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Question Number 160281 by amin96 last updated on 27/Nov/21

∫_0 ^1 ((ln^2 (1−x)lnx)/x)dx=?

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

Answered by TheSupreme last updated on 27/Nov/21

∫_0 ^1 ln^2 (1−x) ((ln(x))/x)dx= =(1/2)ln^2 (x)ln^2 (1−x)+∫_0 ^1 ln^2 (x)((ln(1−x))/(1−x))dx 1−x=u → x=1−u → dx=−du =(1/2)ln^2 (x)ln^2 (1−x)+∫_1 ^0 ln^2 (1−u)((ln(u))/u)du =(1/2)ln^2 (x)ln^2 (1−x)−∫_0 ^1 ln^2 (1−u)((ln(u))/u)du 2∫_0 ^1 ln^2 (1−x)((ln(x))/x)dx=(1/2)ln^2 (x)ln^2 (1−x) ∫_0 ^1 ln^2 (1−x)((ln(x))/x)dx=(1/4)ln^2 (x)ln^2 (1−x)∣_0 ^1 lim_(x→0) f(x)=lim_(x→1) (1/4)ln^2 (x)ln^2 (1−x) ∫_0 ^1 ln^2 (1−x)((ln(x))/x)dx=0

$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\:\frac{{ln}\left({x}\right)}{{x}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}} \left({x}\right)\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$$\mathrm{1}−{x}={u}\:\rightarrow\:{x}=\mathrm{1}−{u}\:\rightarrow\:{dx}=−{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)+\int_{\mathrm{1}} ^{\mathrm{0}} {ln}^{\mathrm{2}} \left(\mathrm{1}−{u}\right)\frac{{ln}\left({u}\right)}{{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}} \left(\mathrm{1}−{u}\right)\frac{{ln}\left({u}\right)}{{u}}{du} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\frac{{ln}\left({x}\right)}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\frac{{ln}\left({x}\right)}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left({x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\:{f}\left({x}\right)={li}\underset{{x}\rightarrow\mathrm{1}} {{m}}\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left({x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\frac{{ln}\left({x}\right)}{{x}}{dx}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$

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