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Question Number 134852 by mnjuly1970 last updated on 07/Mar/21

...nice calculus... find ::: 𝛗=∫_0 ^( 1) ((ln(x)ln(1βˆ’x))/(x(1βˆ’x)))dx=?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:\:\:{calculus}... \\ $$$$\:\:\:\:\:\:{find}\:::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right){ln}\left(\mathrm{1}βˆ’{x}\right)}{{x}\left(\mathrm{1}βˆ’{x}\right)}{dx}=? \\ $$$$ \\ $$

Answered by mnjuly1970 last updated on 07/Mar/21

𝛗=∫_0 ^( 1) ((ln(x)ln(1βˆ’x))/(1βˆ’x))+((ln(x)ln(1βˆ’x))/x)dx =[βˆ’(1/2)ln^2 (1βˆ’x)ln(x)]_0 ^1 +(1/2)∫_0 ^( 1) ((ln^2 (1βˆ’x))/x)dx +[(1/2)ln^2 (x)ln(1βˆ’x)]_0 ^1 +(1/2)∫_0 ^( 1) ((ln^2 (x))/(1βˆ’x))dx =ΞΆ(3)+(1/2)∫_0 ^( 1) ((ln^2 (1βˆ’t))/t)dt=2ΞΆ(3)...βœ“βœ“

$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right){ln}\left(\mathrm{1}βˆ’{x}\right)}{\mathrm{1}βˆ’{x}}+\frac{{ln}\left({x}\right){ln}\left(\mathrm{1}βˆ’{x}\right)}{{x}}{dx} \\ $$$$=\left[βˆ’\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}βˆ’{x}\right){ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}βˆ’{x}\right)}{{x}}{dx} \\ $$$$+\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({x}\right){ln}\left(\mathrm{1}βˆ’{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}βˆ’{x}}{dx} \\ $$$$=\zeta\left(\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}βˆ’{t}\right)}{{t}}{dt}=\mathrm{2}\zeta\left(\mathrm{3}\right)...\checkmark\checkmark \\ $$$$\:\: \\ $$$$\:\:\: \\ $$$$ \\ $$$$ \\ $$

Answered by Ñï= last updated on 08/Mar/21

𝛗=∫_0 ^( 1) ((ln(x)ln(1βˆ’x))/(x(1βˆ’x)))dx=∫_0 ^1 ((lnxln(1βˆ’x))/x)+((lnxln(1βˆ’x))/(1βˆ’x))dx =2∫_0 ^1 ((ln(1βˆ’x)lnx)/x) =βˆ’2Li_2 (x)lnx∣_0 ^1 +2∫_0 ^1 ((Li_2 (x))/x)dx =2Li_3 (x)∣_0 ^1 =2Li_3 (1)=2ΞΆ(3)

$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right){ln}\left(\mathrm{1}βˆ’{x}\right)}{{x}\left(\mathrm{1}βˆ’{x}\right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnxln}\left(\mathrm{1}βˆ’{x}\right)}{{x}}+\frac{{lnxln}\left(\mathrm{1}βˆ’{x}\right)}{\mathrm{1}βˆ’{x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}βˆ’{x}\right){lnx}}{{x}} \\ $$$$=βˆ’\mathrm{2}{Li}_{\mathrm{2}} \left({x}\right){lnx}\mid_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({x}\right)}{{x}}{dx} \\ $$$$=\mathrm{2}{Li}_{\mathrm{3}} \left({x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{2}{Li}_{\mathrm{3}} \left(\mathrm{1}\right)=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$

Commented by mnjuly1970 last updated on 08/Mar/21

thanks alot ...

$${thanks}\:{alot}\:... \\ $$

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