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Question Number 204715 by 073 last updated on 26/Feb/24
prove that ∫_0 ^1 ((ln^2 (1−x))/x)dx=2ζ(3)
$${prove}\:{that} \\ $$$$\overset{\mathrm{1}} {\int}_{\mathrm{0}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$
Answered by witcher3 last updated on 26/Feb/24
−ln(1−x)=y ⇒x=1−e^(−y) ⇒dx=e^y dy ⇔∫_0 ^∞ ((y^2 e^(−y) )/(1−e^(−y) ))dy =∫_0 ^∞ (y^(3−1) /(e^y −1))dy=Γ(3)ζ(3)=2ζ(3) Γ(z)ζ(z)=∫_0 ^∞ (t^(z−1) /(e^t −1))dt
$$−\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{y} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{1}−\mathrm{e}^{−\mathrm{y}} \Rightarrow\mathrm{dx}=\mathrm{e}^{\mathrm{y}} \mathrm{dy} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{y}^{\mathrm{2}} \mathrm{e}^{−\mathrm{y}} }{\mathrm{1}−\mathrm{e}^{−\mathrm{y}} }\mathrm{dy}\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{y}^{\mathrm{3}−\mathrm{1}} }{\mathrm{e}^{\mathrm{y}} −\mathrm{1}}\mathrm{dy}=\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right)=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\Gamma\left(\mathrm{z}\right)\zeta\left(\mathrm{z}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{z}−\mathrm{1}} }{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}\mathrm{dt} \\ $$$$ \\ $$
Answered by Mathspace last updated on 28/Feb/24
I=∫_0 ^1 ((ln^2 (1−x))/x)dx (1−x=t) ⇒ I=−∫_0 ^1 ((ln^2 t)/(1−t))(−dt)=∫_0 ^1 ((ln^2 t)/(1−t))dt =∫_0 ^1 ln^2 (t)Σ_(n=0) ^∞ t^n dt =Σ_(n=0) ^∞ ∫_0 ^1 t^n ln^2 (t)dt A_n =∫_0 ^1 t^n ln^2 t dt=_(by parts) [(t^(n+1) /(n+1))ln^2 t]_0 ^1 −2∫_0 ^1 (t^(n+1) /(n+1))((lnt)/t)dt =0−(2/(n+1))∫_0 ^1 t^n ln(t)dt =((−2)/(n+1)){[(t^(n+1) /(n+1))lnt]_0 ^1 −∫_0 ^1 (t^(n+1) /(n+1))(dt/t)} =(2/((n+1)^2 ))∫_0 ^1 t^n dt =(2/((n+1)^3 )) =A_n I=Σ_(n=0) ^∞ A_n =2Σ_(n=0) ^∞ (1/((n+1)^3 )) =2Σ_(n=1) ^∞ (1/n^3 )=2ξ(3) ξ(3)∼1,2 ⇒ I∼2,4
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}\:\:\left(\mathrm{1}−{x}={t}\right)\:\Rightarrow \\ $$$${I}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} {t}}{\mathrm{1}−{t}}\left(−{dt}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} {t}}{\mathrm{1}−{t}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}} \left({t}\right)\sum_{{n}=\mathrm{0}} ^{\infty} {t}^{{n}} {dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} {ln}^{\mathrm{2}} \left({t}\right){dt} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} {ln}^{\mathrm{2}} {t}\:{dt}=_{{by}\:{parts}} \\ $$$$\left[\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{ln}^{\mathrm{2}} {t}\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\frac{{lnt}}{{t}}{dt} \\ $$$$=\mathrm{0}−\frac{\mathrm{2}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{n}} {ln}\left({t}\right){dt} \\ $$$$=\frac{−\mathrm{2}}{{n}+\mathrm{1}}\left\{\left[\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{lnt}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\frac{{dt}}{{t}}\right\} \\ $$$$=\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{n}} {dt} \\ $$$$=\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:={A}_{{n}} \\ $$$${I}=\sum_{{n}=\mathrm{0}} ^{\infty} {A}_{{n}} =\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\mathrm{2}\xi\left(\mathrm{3}\right) \\ $$$$\xi\left(\mathrm{3}\right)\sim\mathrm{1},\mathrm{2}\:\Rightarrow \\ $$$${I}\sim\mathrm{2},\mathrm{4} \\ $$

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