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

Step 3 of 6: The Beukers–Calabi–Kolk substitution

In plain words

The two auxiliary angles $u$ and $v$ act like knobs on an old radio: turning them sweeps out every point of a triangular region $\Delta$ exactly once, and the sine/cosine combination is engineered so that the corners of the unit square — including the awkward corner where the integrand blows up — land precisely on the corners of that triangle.

x=\frac{\sin u}{\cos v},\quad y=\frac{\sin v}{\cos u},\qquad (u,v)\in\Delta=\Big\{u,v>0,\ u+v<\frac{\pi}{2}\Big\}

Sine and cosine of an angle on the unit circle — the two building blocks of the substitution

Detailed analysis

This substitution sends $u=0$ to $x=0$ and $v=0$ to $y=0$ , while the far edge $u+v=\pi/2$ corresponds to $x=1$ or $y=1$ — so it maps the open unit square exactly onto the open triangle $\Delta$ with two legs of length $\pi/2$ .

Terms in this step

Change of variables: Replacing the integration variables $x,y$ by new variables $u,v$ related by a formula, so as to turn a hard integral into an easier one over a different region.

Knowledge used in this step

Trigonometry