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 $\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n^2}=\dfrac{\pi^2}{12}$.
Explanation of Method
Dividing the $\ln(1+x)$ series by $x$ and integrating term by term over $[0,1]$ produces exactly the given series, so the answer follows immediately from the supplied value.
- Divide the $\ln(1+x)$ series by $x$.
- Integrate term by term over $[0,1]$.
- Match the resulting series to the given sum.
Step 1 · Divide by $x$
$$ \frac{\ln(1+x)}x = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}nx^{n-1} = \sum_{m=0}^{\infty}\frac{(-1)^m}{m+1}x^m. $$
Step 2 · Integrate over $[0,1]$
$$ \int_0^1\frac{\ln(1+x)}x\,dx = \sum_{m=0}^{\infty}\frac{(-1)^m}{(m+1)^2}\int_0^1x^m\cdot(m+1)\,dx\Bigg/(m+1) = \sum_{m=0}^{\infty}\frac{(-1)^m}{(m+1)^2}. $$
Step 3 · Match to the given series
Reindexing with $n=m+1$: $$ \sum_{m=0}^{\infty}\frac{(-1)^m}{(m+1)^2} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12}. $$
Summary Table
| Property | Value |
|---|---|
| Result | $\pi^2/12\approx0.822467$ |
| Related constant | Dirichlet eta, $\eta(2)$ |
| Related special value | $-\operatorname{Li}_2(-1)$ |
Additional Notes
- This is precisely $-\operatorname{Li}_2(-1)$, the negative dilogarithm at $-1$, connecting directly to Problem 264's general dilogarithm series and confirming the boundary value found there.
- The given sum $\eta(2)=\pi^2/12$ follows from $\eta(s)=(1-2^{1-s})\zeta(s)$ at $s=2$: $\eta(2)=(1-\tfrac12)\cdot\tfrac{\pi^2}6=\tfrac{\pi^2}{12}$, tying this integral to the Basel problem.
- This closes the loop with Problem 264: that problem derived the general series $\int_0^x\ln(1+t)/t\,dt=\sum(-1)^{n-1}x^n/n^2$; setting $x=1$ there reproduces exactly this result.