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Question Number 34129 by abdo imad last updated on 01/May/18
find the value of ∫_0 ^1 ln(x)ln(1+x)dx .
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right){dx}\:. \\ $$
Commented by math khazana by abdo last updated on 02/May/18
we have proved that ∫_0 ^1 ln(x)ln(1+x)dx = Σ_(n=1) ^∞ (((−1)^n )/(n(n+1)^2 )) let put S_n = Σ_(k=1) ^n (((−1)^k )/(k(k+1)^2 )) and 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 ⇒ 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 )) S_n = Σ_(k=1) ^n (((−1)^k )/k) −Σ_(k=1) ^n (((−1)^k )/(k+1)) −Σ_(k=1) ^n (((−1)^k )/((k+1)^2 )) Σ_(k=1) ^n (((−1)^k )/k) →Σ_(k=1) ^∞ (((−1)^k )/k) =−ln(2) Σ_(k=1) ^n (((−1)^k )/(k+1)) =Σ_(k=1) ^(n+1) (((−1)^(k−1) )/k) −1 →Σ_(k=1) ^∞ (((−1)^(k−1) )/k)−1 =ln(2) −1 Σ_(k=1) ^n (((−1)^k )/((k+1)^2 )) = Σ_(k=1) ^(n+1) (((−1)^(k−1) )/k^2 ) −1 → Σ_(k=1) ^∞ (((−1)^(k−1) )/k^2 ) = −Σ_(n=1) ^∞ (1/((2n)^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(π^2 /8) −(1/4) (π^2 /6) = ((3π^2 −π^2 )/(24)) = (π^2 /(12)) S_n = −ln(2)−ln(2)+1 −(π^2 /(12)) +1 S_n = 2 −2ln(2) −(π^2 /(12)) .
$${we}\:{have}\:{proved}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right){dx}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:{and}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{{a}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}={lim}_{{x}\rightarrow\mathrm{0}} \:{xF}\left({x}\right)\:=\:\mathrm{1} \\ $$$${c}={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} \:{F}\left({x}\right)=\:−\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)=\:\frac{\mathrm{1}}{\mathrm{4}}\:=\:\mathrm{1}\:+\frac{{b}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:\mathrm{1}=\mathrm{4}\:+\mathrm{2}{b}\:−\mathrm{1}=\:\mathrm{3}+\mathrm{2}{b} \\ $$$$\Rightarrow\mathrm{2}{b}=−\mathrm{2}\Rightarrow\:{b}=−\mathrm{1}\:\Rightarrow\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:\rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}\:−\mathrm{1}\:\rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}−\mathrm{1} \\ $$$$={ln}\left(\mathrm{2}\right)\:−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}^{\mathrm{2}} }\:−\mathrm{1} \\ $$$$\rightarrow\:\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}^{\mathrm{2}} }\:=\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\:\frac{\mathrm{3}\pi^{\mathrm{2}} \:−\pi^{\mathrm{2}} }{\mathrm{24}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$${S}_{{n}} \:\:=\:−{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{2}\right)+\mathrm{1}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\mathrm{1} \\ $$$${S}_{{n}} \:=\:\mathrm{2}\:−\mathrm{2}{ln}\left(\mathrm{2}\right)\:\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 02/May/18
S_∞ = 2 −2ln(2) −(π^2 /(12)) .
$${S}_{\infty} =\:\mathrm{2}\:−\mathrm{2}{ln}\left(\mathrm{2}\right)\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/May/18
Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18
Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18
from graph it is clear that area of the curve is greater than zero but less than 0.5 so 0.5 >∫_0 ^1 lnx.ln(1+x)>0
$${from}\:{graph}\:{it}\:{is}\:{clear}\:{that} \\ $$$${area}\:{of}\:{the}\:{curve}\:{is}\:{greater}\:{than}\:{zero}\: \\ $$$${but}\:{less}\:{than}\:\mathrm{0}.\mathrm{5}\: \\ $$$${so}\:\mathrm{0}.\mathrm{5}\:>\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{lnx}.{ln}\left(\mathrm{1}+{x}\right)>\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 02/May/18
sir tanmay the value of ∫_0 ^1 ln(x)ln(1+x)dx must be negative because ln(x)≤0 for x∈]0,1].
$${sir}\:{tanmay}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right){dx}\:{must}\: \\ $$$$\left.{b}\left.{e}\:{negative}\:{because}\:{ln}\left({x}\right)\leqslant\mathrm{0}\:{for}\:{x}\in\right]\mathrm{0},\mathrm{1}\right]. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/May/18
yes true but see the graph..area under the curve is negative area so ∫_0 ^1 lnx.ln(1+x)dx is the area under the curve but negative area so 0.5>∣∫_0 ^1 lnx.ln(1+x)dx∣>0
$${yes}\:{true}\:{but}\:{see}\:{the}\:{graph}..{area}\:{under}\: \\ $$$${the}\:{curve}\:{is}\:{negative}\:{area}\:{so} \\ $$$$\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{lnx}.{ln}\left(\mathrm{1}+{x}\right){dx}\:{is}\:{the}\:{area}\:{under}\:{the}\:{curve} \\ $$$${but}\:{negative}\:{area}\: \\ $$$${so}\:\mathrm{0}.\mathrm{5}>\mid\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{lnx}.{ln}\left(\mathrm{1}+{x}\right){dx}\mid>\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/May/18
correction pls area under the curve is negetive area so 0.5>∣∫_0 ^1 lnx.ln(1+x)dx∣>0
$${correction}\:{pls} \\ $$$${area}\:{under}\:{the}\:{curve}\:{is}\:{negetive}\:{area}\:{so}\: \\ $$$$\:\:\:\mathrm{0}.\mathrm{5}>\mid\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{lnx}.{ln}\left(\mathrm{1}+{x}\right){dx}\mid>\mathrm{0} \\ $$$$ \\ $$
Answered by MJS last updated on 02/May/18
∫f′g=fg−∫fg′ f′=ln(x+1) ⇒ f=(x+1)(ln(x+1)−1) g=lnx ⇒ g′=(1/x) fg=(x+1)(ln(x+1)−1)lnx ∫fg′=∫((x+1)/x)(ln(x+1)−1)dx= =∫ ((ln(x+1))/x)dx+∫ln(x+1)dx−∫(1/x)dx−∫1dx= =∫ ((ln(x+1))/x)dx+(x+1)(ln(x+1)−1)−lnx−x= =∫ ((ln(x+1))/x)dx+(x+1)ln(x+1)−lnx−2x ∫_0 ^1 lnx ln(x+1)dx= =[(x+1)(lnx−1)ln(x+1)−x(lnx−2)]_0 ^1 −∫_0 ^1 ((ln(x+1))/x)dx= lim_(x→0^+ ) xlnx=lim_(x→0^+ ) ((lnx)/(1/x))= [l′Hopital] = =lim_(x→0^+ ) ((1/x)/(−(1/x^2 )))=lim_(x→0^+ ) −x=0 ⇒ ⇒lim_(x→0^+ ) (x+1)(lnx−1)ln(x+1)−x(lnx−2)=0 =2−2ln2−∫_0 ^1 ((ln(x+1))/x)dx ∫ ((ln(x+1))/x)dx= u=−x → dx=−du =−∫−((ln(1−u))/u)du=−Li_2 (u)=−Li_2 (−x) polylogarithm Li_s (x)=Σ_(n=1) ^∞ (x^n /n^s ) −(Li_2 (−1)−Li_2 (0))=−Σ_(n=1) ^∞ (((−1)^n )/n^2 )+Σ_(n=1) ^∞ (0^n /n^2 )= =−Σ_(n=1) ^∞ (((−1)^n )/n^2 )=(π^2 /(12)) ∫_0 ^1 lnx ln(x+1)dx=2−2ln2−(π^2 /(12))≈−.208761
$$\int{f}'{g}={fg}−\int{fg}' \\ $$$${f}'=\mathrm{ln}\left({x}+\mathrm{1}\right)\:\Rightarrow\:{f}=\left({x}+\mathrm{1}\right)\left(\mathrm{ln}\left({x}+\mathrm{1}\right)−\mathrm{1}\right) \\ $$$${g}=\mathrm{ln}{x}\:\Rightarrow\:{g}'=\frac{\mathrm{1}}{{x}} \\ $$$${fg}=\left({x}+\mathrm{1}\right)\left(\mathrm{ln}\left({x}+\mathrm{1}\right)−\mathrm{1}\right)\mathrm{ln}{x} \\ $$$$\int{fg}'=\int\frac{{x}+\mathrm{1}}{{x}}\left(\mathrm{ln}\left({x}+\mathrm{1}\right)−\mathrm{1}\right){dx}= \\ $$$$=\int\:\frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}}{dx}+\int\mathrm{ln}\left({x}+\mathrm{1}\right){dx}−\int\frac{\mathrm{1}}{{x}}{dx}−\int\mathrm{1}{dx}= \\ $$$$=\int\:\frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}}{dx}+\left({x}+\mathrm{1}\right)\left(\mathrm{ln}\left({x}+\mathrm{1}\right)−\mathrm{1}\right)−\mathrm{ln}{x}−{x}= \\ $$$$=\int\:\frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}}{dx}+\left({x}+\mathrm{1}\right)\mathrm{ln}\left({x}+\mathrm{1}\right)−\mathrm{ln}{x}−\mathrm{2}{x} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{ln}{x}\:\mathrm{ln}\left({x}+\mathrm{1}\right){dx}= \\ $$$$=\left[\left({x}+\mathrm{1}\right)\left(\mathrm{ln}{x}−\mathrm{1}\right)\mathrm{ln}\left({x}+\mathrm{1}\right)−{x}\left(\mathrm{ln}{x}−\mathrm{2}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{x}\mathrm{ln}{x}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{ln}{x}}{\frac{\mathrm{1}}{{x}}}=\:\left[\mathrm{l}'\mathrm{Hopital}\right]\:= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\frac{\mathrm{1}}{{x}}}{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}−{x}=\mathrm{0}\:\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left({x}+\mathrm{1}\right)\left(\mathrm{ln}{x}−\mathrm{1}\right)\mathrm{ln}\left({x}+\mathrm{1}\right)−{x}\left(\mathrm{ln}{x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$=\mathrm{2}−\mathrm{2ln2}−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}}{dx} \\ $$$$ \\ $$$$\int\:\frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:{u}=−{x}\:\rightarrow\:{dx}=−{du} \\ $$$$=−\int−\frac{\mathrm{ln}\left(\mathrm{1}−{u}\right)}{{u}}{du}=−\mathrm{Li}_{\mathrm{2}} \left({u}\right)=−\mathrm{Li}_{\mathrm{2}} \left(−{x}\right) \\ $$$$ \\ $$$$\mathrm{polylogarithm}\:\mathrm{Li}_{{s}} \left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{{s}} } \\ $$$$−\left(\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)−\mathrm{Li}_{\mathrm{2}} \left(\mathrm{0}\right)\right)=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{0}^{{n}} }{{n}^{\mathrm{2}} }= \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{ln}{x}\:\mathrm{ln}\left({x}+\mathrm{1}\right){dx}=\mathrm{2}−\mathrm{2ln2}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\approx−.\mathrm{208761} \\ $$

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