Answer: The Dirichlet Beta Integral

Key Idea / Intuition

Near $x=1$ the denominator $x^2-1$ vanishes, but so does the numerator $\ln x$, so the integrand is actually continuous there — no real singularity. The trick is to expand $\frac{1}{1-x^2}$ as a geometric series in $x^2$, interchange sum and integral, and recognise the resulting series as a classical constant: $\frac{\pi^2}{8}$.

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Formal Proof / Solution

Step 1: Rewrite the sign.

Note $x^2 - 1 < 0$ on $(0,1)$ and $\ln x < 0$ on $(0,1)$, so the integrand is positive. Write:

$I = \int_0^1 \frac{\ln x}{x^2-1}\,dx = \int_0^1 \frac{-\ln x}{1-x^2}\,dx$

Step 2: Expand as a geometric series.

For $0 \le x < 1$:

$\frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n}$

So:

$I = -\int_0^1 \ln x \sum_{n=0}^{\infty} x^{2n}\,dx = \sum_{n=0}^{\infty} \left(-\int_0^1 x^{2n}\ln x\,dx\right)$

The interchange is justified by the monotone convergence theorem (all terms are positive).

Step 3: Evaluate each term.

For any $\alpha > -1$:

$\int_0^1 x^\alpha \ln x\,dx = \frac{-1}{(\alpha+1)^2}$

(Differentiate $\int_0^1 x^\alpha dx = \frac{1}{\alpha+1}$ with respect to $\alpha$.)

With $\alpha = 2n$:

$-\int_0^1 x^{2n}\ln x\,dx = \frac{1}{(2n+1)^2}$

Step 4: Sum the series.

$I = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \cdots$

This is the Leibniz/Dirichlet beta sum. Since $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ and the even terms contribute $\frac{1}{4}\cdot\frac{\pi^2}{6} = \frac{\pi^2}{24}$:

$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{\pi^2}{8}$

Result:

$\boxed{I = \dfrac{\pi^2}{8}}$