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

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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 ter$m$$s$ of the dilogarith$m$ function.2012-10-17