Is the Beukers-Kolk-Calabi substitution incorrect?

Question

Consider the sum following sum:

$$ I=\sum_{i = 0}^{\infty}\frac{(-1)^{i}}{(2i+1)^{2}(2i+2)}. $$

Clearly this can be transformed into a triple integral:

$$ I=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{x}{1+x^{2}y^{2}z^{2}}dxdydz.$$

However, because of symmetry I can rewrite the integral as follows:

$$ I=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{y}{1+x^{2}y^{2}z^{2}}dxdydz=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{z}{1+x^{2}y^{2}z^{2}}dxdydz.$$

Now apply the following substitution to each of the integrals:

$$ x = \frac{\sin(u)}{\cos(v)},\space y = \frac{\sin(v)}{\cos(w)},\space z = \frac{\sin(w)}{\cos(u)} $$

and you will find out that suddenly, those integrals are not mutually equal. Why is that so? The same substitution can be used to evaluate, for instance, the alternating sum of the reciprocals of odd cubes, in which case it yields a correct result.

$$ \sum_{i = 0}^{\infty}\frac{(-1)^{i}}{(2i+1)^{3}}=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{1}{1+x^{2}y^{2}z^{2}}dxdydz = \frac{\pi^{3}}{32}. $$

Answer 1

The conditions are that $u$, $v$ and $w$ are all positive, while $u+v$, $v+w$ and $w+u$ are all less than $\pi/2$, so the correct way of integrating is $$ \int_{u=0}^{\pi/2} \int_{v=0}^{\pi/2-u} \int_{w=0}^{\min(\pi/2-u,\pi/2-v)} \cdots $$

Answer 2

To answer your question in general, one can evaluate the sum $S(k)=\sum_{n \ge 0} \frac{(-1)^{nk}}{(2n+1)^k}$ for $k \in \mathbb{N}$ via evaluating the multidimensional integral $$I_k=\int_{(0,1)^k} \frac{1}{1-(-1)^k x_1^2 \ \dots \ x_k^2} \ dx_k \ \dots \ dx_1.$$

With some rigorous justification, we can use Calabi's change of variables $x_i=\frac{\sin(u_i)}{\cos(u_{i+1})}$ for $1 \le i \le k,$ with $u_{k+1} :=u_1$ to evaluate this integral $I_k.$ Calabi's change of variables has Jacobian determinant: $$\left |\frac{\partial(x_1, \ \dots \ ,x_k)}{\partial (u_1, \ \dots \ , u_k)} \right|=\left(\frac{\pi}{2} \right)^k \left(1-(-1)^k x_1^2 \ \dots \ x_k^2\right),$$ and diffeomorphically maps the open cube $(0,1)^k$ to the convex polytope $$\Delta^k = \left \lbrace (u_1, \ \dots \, u_k): u_i+u_{i+1} < 1, u_i>0, i \in \lbrace 1, \ \dots \ , k \rbrace \right \rbrace.$$ Thus, $$S(k)= \left( \frac{\pi}{2} \right)^k \text{Volume}(\Delta^k).$$

There is a general way to compute the volume of $\Delta^k,$ which amounts to dissecting $\Delta^k$ into a disjoint union consisting of the open cube $(0,1/2)^k$ and simplices in $\mathbb{R}^k$ based on combinatorial arguments. See this joint paper by Daniele Ritelli and myself: https://www.ams.org/journals/qam/2018-76-03/S0033-569X-2018-01499-3/

I used a special case $k=4$ of the general argument to answer this question: https://math.stackexchange.com/a/1794967/169367.

When it is all said and done, the result is the following: $$S(k)= \left(\frac{\pi}{4} \right)^k +\left(\frac{\pi}{4} \right)^k \sum_{n=1}^{ \left \lfloor \frac{k}{2} \right \rfloor} \sum_{\substack{(r_1, \dots, r_n) \in [k]^n: \\ |r_p-r_q| \notin \lbrace 0,1,k-1 \rbrace, \\ p,q \in [n]} } \prod_{i=1}^{n} \frac{1}{i+\sum_{j=1}^{i} \alpha_j},$$ where $[m]:= \lbrace 1, \dots, m \rbrace$ and $$\alpha_j=2- \delta(k,2) - \sum_{m=1}^{j-1} \delta(|r_m-r_j|,2)+\delta(|r_m-r_j|,k-2)$$ and $\delta(a,b)$ is the Kronecker Delta Function. In particular, the inner sum in the second term of that gargantuan formula above is taken over all tuples $(r_1, \dots, r_n) \in [k]^ n$ having cyclically pairwise nonconsecutive entries.