MathLabs

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

Step 2 of 6: Odd squares as a double integral
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In plain words

The expression 11−t\frac{1}{1-t} is the sum of an endlessly doubling-back geometric series 1+t+t2+⋯1+t+t^2+\cdots; setting t=x2y2t=x^2y^2 and integrating each power separately over the unit square turns an algebraic sum into a stack of ordinary calculus integrals, one per term, that happen to reassemble into exactly the odd-square sum from Step 1.

∫01 ⁣ ⁣∫01dx dy1−x2y2=∑k=0∞(∫01x2kdx)(∫01y2kdy)=∑k=0∞1(2k+1)2\int_0^1\!\!\int_0^1 \frac{dx\,dy}{1-x^2y^2}=\sum_{k=0}^{\infty}\left(\int_0^1 x^{2k}dx\right)\left(\int_0^1 y^{2k}dy\right)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}
A paraboloid over the unit square, illustrating a double integral as the volume beneath a surface (the actual integrand 1/(1−x2y2)1/(1-x^2y^2) only blows up at the corner (1,1)(1,1))
A 3D plot of the generic paraboloid z = x² + y² over the unit square [0,1]×[0,1], used here only to illustrate the general idea that a double integral computes the volume under a surface. The actual integrand of this proof, 1/(1−x²y²), is finite everywhere in the square except at the single corner (1,1), where it diverges mildly.
Detailed analysis

Expand 1/(1−x2y2)1/(1-x^2y^2) as a geometric series in (xy)2(xy)^2 and integrate term by term over the unit square; every term is positive, so swapping the infinite sum and the integral is legitimate. The result is exactly the odd-square sum from Step 1.

Terms in this step
Geometric series
A series of the form 1+t+t2+t3+⋯1+t+t^2+t^3+\cdots; for ∣t∣<1|t|<1 it converges to 11−t\frac{1}{1-t}.
Improper integral
An integral where the integrand is unbounded somewhere in the region (here, near the corner (1,1)(1,1)), so the integral is defined as a limit rather than an ordinary Riemann sum.
Knowledge used in this step
Multiple integralsPower series and Taylor series
Common mistake. The integrand is unbounded near the corner (1,1)(1,1); the integral must be understood as improper, and convergence should be checked (each term ∫01x2kdx∫01y2kdy=1/(2k+1)2\int_0^1x^{2k}dx\int_0^1y^{2k}dy=1/(2k+1)^2 is finite, and the resulting series converges) rather than assumed.