An intergral with variable upper limit

Question

Let $$\psi \left(x \right)=\int_{0}^{x}\frac{\ln(1-t)}{t}dt,x\in (0,1).$$ Show $$\forall x\in (0,1), \psi\left(x \right)=?$$

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I return the old variable $t$ by the substitution $s=ln(1-t)$,and then get $$\int_{0}^{\ln(1-x)}\frac{se^{s}}{e^{s}-1}ds.$$ but seemingly,this integration is not easy for me to solve as well as the first one.

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NOTE :

In fact， I want use the result of this question to find the sum functions of power series :$$\sum_{n=1}^{\infty}\frac{{x}^{n}}{n^{2}},\left(0<x<1 \right).$$

Answer 1

Hint: Expand $\ln(1-t)$ into its well-known Taylor series, then reverse the order of summation and integration. The result will be a polylogarithmic series.

Answer 2

Just expanding Lucian's answer, over $[0,1)$ we have: $$\log(1-t) = -\sum_{j=1}^{+\infty}\frac{t^j}{j},$$ hence: $$\int_{0}^{x}\frac{\log(1-t)}{t}\,dt=-\sum_{j=1}^{+\infty}\frac{x^j}{j^2}=-\operatorname{Li}_2(x).$$