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Question Number 206592 by NasaSara last updated on 19/Apr/24
Answered by namphamduc last updated on 20/Apr/24
∫_0 ^1 ((ln(x))/(1−x))dx=(4/3)∫_0 ^1 ((ln(x))/(1−x))dx−(1/3)∫_0 ^1 ((ln(x))/(1−x))dx x→x^2 ⇒∫_0 ^1 ((ln(x))/(1−x))dx=(4/3)∫_0 ^1 ((ln(x))/(1−x))dx−(4/3)∫_0 ^1 ((xln(x))/(1−x^2 ))dx=(4/3)∫_0 ^1 ((ln(x))/(1−x^2 ))dx=(2/3)∫_0 ^∞ ((ln(x))/(1−x^2 )) Let : I(a)=∫_0 ^∞ ((ln(1−a^2 +a^2 x^2 ))/(1−x^2 ))dx,I(0)=0,I(1)=2∫_0 ^∞ ((ln(x))/(1−x^2 ))dx I′(a)=−∫_0 ^∞ ((2a)/(a^2 x^2 +1−a^2 ))dx=−(π/( (√(1−a^2 )))) ⇒I(1)=−π∫_0 ^1 (1/( (√(1−a^2 ))))da=−πsin^(−1) (a)∣_0 ^1 =−π.(π/2)=−(π^2 /2) ⇒∫_0 ^1 ((ln(x))/(1−x))dx=(1/3)I(1)=−(π^2 /2).(1/3)=−(π^2 /6) ⇒∫_0 ^1 ((ln(x))/(x−1))dx=(π^2 /6)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$${x}\rightarrow{x}^{\mathrm{2}} \Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}−\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{Let}\::\:{I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{1}−{a}^{\mathrm{2}} +{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx},{I}\left(\mathrm{0}\right)=\mathrm{0},{I}\left(\mathrm{1}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$${I}'\left({a}\right)=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}−{a}^{\mathrm{2}} }{dx}=−\frac{\pi}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }} \\ $$$$\Rightarrow{I}\left(\mathrm{1}\right)=−\pi\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{da}=−\pi\mathrm{sin}^{−\mathrm{1}} \left({a}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} =−\pi.\frac{\pi}{\mathrm{2}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}=\frac{\mathrm{1}}{\mathrm{3}}{I}\left(\mathrm{1}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{3}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)}{{x}−\mathrm{1}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$
Commented by NasaSara last updated on 20/Apr/24
thank you
$${thank}\:{you} \\ $$
Answered by Berbere last updated on 20/Apr/24
∫_0 ^1 ((ln(x))/(1−x))dx=(∂/∂s)∫_0 ^1 ((x^s −1)/(x−1))dx∣_(s=0) =(∂/∂s).−∫_0 ^1 ((x^s −1)/(1−x))dx =(∂/∂s)Ψ(1+s)=Ψ′(1+s)∣_(s=0) =Ψ^((1)) (1)=ζ(2)=(π^2 /6)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}=\frac{\partial}{\partial{s}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{s}} −\mathrm{1}}{{x}−\mathrm{1}}{dx}\mid_{{s}=\mathrm{0}} =\frac{\partial}{\partial{s}}.−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{s}} −\mathrm{1}}{\mathrm{1}−{x}}{dx} \\ $$$$=\frac{\partial}{\partial{s}}\Psi\left(\mathrm{1}+{s}\right)=\Psi'\left(\mathrm{1}+{s}\right)\mid_{{s}=\mathrm{0}} =\Psi^{\left(\mathrm{1}\right)} \left(\mathrm{1}\right)=\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$
Commented by NasaSara last updated on 20/Apr/24
thanks
$${thanks} \\ $$

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