{1}} }−\mathrm{2ln2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{2}^{\mathrm{n}+\mathrm{1}} } \\ $$$$+\mathrm{2}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\:−\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} \:\mathrm{2}^{\mathrm{n}+\mathrm{1}} } \\ $$$$=−\mathrm{ln}^{\mathrm{2}} \mathrm{2}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n2}^{\mathrm{n}} }−\mathrm{2ln2}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} \mathrm{2}^{\mathrm{n}} }\:+\mathrm{2}\xi\left(\mathrm{3}\right)−\mathrm{2}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} \:\mathrm{2}^{\mathrm{n}} } \\ $$$$=−\mathrm{ln}^{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{2}\xi\left(\mathrm{3}\right)\:−\mathrm{2ln}\left(\mathrm{2}\right)\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} \mathrm{2}^{\mathrm{n}} }\:−\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} \:\mathrm{2}^{\mathrm{n}} } \\ $$$$\mathrm{we}\:\mathrm{have}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\:=−\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\:\Rightarrow \\ $$$$\int\Sigma\:\:\frac{\mathrm{x}^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\mathrm{dx}\:=−\int\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:+\mathrm{c}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} \:}\:=−\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{t}\right)}{\mathrm{t}}\:\mathrm{dt}\:\left(\mathrm{c}=\mathrm{0}\right) \\ $$$$…\mathrm{be}\:\mathrm{continued}… \\ $$

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