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Question Number 78620 by mathmax by abdo last updated on 19/Jan/20
explicit f(x) =∫_0 ^(+∞) ln(1−xe^(−t) )dt with ∣x∣<1
explicitf(x)=∫0+∞ln(1−xe−t)dtwith∣x∣<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)
ln(1−xe−t)=−Σ(xe−t)kk∫0+∞ln(1−xe−t)dt=∫0+∞−Σxke−ktkdt=Σ[0+∞e−ktk2.xk]=−∑k⩾1xkk2=∫0xln(1−t)tdx=Li2(x)

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