In Aigner, Ziegler: Proofs_from_THE_BOOK they explain an elegant derivation of $\zeta(2)$ given by Beukers, Calabi and Kolk. One starts from the integral $$ J = \int_0^1 \int_0^1 \frac{1}{1-x^2y^2}dxdy = \frac{3}{4}\zeta(2) $$ To compute the integral they use the following non-trivial coordinate transform $$ x = \frac{\sin u}{\cos v}, y = \frac{\sin v}{\cos u} $$ The reason for this transform is that the Jacobi determinate of this transformation turns out to be $$ |D| = 1-\frac{\sin^2u\sin^2v}{\cos^2u\cos^2v} = 1 - x^2y^2 $$ So after the transformation the integral becomes magically $$ \int_{f_1(u,v)}^{f_2(u,v)} \int_{g_1(u,v)}^{g_2(u,v)} 1 dudv $$ Now they argue that the limits of the integration are $$ f_1(u,v) = g_1(u,v) = 0 \\ f_2(u,v) = \frac{\pi}{2}\\ g_2(u,v) = \frac{\pi}{2}-v $$ so that the integral turns out to be $$ J = \frac{\pi^2}{8} $$ But I have problems understanding how one comes up with the limits of integration for such a coordinate transform (especially I wonder about the "$-v$"). Can someone give maybe a step-by-step solution for computing the limits of integration here? I would also be interested how the limits would look like, if I would want to compute $\zeta(4)$ from the corresponding quadruple integral.


1 Answer 1


First note that the limits are not functions of two variables but in fact we have for every double integral $$ J = \int_{v=f_1}^{v=f_2} \int_{u=g_1(v)}^{u=g_2(v)} 1 dudv $$ So the outer limits are constants, and the limits of the inner integral are functions of $v$ only. To compute the limits in a systematic way, I suggest the following "algorithm":

  1. From the limits of the integral form the following system of inequalities $$ \begin{align} 0 \leq x \leq 1 && 0 \leq y \leq 1 \end{align} $$
  2. Substitute the coordinate transform and solve for the inner integration variable $u$: $$ \begin{align} 0 \leq \frac{\sin u}{\cos v} \leq 1 && 0 \leq \frac{\sin v}{\cos u} \leq 1 \\ 0 \leq u \leq \arcsin(\cos v) && 0 \leq v \leq \arcsin(\cos u) \\ 0 \leq u \leq \arcsin(\sin (\frac{\pi}{2}-v) && 0 \leq v \leq \arcsin(\sin (\frac{\pi}{2}-u) \\ 0 \leq u \leq \frac{\pi}{2}-v && 0 \leq u \leq \frac{\pi}{2}-v \end{align} $$ From this you can read off $g_1(u,v)=0$ and $g_2(u,v)=\frac{\pi}{2}-v$.
  3. Repeat step two, but this time solve for the outer integration variable $v$ and set $u=0$ at the end, because the outer limits $f_1$ and $f_2$ must be independent of $u$: $$ \begin{align} \ldots && \ldots \\ 0 \leq v \leq \frac{\pi}{2}-u && 0 \leq v \leq \frac{\pi}{2}-u \\ 0 \leq v \leq \frac{\pi}{2} && 0 \leq v \leq \frac{\pi}{2} \end{align} $$ From the last line you can read off the outer limits $f_1=0$ and $f_2=\frac{\pi}{2}$.

P.S.: I'm answering my own question here. So take this with a grain of salt.

  • $\begingroup$ If there a source that you can point to that explains how the limit are computed after a change of variables? I am looking for a procedure that can be applied "mechanically" in multiple dimensions (>>3) without having to "draw" or "infer"etc. Many thanks $\endgroup$
    – Confounded
    Commented Nov 15, 2021 at 0:42

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