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Question Number 184098 by paul2222 last updated on 02/Jan/23

∫_0 ^1 ((ln^2 (1−x))/(1−x))dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}−\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$

Commented by paul2222 last updated on 02/Jan/23

let 1−x=u ∫_0 ^1 ((ln^2 (u))/u)du let ln(u)=−k du=−e^(−k) dk ∫_0 ^∞ (k^2 /e^(−k) )e^(−k) dk=∫_0 ^∞ k^2 dk (the integral diverges)

$$\boldsymbol{\mathrm{let}}\:\mathrm{1}−\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{u}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{u}}\right)}{\boldsymbol{\mathrm{u}}}\boldsymbol{\mathrm{du}} \\ $$$$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{u}}\right)=−\boldsymbol{\mathrm{k}}\:\boldsymbol{\mathrm{du}}=−\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{k}}} \boldsymbol{\mathrm{dk}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{k}}^{\mathrm{2}} }{\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{k}}} }\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{k}}} \boldsymbol{\mathrm{dk}}=\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{k}}^{\mathrm{2}} \boldsymbol{\mathrm{dk}} \\ $$$$\left(\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{integral}}\:\boldsymbol{\mathrm{diverges}}\right) \\ $$

Commented by mokys last updated on 03/Jan/23

e^y = 1 − x → e^y dy = −dx x = 1 → y = −∞ , x = 0 → y = 0 ∫_(−∞) ^( 0) y^2 dy so is diverge

$${e}^{{y}} \:=\:\mathrm{1}\:−\:{x}\:\rightarrow\:{e}^{{y}} {dy}\:=\:−{dx} \\ $$$$ \\ $$$$\:{x}\:=\:\mathrm{1}\:\rightarrow\:{y}\:=\:−\infty\:\:\:,\:\:{x}\:=\:\mathrm{0}\:\rightarrow\:{y}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\:\int_{−\infty} ^{\:\mathrm{0}} \:{y}^{\mathrm{2}} \:{dy}\:{so}\:{is}\:{diverge}\: \\ $$

Answered by Ml last updated on 03/Jan/23

∫_0 ^1 (((ln(1−x))^2 )/((1−x)))dx { ((u=ln(1−x))),((du=−(1/((1−x)))dx)) :} , dx=−(1−x)du ∫_0 ^1 ((u^2 (−(1−x))du)/((1−x)))=−∫_0 ^1 u^2 du=−(1/3)u^3 ⇒ −(1/3)(ln(1−x))^3 ∣_0 ^1 =∞

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{x}\right)}\mathrm{dx} \\ $$$$\begin{cases}{\mathrm{u}=\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}\\{\mathrm{du}=−\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)}\mathrm{dx}}\end{cases}\:,\:\mathrm{dx}=−\left(\mathrm{1}−\mathrm{x}\right)\mathrm{du} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{u}^{\mathrm{2}} \:\left(−\left(\mathrm{1}−\mathrm{x}\right)\right)\mathrm{du}}{\left(\mathrm{1}−\mathrm{x}\right)}=−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{u}^{\mathrm{2}} \mathrm{du}=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{u}^{\mathrm{3}} \\ $$$$\Rightarrow\:−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\right)^{\mathrm{3}} \mid_{\mathrm{0}} ^{\mathrm{1}} =\infty \\ $$

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