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

Step 5 of 6: The integral is the area of a triangle

In plain words

Once the awkward denominator has cancelled, what is left to integrate is simply the constant function $1$ — and integrating $1$ over a region is just a fancy way of asking for that region's area, which for a right triangle is half the product of its two legs.

\int_0^1\!\!\int_0^1 \frac{dx\,dy}{1-x^2y^2}=\iint_{\Delta} du\,dv=\operatorname{Area}(\Delta)=\frac12\left(\frac{\pi}{2}\right)^2=\frac{\pi^2}{8}

Detailed analysis

After the substitution the integrand is identically $1$ , so the double integral is simply the area of $\Delta$ : a right triangle with both legs equal to $\pi/2$ .

Knowledge used in this step

Multiple integrals