By expanding the integrand in a power series, calculate

By expanding the integrand in a power series, calculate $\displaystyle\int_0^1\frac{\ln(1-x)}x\,dx$, given that $\sum_{n=1}^{\infty}\dfrac1{n^2}=\dfrac{\pi^2}6$.

Step 1 · Divide by $x$

$$ \frac{\ln(1-x)}x = -\sum_{n=1}^{\infty}\frac{x^{n-1}}n. $$

Step 2 · Integrate over $[0,1]$

$$ \int_0^1\frac{\ln(1-x)}x\,dx = -\sum_{n=1}^{\infty}\frac1n\int_0^1x^{n-1}\,dx = -\sum_{n=1}^{\infty}\frac1{n^2}. $$

Step 3 · Apply the given result

$$ \int_0^1\frac{\ln(1-x)}x\,dx = -\frac{\pi^2}6. $$ This is $-\zeta(2)=-\operatorname{Li}_2(1)$, the negative of Euler's famous 1734 solution to the Basel problem.

Summary Table

Additional Notes

• This is the mirror image of Problem 300: there $\ln(1+x)/x$ integrated to $\pi^2/12$ (half the Basel sum, via the alternating series), while here $\ln(1-x)/x$ integrates to $-\pi^2/6$ (the full, non-alternating Basel sum, negated).
• Since $\pi^2/6=2\cdot\pi^2/12$, the two results are related by exactly a factor of $-2$ — a reflection of the general dilogarithm identity $\operatorname{Li}_2(-1)=-\tfrac12\operatorname{Li}_2(1)$.
• The Basel problem, $\sum1/n^2=\pi^2/6$, is one of the most celebrated results in analysis; this integral representation as $-\operatorname{Li}_2(1)$ is a standard route to (re)deriving it independently, e.g. via Fourier series or double-integral techniques.