Menu Close

find-0-1-ln-x-ln-1-x-dx-




Question Number 61979 by maxmathsup by imad last updated on 13/Jun/19
find ∫_0 ^1 ln(x)ln(1+x) dx
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right)\:{dx} \\ $$
Commented by maxmathsup by imad last updated on 13/Jun/19
let A =∫_0 ^1 ln(x)ln(1+x)dx  we have ln^′ (1+x) =(1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n  ⇒  ln(1+x) =Σ_(n=0) ^∞  (((−1)^n x^(n+1) )/(n+1)) +c    (c=0) ⇒ln(1+x) =Σ_(n=1) ^∞  (((−1)^(n−1) x^n )/n)  with ∣x∣<1 ⇒ A =∫_0 ^1 (Σ_(n=1) ^∞  (((−1)^(n−1) x^n )/n))ln(x)dx  =Σ_(n=1) ^∞  (((−1)^(n−1) )/n) ∫_0 ^1  x^n ln(x)dx  let  A_n =∫_0 ^1  x^n ln(x)dx by parts  A_n =[(1/(n+1))x^(n+1) ln(x)]_(x→0) ^1  −∫_0 ^1  (1/(n+1))x^(n+1)  (dx/x) =−(1/(n+1)) ∫_0 ^1  x^n dx =−(1/((n+1)^2 )) ⇒  A =−Σ_(n=1) ^∞  (((−1)^(n−1) )/n) (1/((n+1)^2 )) =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 ))  lim_(x→+∞) xF(x) =0 =1+b ⇒b =−1 ⇒F(x) =(1/x) −(1/(x+1)) −(1/((x+1)^2 )) ⇒  A =Σ_(n=1) ^∞  (((−1)^n )/n) −Σ_(n=1) ^∞  (((−1)^n )/(n+1)) −Σ_(n=1) ^∞  (((−1)^n )/((n+1)^2 ))  Σ_(n=1) ^∞  (((−1)^n )/n) =−ln(2)  Σ_(n=1) ^∞  (((−1)^n )/(n+1)) =Σ_(n=2) ^∞  (((−1)^(n−1) )/n) =Σ_(n=1) ^∞  (((−1)^(n−1) )/n) −1 =ln(2)−1  Σ_(n=1) ^∞   (((−1)^n )/((n+1)^2 )) =Σ_(n=2) ^∞  (((−1)^(n−1) )/n^2 ) =Σ_(n=1) ^∞  (((−1)^(n−1) )/n^2 ) −1  let δ(x) =Σ_(n=1) ^∞  (((−1)^n )/n^x )   and ξ(x) =Σ_(n=1) ^∞ (1/n^x )      (x>1) we have proved   δ(x) =(2^(1−x) −1)ξ(x) ⇒Σ_(n=1) ^∞  (((−1)^n )/n^2 ) =(2^(1−2) −1)ξ(2) =−(1/2)(π^2 /6) =−(π^2 /(12)) ⇒  Σ_(n=1) ^∞  (((−1)^n )/((n+1)^2 )) =(π^2 /(12)) −1 ⇒ A =−ln(2)−ln(2)+1−(π^2 /(12)) +1 ⇒  A =2−2ln(2)−(π^2 /(12)) .
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right){dx}\:\:{we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:+{c}\:\:\:\:\left({c}=\mathrm{0}\right)\:\Rightarrow{ln}\left(\mathrm{1}+{x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {x}^{{n}} }{{n}} \\ $$$${with}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {x}^{{n}} }{{n}}\right){ln}\left({x}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {ln}\left({x}\right){dx}\:\:{let}\:\:{A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {ln}\left({x}\right){dx}\:{by}\:{parts} \\ $$$${A}_{{n}} =\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)\right]_{{x}\rightarrow\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:\frac{{dx}}{{x}}\:=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {dx}\:=−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${A}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:{decompose}\:{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{F}\left({x}\right)\:=\frac{{a}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{0}} \:{xF}\left({x}\right)\:=\mathrm{1} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)\:=−\mathrm{1}\:\Rightarrow{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)\:=\mathrm{0}\:=\mathrm{1}+{b}\:\Rightarrow{b}\:=−\mathrm{1}\:\Rightarrow{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${A}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:−\mathrm{1}\:={ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:−\mathrm{1} \\ $$$${let}\:\delta\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:\:\:{and}\:\xi\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{{x}} }\:\:\:\:\:\:\left({x}>\mathrm{1}\right)\:{we}\:{have}\:{proved}\: \\ $$$$\delta\left({x}\right)\:=\left(\mathrm{2}^{\mathrm{1}−{x}} −\mathrm{1}\right)\xi\left({x}\right)\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:−\mathrm{1}\:\Rightarrow\:{A}\:=−{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{2}\right)+\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\mathrm{1}\:\Rightarrow \\ $$$${A}\:=\mathrm{2}−\mathrm{2}{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:. \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *