Question Number 160734 by mnjuly1970 last updated on 05/Dec/21
![# Advanced Calculus # Φ = ∫_0 ^( 1) (((ln^ ( (1/(1− x)) ))/x) )^( 3) dx =^? 3 ( ζ (2 ) + ζ (3 )) −−−− solution−−−− Φ =^(I.B.P) [ (( 1)/(2x^( 2) )) ln^( 3) ( 1−x)]_0 ^1 +(3/2) ∫_0 ^( 1) (( ln^( 2) (1− x ))/(x^( 2) (1 − x ))) dx = (1/2) lim_( ξ →1^(− ) ) ((ln^( 3) ( 1− ξ ))/ξ^( 2) ) +(3/2)[∫_0 ^( 1) (( ln^( 2) ( 1− x ))/x)dx = 2 ζ (3)] + (3/2)[∫_0 ^( 1) (( ln^( 2) ( 1−x))/x^( 2) ) dx = (π^( 2) /3) = 2ζ (2 )] +(3/2)∫_0 ^( 1) (( ln^( 2) (1− x))/(1−x)) dx} =(1/2) lim_( ξ →1^( −) ) {((ln^( 3) ( 1−ξ ))/ξ^( 2) ) −ln^( 3) (1− ξ ) } +(3/2) (2ζ (3 )) +(3/2) ( 2ζ (2 )) = 3( ζ (3 ) + 3ζ (2 ) ) ■ m.n](Q160734.png)
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:#\:\mathrm{Advanced}\:\:\:\mathrm{Calculus}\:# \\ $$$$\:\:\:\:\:\:\:\:\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{ln}^{\:} \:\left(\:\frac{\mathrm{1}}{\mathrm{1}−\:{x}}\:\:\right)}{{x}}\:\right)^{\:\mathrm{3}} {dx}\:\overset{?} {=}\:\mathrm{3}\:\left(\:\zeta\:\left(\mathrm{2}\:\right)\:+\:\zeta\:\left(\mathrm{3}\:\right)\right) \\ $$$$\:\:\:\:\:\:−−−−\:\:{solution}−−−− \\ $$$$\:\:\:\:\:\:\:\:\Phi\:\overset{\mathrm{I}.\mathrm{B}.\mathrm{P}} {=}\:\left[\:\frac{\:\mathrm{1}}{\mathrm{2}{x}^{\:\mathrm{2}} }\:{ln}^{\:\mathrm{3}} \left(\:\mathrm{1}−{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{3}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \:\left(\mathrm{1}−\:{x}\:\right)}{{x}^{\:\mathrm{2}} \:\left(\mathrm{1}\:−\:{x}\:\right)}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{lim}_{\:\xi\:\rightarrow\mathrm{1}^{−\:} } \:\frac{{ln}^{\:\mathrm{3}} \left(\:\mathrm{1}−\:\xi\:\right)}{\xi^{\:\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{3}}{\mathrm{2}}\left[\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\:\mathrm{1}−\:{x}\:\right)}{{x}}{dx}\:=\:\mathrm{2}\:\zeta\:\left(\mathrm{3}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\:\frac{\mathrm{3}}{\mathrm{2}}\left[\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\:\mathrm{1}−{x}\right)}{{x}^{\:\mathrm{2}} }\:{dx}\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{3}}\:=\:\mathrm{2}\zeta\:\left(\mathrm{2}\:\right)\right]\: \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−\:{x}\right)}{\mathrm{1}−{x}}\:{dx}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\:{lim}_{\:\xi\:\rightarrow\mathrm{1}^{\:−} } \left\{\frac{{ln}^{\:\mathrm{3}} \left(\:\mathrm{1}−\xi\:\right)}{\xi^{\:\mathrm{2}} }\:\:−{ln}^{\:\mathrm{3}} \left(\mathrm{1}−\:\xi\:\right)\:\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{3}}{\mathrm{2}}\:\left(\mathrm{2}\zeta\:\left(\mathrm{3}\:\right)\right)\:\:+\frac{\mathrm{3}}{\mathrm{2}}\:\left(\:\mathrm{2}\zeta\:\left(\mathrm{2}\:\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{3}\left(\:\:\:\zeta\:\left(\mathrm{3}\:\right)\:+\:\mathrm{3}\zeta\:\left(\mathrm{2}\:\right)\:\:\right)\:\:\:\:\:\:\:\blacksquare\:\:\:{m}.{n}\:\:\: \\ $$$$ \\ $$
Commented by Gbenga last updated on 06/Dec/21
$$\boldsymbol{{can}}\:\boldsymbol{{you}}\:\boldsymbol{{teach}}\:\boldsymbol{{me}}\:\boldsymbol{{this}}\:\boldsymbol{{sir}} \\ $$