# $\int_0^1 {\frac{{\ln (1 - x)}}{x}}$ without power series

Is there a way to calculate $$\int_0^1{ \ln (1 - x)\over x}\;dx$$ without using power series?

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## 1 Answer

A related problem. Using the change of variables $x=1-e^{-t}$ and taking advatage of the fact that

$$\Gamma(s)\zeta(s) = \int_{0}^{\infty} \frac{t^{s-1}}{e^{t}-1}\,,$$

the value of the integral follows

$$-\int_{0}^{\infty} \frac{t}{e^{t}-1} \,dt = -\zeta(2) = -\frac{\pi^2}{6} \,.$$

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You can get the indefinite integral in terms of the dilogarithm function. – Mhenni Benghorbal Oct 17 '12 at 3:37