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Question Number 109136 by EmericGent last updated on 21/Aug/20

∫_0 ^(1/2) ((ln(1-t)ln(t))/t) dt I′m about to give up

$$\int_{\mathrm{0}} ^{\mathrm{1}/\mathrm{2}} \frac{{ln}\left(\mathrm{1}-{t}\right){ln}\left({t}\right)}{{t}}\:{dt} \\ $$$${I}'{m}\:{about}\:{to}\:{give}\:{up} \\ $$

Answered by Sarah85 last updated on 21/Aug/20

∫((ln (1−t) ln t)/t)dt by parts f′=((ln (1−t))/t) ⇒ f=−Li_2 t g=ln t ⇒ g′=(1/t) −Li_2 t ln t −∫−((Li_2 t)/t)dt= =Li_3 t −Li_2 t ln t +C ⇒ answer is (7/8)ζ(3)−(1/3)(ln 2)^3

$$\int\frac{\mathrm{ln}\:\left(\mathrm{1}−{t}\right)\:\mathrm{ln}\:{t}}{{t}}{dt} \\ $$$$\mathrm{by}\:\mathrm{parts} \\ $$$${f}'=\frac{\mathrm{ln}\:\left(\mathrm{1}−{t}\right)}{{t}}\:\Rightarrow\:{f}=−\mathrm{Li}_{\mathrm{2}} \:{t} \\ $$$${g}=\mathrm{ln}\:{t}\:\Rightarrow\:{g}'=\frac{\mathrm{1}}{{t}} \\ $$$$−\mathrm{Li}_{\mathrm{2}} \:{t}\:\mathrm{ln}\:{t}\:−\int−\frac{\mathrm{Li}_{\mathrm{2}} \:{t}}{{t}}{dt}= \\ $$$$=\mathrm{Li}_{\mathrm{3}} \:{t}\:−\mathrm{Li}_{\mathrm{2}} \:{t}\:\mathrm{ln}\:{t}\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{3}} \\ $$

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