An interesting solution of the Basel problem is given in https://www.tandfonline.com/doi/abs/10.1080/00029890.2018.1452473 (Basel Problem: A Solution Motivated by the Power of a Point, by Kapil R. Shenvi Pause).

In the integral $$\zeta(2)=\int_0^1\int_0^1\frac{dx\,dy}{1-xy}$$ let's make a change of variables $$x=\cos{\phi}-\tan{\theta}\sin{\phi},\;\;y=\cos{\phi}+\tan{\theta}\sin{\phi}. \tag{1}$$ The Jacobian of this transformation is twice the inverse of $1/(1-xy)$ and hence $$\zeta(2)=2\int\int d\theta\,d\phi.$$ The integration domain in the last integral is determined by the conditions $0\le x,y\le 1$, which gives $$-\frac{\phi}{2}\le\theta\le\frac{\phi}{2},\;\;\;\phi-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}-\phi. \tag{2}$$ Geometrically (2) is a quadrilateral in the $(\theta,\phi)$ plane with mutually orthogonal diagonals of lengths $\pi/3$ and $\pi/2$. Therefore $$\zeta(2)=2\;\frac{1}{2}\frac{\pi}{2}\frac{\pi}{3}=\frac{\pi^2}{6}.$$ Very nice, isn't it?

In the paper the transformation of variables (1) is motivated by an interpretation of $1/(1-xy)$ in terms of inversive geometry. Can this approach be extended to $$\zeta(n)=\int_0^1\cdots \int_0^1\frac{dx_1\cdots dx_n}{1-x_1\cdots x_n}?$$ In particular can we calculate in this manner $\zeta(2n)$? What is the relation (if any) of (1) with the Beukers-Kolk-Calabi change of variables? For the latter see https://pdfs.semanticscholar.org/35be/01e63c0bfd32b82c97d58ccc9c35471c3617.pdf

P.S. The Beukers-Kolk-Calabi change of variables is $$x=\frac{\sin{\phi}}{\cos{\theta}},\;\;\;y=\frac{\sin{\theta}}{\cos{\phi}}.$$ Its Jacobian is $1-x^2y^2$ and it is applied to the integral $$\zeta(2)=\frac{4}{3}\int_0^1\int_0^1\frac{dx\,dy}{1-x^2y^2}.$$ In this case the integration domain becomes isosceles right triangle $0\le \phi,\;0\le \theta,\;\phi+\theta\le \pi/2$.

  • 1
    $\begingroup$ Wouldn't you rather calculate $\zeta(2n+1)$? $\endgroup$ – Gerry Myerson Jul 16 '18 at 12:03
  • $\begingroup$ I mean for $\zeta(2n)$ we can calculate the corresponding volumes in terms of powers of $\pi$ and rational numbers. For $\zeta(2n+1)$ one can get some new integral representations (maybe) but I doubt the volumes can be calculated completely in this case. $\endgroup$ – Zurab Silagadze Jul 16 '18 at 12:12
  • 1
    $\begingroup$ I am a bit confused; haven't you explored and answered this question yourself in ias.ac.in/article/fulltext/reso/020/09/0822-0843 ? $\endgroup$ – Carlo Beenakker Jul 16 '18 at 12:28
  • $\begingroup$ In that article I used Beukers-Kolk-Calabi change of variables. The change of variables by Shenvi Pause is quite different. I wonder if there is some connection between them. $\endgroup$ – Zurab Silagadze Jul 16 '18 at 12:34
  • $\begingroup$ Indeed that's a very nice article. For the record: your related answer mathoverflow.net/a/270355/29783 and the threads linked to that question. $\endgroup$ – Wolfgang Jul 26 '18 at 8:08

It doesn't quite answer your questions, but there is another change of variables that Zagier and Kontsevich use to evaluate

$$\int_{0}^{1} \int_{0}^{1} \frac{1}{4\sqrt{xy} \ (1-xy)} \ dx \ dy.$$ (See page 8 of https://www.maths.ed.ac.uk/~v1ranick/papers/kontzagi.pdf). The above integral one hand is equal to the series $$\sum_{n=0}^{\infty} \frac{1}{4(n+1/2)^2}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$$ and on the other hand equal to $\pi^2/8$ using the change of variables $$x=\frac{\xi^2(1+\eta^2)}{1+\xi^2}, \quad y=\frac{\eta^2(1+\xi^2)}{1+\eta^2}.$$ It turns out the Jacobian Determinant for this transformation is $$\left|\frac{\partial(x,y)}{\partial(\xi,\eta)} \right|=\frac{4\sqrt{xy}(1-xy)}{(1+\xi^2)(1+\eta^2)}.$$ Like the Shenvi Pause change of variables and the Eugenio Calabi change of variables that you mention, the Jacobian Determinant contains the denominator of the integrand. Perhaps making some kind of trigonometric substitution for $\eta$ and $\xi$ could recover the Shenvi Pause change of variables.

There is, however, a $k$-dimensional generalization for the above change of variables. To evaluate $$\int_{(0,1)^k} \frac{1}{2^k\sqrt{x_1 \dots x_k}(1-(-1)^k x_1 \dots x_k)} \ dx_1 \dots \ dx_k,$$ which is equal to $$\sum_{n=0}^{\infty} \frac{(-1)^{nk}}{2^k(n+1/2)^k}=\sum_{n=0}^{\infty} \frac{(-1)^{nk}}{(2n+1)^k},$$ put $$x_i= \frac{\xi^2_i(1+\xi^2_{i+1})}{1+\xi^2_i}, \quad i \in \lbrace 1, \dots, k \rbrace,$$ where $\xi^2_{k+1}:=\xi^2_{1}.$ It turns out $$\left|\frac{\partial(x_1, \dots, x_k)}{\partial(\xi_1, \dots, \xi_k)} \right|=\frac{2^k \sqrt{x_1 \dots x_k}(1-(-1)^k x_1 \dots x_k)}{(1+\xi_1^2) \dots (1+\xi_k^2)}.$$

It is quite similar to Calabi's change of variables because of the structure (cyclical indexing) and the Jacobian Determinant once again containing the denominator of the integrand. In my published article (http://www.ams.org/journals/qam/0000-000-00/S0033-569X-2018-01499-3/), I discuss how the generalized Calabi's change of variables and the generalized Zagier/Kontsevich's change of variables lead to integrals whose evaluations are remarkably similar and a non-traditional representation of $\zeta(k)$ for even $k.$

Some trigonometric substitution for $\xi_i$ might provide a generalized version of the Shenvi Pause's change of variables.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.