By expanding the integrand in a power series, calculate
By expanding the integrand in a power series, calculate $\displaystyle\int_0^1\ln x\ln(1-x)\,dx$.
Explanation of Method
Expanding only $\ln(1-x)$ in series leaves a factor of $\ln x$ inside each term; each resulting integral $\int_0^1x^n\ln x\,dx$ has a simple closed form via integration by parts, turning the whole problem into a partial-fractions sum.
- Expand $\ln(1-x)$ and factor out $\ln x$.
- Evaluate $\int_0^1x^n\ln x\,dx$ by parts.
- Sum the resulting series via partial fractions.
Step 1 · Expand $\ln(1-x)$
$$ \ln x\ln(1-x) = -\ln x\sum_{n=1}^{\infty}\frac{x^n}n. $$
Step 2 · Evaluate $\int_0^1x^n\ln x\,dx$
By parts ($u=\ln x$, $dv=x^n\,dx$), with the boundary terms vanishing at both ends: $$ \int_0^1x^n\ln x\,dx = -\frac1{n+1}\int_0^1x^n\,dx = -\frac1{(n+1)^2}. $$
Step 3 · Sum the resulting series
$$ \int_0^1\ln x\ln(1-x)\,dx = -\sum_{n=1}^{\infty}\frac1n\left(-\frac1{(n+1)^2}\right) = \sum_{n=1}^{\infty}\frac1{n(n+1)^2}. $$ Partial fractions give $\dfrac1{n(n+1)^2}=\dfrac1n-\dfrac1{n+1}-\dfrac1{(n+1)^2}$ (check at $n=1$: $\frac14=1-\frac12-\frac14$ ✓). The first two terms telescope to $1$; the third sums to $\sum_{m=2}^{\infty}\frac1{m^2}=\frac{\pi^2}6-1$. So: $$ \int_0^1\ln x\ln(1-x)\,dx = 1-\left(\frac{\pi^2}6-1\right) = 2-\frac{\pi^2}6. $$
Summary Table
| Property | Value |
|---|---|
| Series | $\sum1/[n(n+1)^2]$ |
| Result | $2-\pi^2/6\approx0.3551$ |
Additional Notes
- The building block $\int_0^1x^n\ln x\,dx=-1/(n+1)^2$ recurs throughout this family of "logarithm times power" integrals — it is exactly the negative reciprocal square, a direct consequence of differentiating $\int_0^1x^n\,dx=1/(n+1)$ with respect to $n$.
- This result is a classical example of an Euler-type sum, connecting elementary partial fractions to the Basel constant $\pi^2/6$ from Problem 301.
- By symmetry $x\leftrightarrow1-x$, the companion integral $\int_0^1\ln(1-x)\ln x\,dx$ is identical to this one — the integrand is already symmetric under that swap, so no new computation is needed for the mirrored version.