Answer: The Integral That Knows Its Own Interval

Key Idea / Intuition

At first glance, the integrand has a singularity at both $x = 0$ (where $\ln x \to -\infty$) and $x = 1$ (where $x - 1 \to 0$). But both are removable in the sense that the integrand stays integrable — the two singularities "cooperate." The key trick is to expand $\frac{1}{1-x}$ as a geometric series and integrate term by term, turning the integral into a famous sum: $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$.

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Formal Proof / Solution

Step 1: Rewrite the integrand.

Note that $\frac{\ln x}{x-1} = \frac{-\ln x}{1-x}$, and for $0 < x < 1$ we have the geometric series: $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n.$

So: $\frac{\ln x}{x-1} = -\ln x \cdot \sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty} (-x^n \ln x).$

Step 2: Check that term-by-term integration is valid.

Each term $-x^n \ln x \geq 0$ on $(0,1)$ (since $\ln x \leq 0$ there). By the Monotone Convergence Theorem, we may interchange sum and integral: $I = \sum_{n=0}^{\infty} \int_0^1 (-x^n \ln x)\, dx.$

Step 3: Compute each term.

For fixed $n \geq 0$: $\int_0^1 x^n (-\ln x)\, dx.$

Use the substitution $x = e^{-t}$, $dx = -e^{-t}\,dt$: $\int_0^\infty e^{-nt} \cdot t \cdot e^{-t}\, dt = \int_0^\infty t\, e^{-(n+1)t}\, dt = \frac{1}{(n+1)^2},$

where we used $\int_0^\infty t\, e^{-at}\, dt = \frac{1}{a^2}$ for $a > 0$.

Step 4: Sum the series.

$I = \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} = \sum_{m=1}^{\infty} \frac{1}{m^2} = \frac{\pi^2}{6}.$

Conclusion: $\boxed{I = \int_0^1 \frac{\ln x}{x-1}\,dx = \frac{\pi^2}{6}.}$

Why this is beautiful: The integral secretly encodes all the information of the Basel problem. Both singularities at $x=0$ and $x=1$ are integrable (the integrand extends continuously to the value $1$ at $x=1$ and vanishes at $x=0$), and the geometric series expansion converts the analytic problem into an arithmetic one.