Question Number 27909 by mondodotto@gmail.com last updated on 16/Jan/18
Commented by abdo imad last updated on 16/Jan/18
![we have (d/dx)ln(1−x)= ((−1)/(1−x))=−Σ_(n=0) ^∝ x^n ⇒ ln(1−x)= −Σ_(n=0) ^∝ (x^(n+1) /(n+1)) =−Σ_(n=1) ^∝ (x^n /n) +λ and λ=f(0)=0 so ln(1−x)=−Σ_(n=1) ^∝ (x^n /n) and ∫_0 ^1 ln(x)ln(1−x)dx= −∫_0 ^1 (Σ_(n=1) ^∝ (x^n /n))lnx dx =−Σ_(n=1) ^(∝ ) (1/n) ∫_0 ^1 x^n lnxdx but integration by parts give ∫_0 ^1 x^n lnxdx= [ (1/(n+1)) x^(n+1) ln(x)]_0 ^1 −∫_0 ^1 (1/(n+1)) x^(n+1) (dx/x) = −(1/((n+1)^2 )) ⇒ I= Σ_(n=1) ^∝ (1/(n(n+1)^2 )) let decompose F(x)= (1/(x(x+1)^2 ))= (a/x) + (b/(x+1)) + (c/((x+1)^2 )) a=lim_(x−>0) xF(x)=1 c= lim_(x−>−1) (x+1)^2 F(x)=−1 so F(x)= (1/x) +(b/(x+1)) − (1/((x+1)^2 )) F(1)= (1/4)=1 +(b/2) −(1/4)⇒1+(b/2) = (1/2) ⇒b=−1 F(x)= (1/x) −(1/(x+1)) −(1/((x+1)^2 )) and I=lim_(n−>∝) S_n / S_n = Σ_(k=1) ^n (1/(k(k+1)^2 )) = Σ_(k=1) ^n (1/k) −Σ_(k=1) ^n (1/(k+1)) −Σ_(k=1) ^n (1/((k+1)^2 )) but Σ_(k=1) ^n = H_n ^ and Σ_(k=1) ^n (1/((k+1)^2 )) = Σ_(k=1) ^(n+1) (1/k^2 ) −1 Σ_(k=1) ^n (1/(k+1)) = Σ_(k=2) ^(n+1) (1/k) = H_(n+1) −1 S_n = H_n −H_(n+1) +1 − Σ_(k=1) ^(n+1) (1/k^2 ) +1 lim_(n−>∝) H_n −H_(n+1) =0 lim_(n−>∝) Σ_(k=1) ^(n+1) (1/k^2 )= (π^2 /6) lim_(n−>∝ ) S_n = 2−(π^2 /6) I= 2−(π^2 /6) .](https://www.tinkutara.com/question/Q27911.png)
$${we}\:{have}\:\frac{{d}}{{dx}}{ln}\left(\mathrm{1}−{x}\right)=\:\frac{−\mathrm{1}}{\mathrm{1}−{x}}=−\sum_{{n}=\mathrm{0}} ^{\propto} \:{x}^{{n}} \\ $$$$\Rightarrow\:{ln}\left(\mathrm{1}−{x}\right)=\:−\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=−\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{{x}^{{n}} }{{n}}\:+\lambda\:\:\:{and}\:\lambda={f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${so}\:{ln}\left(\mathrm{1}−{x}\right)=−\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{{x}^{{n}} }{{n}}\:\:{and} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right){dx}=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{{x}^{{n}} }{{n}}\right){lnx}\:{dx} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\propto\:} \:\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \:{lnxdx}\:\:{but}\:{integration}\:{by}\:{parts}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {lnxdx}=\:\left[\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}+\mathrm{1}} \:\frac{{dx}}{{x}} \\ $$$$=\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\Rightarrow\:\:{I}=\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\:\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\:\:\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }=\:\frac{{a}}{{x}}\:+\:\frac{{b}}{{x}+\mathrm{1}}\:+\:\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}={lim}_{{x}−>\mathrm{0}} {xF}\left({x}\right)=\mathrm{1} \\ $$$${c}=\:{lim}_{{x}−>−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=−\mathrm{1}\:\:{so}\: \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:−\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)=\:\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{1}\:+\frac{{b}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\mathrm{1}+\frac{{b}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{b}=−\mathrm{1} \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:{and}\:{I}={lim}_{{n}−>\propto} {S}_{{n}} \:\:/ \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${but}\:\sum_{{k}=\mathrm{1}} ^{{n}} =\:{H}_{{n}} ^{} \:\:{and}\:\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\:\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}}\:=\:{H}_{{n}+\mathrm{1}} \:\:\:−\mathrm{1} \\ $$$${S}_{{n}} =\:{H}_{{n}} −{H}_{{n}+\mathrm{1}} \:\:+\mathrm{1}\:−\:\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:+\mathrm{1} \\ $$$${lim}_{{n}−>\propto} \:{H}_{{n}} \:\:−{H}_{{n}+\mathrm{1}} =\mathrm{0} \\ $$$${lim}_{{n}−>\propto} \:\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:\:{lim}_{{n}−>\propto\:} \:{S}_{{n}} \:\:=\:\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\: \\ $$$${I}=\:\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:. \\ $$