Question Number 45232 by maxmathsup by imad last updated on 10/Oct/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right){dx}\:. \\ $$
Commented by maxmathsup by imad last updated on 11/Oct/18
![let I = ∫_0 ^1 ln(x)ln(1+x)dx we have for ∣x∣<1 ln^′ (1+x)=Σ_(n=0) ^∞ (−1)^n x^n ⇒ln(1+x)=Σ_(n=0) ^∞ (((−1)^n )/(n+1))x^(n+1) +λ x=0 ⇒λ=0 ⇒ln(1+x) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n ⇒ I =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) ∫_0 ^1 x^n ln(x)dx by parts A_n =∫_0 ^1 x^n ln(x)dx =[(1/(n+1))x^(n+1) ln(x)]_(x→0) ^1 −∫_0 ^1 (1/((n+1))) x^n dx =−(1/((n+1)^2 )) ⇒I =Σ_(n=1) ^∞ (((−1)^n )/(n(n+1)^2 )) let decompose F(x)=(1/(x(x+1)^2 )) F(x)=(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 ⇒F(x)=(1/x) +(b/(x+1)) −(1/((x+1)^2 )) F(1) =(1/4) =1 +(b/2) −(1/4) ⇒1 =4+2b−1 =3+2b ⇒2b=−2 ⇒b=−1 ⇒ F(x) =(1/x) −(1/(x+1)) −(1/((x+1)^2 )) ⇒ I =Σ_(n=1) ^∞ (((−1)^n )/n) −Σ_(n=1) ^∞ (((−1)^n )/(n+1)) −Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) but Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n) = ln(2)−1 Σ_(n=1) ^∞ (1/((n+1)^2 )) =Σ_(n=2) ^∞ (((−1)^n )/n^2 ) = Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +1 let find Σ_(n=1) ^∞ (((−1)^n )/n^2 ) we have Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =Σ_(n=1) ^∞ (1/((2n)^2 )) −Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(1/4) (π^2 /6) −Σ_(n=0) ^∞ (1/((2n+1)^2 )) but Σ_(n=1) ^∞ (1/n^2 ) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4) Σ (1/n^2 ) =(3/4) (π^2 /6) =(π^2 /8) ⇒Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(π^2 /(24)) −(π^2 /8) =−(π^2 /(12)) ⇒ ∫_0 ^1 ln(x)ln(1+x)dx =−ln(2)−ln(2)+1 −(−(π^2 /(12))+1) =(π^2 /(12)) −2ln(2) .](Q45324.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right){dx}\:\:{we}\:{have}\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$$${ln}^{'} \left(\mathrm{1}+{x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \:\Rightarrow{ln}\left(\mathrm{1}+{x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:+\lambda \\ $$$${x}=\mathrm{0}\:\Rightarrow\lambda=\mathrm{0}\:\Rightarrow{ln}\left(\mathrm{1}+{x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{{n}} \:\Rightarrow \\ $$$${I}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {ln}\left({x}\right){dx}\:\:{by}\:{parts} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {ln}\left({x}\right){dx}\:=\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)\right]_{{x}\rightarrow\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\:{x}^{{n}} {dx} \\ $$$$=−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{I}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{0}} {xF}\left({x}\right)=\mathrm{1} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=−\mathrm{1}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\:=\mathrm{1}\:+\frac{{b}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\mathrm{1}\:=\mathrm{4}+\mathrm{2}{b}−\mathrm{1}\:=\mathrm{3}+\mathrm{2}{b}\:\Rightarrow\mathrm{2}{b}=−\mathrm{2}\:\Rightarrow{b}=−\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{but} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\:=\:{ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:+\mathrm{1}\:\:{let}\:{find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$${we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:{but}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}\:\Sigma\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right){dx}\:=−{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{2}\right)+\mathrm{1}\:−\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\mathrm{1}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:−\mathrm{2}{ln}\left(\mathrm{2}\right)\:. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18
$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}\right)×{ln}\left({x}\right){dx}\:\:\approx−\mathrm{0}.\mathrm{21}\left({approx}\right) \\ $$