By expanding the integrand in a power series, calculate
By expanding the integrand in a power series, calculate $\displaystyle\int_0^1\ln\frac{1+x}{1-x}\,dx$.
Explanation of Method
Problem 272 already expanded $\ln\frac{1+x}{1-x}$ as an odd-power series with doubled coefficients; integrating that series term by term over $[0,1]$ produces a telescoping-style combination of odd and even reciprocals that collapses exactly to the alternating harmonic series.
- Recall the series $\ln\frac{1+x}{1-x}=2\sum x^{2n+1}/(2n+1)$.
- Integrate term by term over $[0,1]$.
- Recognize the resulting series as the alternating harmonic series.
Step 1 · The series (from Problem 272)
$$ \ln\frac{1+x}{1-x} = 2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}. $$
Step 2 · Integrate over $[0,1]$
$$ \int_0^1\ln\frac{1+x}{1-x}\,dx = 2\sum_{n=0}^{\infty}\frac1{2n+1}\int_0^1x^{2n+1}\,dx = 2\sum_{n=0}^{\infty}\frac1{(2n+1)(2n+2)}. $$
Step 3 · Recognize the alternating harmonic series
Using $\dfrac1{(2n+1)(2n+2)}=\dfrac1{2n+1}-\dfrac1{2n+2}$: $$ \int_0^1\ln\frac{1+x}{1-x}\,dx = 2\sum_{n=0}^{\infty}\left(\frac1{2n+1}-\frac1{2n+2}\right) = 2\left(1-\frac12+\frac13-\frac14+\cdots\right) = 2\ln2, $$ since the bracketed series is exactly the alternating harmonic series from Problem 289.
Summary Table
| Property | Value |
|---|---|
| Result | $2\ln2\approx1.386294$ |
| Equivalent form | $2\int_0^1\operatorname{artanh}(x)\,dx$ |
Additional Notes
- A direct antiderivative check confirms this: $\int\ln\frac{1+x}{1-x}\,dx=(1+x)\ln(1+x)+(1-x)\ln(1-x)-2+C$, which evaluates to $(2\ln2-2)-(-2)=2\ln2$ over $[0,1]$ — matching the series result exactly, even though the integrand itself blows up logarithmically as $x\to1^-$ (an integrable singularity).
- The reduction to exactly the alternating harmonic series (Problem 289) is no coincidence: $\int_0^1\ln\frac{1+x}{1-x}dx=2\int_0^1\operatorname{artanh}(x)\,dx$, and integrating $\operatorname{artanh}$ termwise naturally regenerates the $\ln2$ series.
- This integral, despite the singularity at $x=1$, converges because $\ln\frac1{1-x}$ diverges only logarithmically — integrable near any endpoint, unlike a power-law singularity.