Question Number 78620 by mathmax by abdo last updated on 19/Jan/20
$${explicit}\:\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{+\infty} {ln}\left(\mathrm{1}−{xe}^{−{t}} \right){dt}\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$
Answered by mind is power last updated on 19/Jan/20
![ln(1−xe^(−t) )=−Σ(((xe^(−t) )^k )/k) ∫_0 ^(+∞) ln(1−xe^(−t) )dt=∫_0 ^(+∞) −Σ((x^k e^(−kt) )/k)dt =Σ[_0 ^(+∞) (e^(−kt) /k^2 ).x^k ]=−Σ_(k≥1) (x^k /k^2 )=∫_0 ^x ((ln(1−t))/t)dx=Li_2 (x)](https://www.tinkutara.com/question/Q78690.png)
$$\mathrm{ln}\left(\mathrm{1}−\mathrm{xe}^{−\mathrm{t}} \right)=−\Sigma\frac{\left(\mathrm{xe}^{−\mathrm{t}} \right)^{\mathrm{k}} }{\mathrm{k}} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \mathrm{ln}\left(\mathrm{1}−\mathrm{xe}^{−\mathrm{t}} \right)\mathrm{dt}=\int_{\mathrm{0}} ^{+\infty} −\Sigma\frac{\mathrm{x}^{\mathrm{k}} \mathrm{e}^{−\mathrm{kt}} }{\mathrm{k}}\mathrm{dt} \\ $$$$=\Sigma\left[_{\mathrm{0}} ^{+\infty} \frac{\mathrm{e}^{−\mathrm{kt}} }{\mathrm{k}^{\mathrm{2}} }.\mathrm{x}^{\mathrm{k}} \right]=−\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{x}^{\mathrm{k}} }{\mathrm{k}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{x}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{t}\right)}{\mathrm{t}}\mathrm{dx}=\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right) \\ $$