Question Number 134852 by mnjuly1970 last updated on 07/Mar/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:\:\:{calculus}… \\ $$$$\:\:\:\:\:\:{find}\:::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right){ln}\left(\mathrm{1}β{x}\right)}{{x}\left(\mathrm{1}β{x}\right)}{dx}=? \\ $$$$ \\ $$
Answered by mnjuly1970 last updated on 07/Mar/21
![π=β«_0 ^( 1) ((ln(x)ln(1βx))/(1βx))+((ln(x)ln(1βx))/x)dx =[β(1/2)ln^2 (1βx)ln(x)]_0 ^1 +(1/2)β«_0 ^( 1) ((ln^2 (1βx))/x)dx +[(1/2)ln^2 (x)ln(1βx)]_0 ^1 +(1/2)β«_0 ^( 1) ((ln^2 (x))/(1βx))dx =ΞΆ(3)+(1/2)β«_0 ^( 1) ((ln^2 (1βt))/t)dt=2ΞΆ(3)...ββ](https://www.tinkutara.com/question/Q134856.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right){ln}\left(\mathrm{1}β{x}\right)}{\mathrm{1}β{x}}+\frac{{ln}\left({x}\right){ln}\left(\mathrm{1}β{x}\right)}{{x}}{dx} \\ $$$$=\left[β\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}β{x}\right){ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}β{x}\right)}{{x}}{dx} \\ $$$$+\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({x}\right){ln}\left(\mathrm{1}β{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}β{x}}{dx} \\ $$$$=\zeta\left(\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}β{t}\right)}{{t}}{dt}=\mathrm{2}\zeta\left(\mathrm{3}\right)…\checkmark\checkmark \\ $$$$\:\: \\ $$$$\:\:\: \\ $$$$ \\ $$$$ \\ $$
Answered by ΓΓ―= last updated on 08/Mar/21
$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right){ln}\left(\mathrm{1}β{x}\right)}{{x}\left(\mathrm{1}β{x}\right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnxln}\left(\mathrm{1}β{x}\right)}{{x}}+\frac{{lnxln}\left(\mathrm{1}β{x}\right)}{\mathrm{1}β{x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}β{x}\right){lnx}}{{x}} \\ $$$$=β\mathrm{2}{Li}_{\mathrm{2}} \left({x}\right){lnx}\mid_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left({x}\right)}{{x}}{dx} \\ $$$$=\mathrm{2}{Li}_{\mathrm{3}} \left({x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{2}{Li}_{\mathrm{3}} \left(\mathrm{1}\right)=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$
Commented by mnjuly1970 last updated on 08/Mar/21
$${thanks}\:{alot}\:… \\ $$