Worked solution: The Beukers–Calabi–Kolk double-integral proof (1993)

Frequently asked questions

Why does the substitution work at all?

Eugenio Calabi found it by looking for a change of variables that would give a constant Jacobian for integrands of this shape; there is no simple reason to expect it in advance, which is why the proof is often called "miraculous".

Does this method generalise beyond ζ(2)?

Yes: Beukers, Calabi and Kolk used higher-dimensional analogues of the same substitution to evaluate ζ(4), ζ(6) and other generalised harmonic series as volumes of polytopes.

Is this proof considered fully rigorous?

Yes — unlike Euler's 1735 argument, every step here (a geometric series, term-by-term integration of positive terms, a change of variables with an explicit Jacobian) is standard, rigorous multivariable calculus.

Source

Frits Beukers, Eugenio Calabi, Johan A. C. Kolk, 1993

Sums of generalized harmonic series and volumes (Nieuw Archief voor Wiskunde, 4th series, vol. 11, no. 3, pp. 217–224)

Step 6 of 6: Back to ζ(2)

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In plain words

The triangle's area was computed with nothing but calculus and geometry, no infinite products in sight; plugging that number back into the equation from Step 1 hands back $\zeta(2)$ , closing the loop from a sum, to an integral, to a triangle, and back to a sum.

\zeta(2)=\frac43\cdot\frac{\pi^2}{8}=\frac{\pi^2}{6}

Detailed analysis

Substituting the triangle's area back into the identity from Step 1 gives $\pi^2/6$ — the same number Euler found in 1735, but reached here purely through multivariable calculus, with no infinite products and no appeal to the Maclaurin series of $\sin x$ .

Knowledge used in this step

Numerical series and convergence tests