---
abstract: 'Among the several proofs known for $\sum_{n=1}^\infty{\!  1/n^2} = {\pi^2/6}$, the one by Beukers, Calabi, and Kolk involves the evaluation of $\,\int_0^1\!\!{\int_0^1{1/(1-x^2 y^2) \, dx} \, dy}$. It starts by showing that this double integral is equivalent to $\frac34 \sum_{n=1}^\infty{\!  1/n^2}$, and then a non-trivial *trigonometric* change of variables is applied which transforms that integral into $\,{\int\!\!\int}_T \: 1 \; du \, dv$, where $T$ is a triangular domain whose area is simply ${\pi^2/8}$. Here in this note, I introduce a hyperbolic version of this change of variables and, by applying it to the above integral, I find exact closed-form expressions for $\int_0^\infty{\left[\sinh^{-1}{(\cosh{u})}-u\right] d u}$, $\,\int_{\alpha}^\infty{\left[u-\cosh^{-1}{(\sinh{u})}\right] d u}$, and $\,\int_{\,\alpha/2}^\infty{\ln{(\tanh{u})} \: d u}$, where $\alpha = \sinh^{-1}(1)$. From the latter integral, I also derive a two-term dilogarithm identity.' 
address: 'Institute of Physics, University of Brasilia, P.O. Box 04455, 70919-970, Brasilia, DF, Brazil'
author:
- 'F. M. S. Lima'
title: 'New definite integrals and a two-term dilogarithm identity'
---

Euler sums ,Riemann zeta function ,Double integrals ,Hyperbolic functions ,Dilogarithm identities

40C10 ,11M06 ,33B30

Introduction
============

The Riemann zeta function $\zeta{(s)}$ is defined, for complex values of $s$, $\Re{(s)}>1$, by $\zeta{(s)} := \sum_{n=1}^\infty{\,1/n^s}$. For integer values of $s$, $s>1$, the first zeta value is $\zeta{(2)} = \sum_{n=1}^\infty{1/n^2}$. Euler was the first to estimate the numerical value of this series to more than $5$ decimal places (1735), as well as to determine an exact closed-form expression for it (1740), as given by [@Euler]:

\[Euler’s result for $\zeta{(2)}$\] \[lem:z2\] $\sum_{n=1}^\infty{\!  1/n^2} = {\pi^2/6}$.

From Euler’s time onward, this result has been proved in several forms, from elementary to complex ones [@Kalman; @Chapman].  Among these proofs, at least two involve the evaluation of a double integral over the unit square and these are the ones in which we are interested here.  Let us briefly describe these two proofs for a better exposition of the hyperbolic approach and its consequences. 

The short proof by Apostol (1983) can be found in Ref.  [@Apostol].  He starts by expanding the integrand of Beukers’ integral $I := \int_0^1\!\!\!{\int_0^1{\!\frac{1}{1-x\,y}~dx} \, dy}$ as a geometric series, in order to show that it is equivalent to $$\int_0^1\!\!\!{\int_0^1{\left(1 +x y +x^2 y^2 +\ldots\right) dx} \, dy} = \int_0^1{\left(1+\frac{y}{2}+\frac{y^2}{3}+\ldots\right) dy} %\nonumber \\
= \sum_{n=1}^\infty{\frac{1}{n^2}} \, ,$$ the interchange of integrals and sums being justifiable by the fact that each integrand is nonnegative and the sums converge absolutely (see the only corollary of Theorem 7.16 in Ref.  [@bk:Rudin]).[^1] By applying a simple linear change of variables corresponding to the rotation of the coordinate axes through the angle $\pi/4$ radians, as given by[^2] $$x = \frac{u-v}{\sqrt{2}} \quad \text{and} \quad y = \frac{u+v}{\sqrt{2}} \, ,$$ directly to the Beukers’ integral, one finds [@Apostol; @bk:TheBook]: $$I = 4 \int_0^{\sqrt{2}/2}\:{\frac{\arctan\!{\left( \frac{u}{\sqrt{2-u^2}}\right)}}{\sqrt{2-u^2}}\, du} %\nonumber \\
+ 4 \int_{\,\sqrt{2}/2}^{\sqrt{2}}{\frac{\arctan\!{\left( \frac{\sqrt{2}-u}{\sqrt{2-u^2}}\right)}}{\sqrt{2-u^2}}\, du} \, .$$ Now, the substitution $\,u = \sqrt{2} \: \sin{\theta}$, followed by the use of the trigonometric identity $\,\arctan{\left(\sec{\theta}-\tan{\theta}\right)} = {\,\pi/4}-{\,\theta/2}\,$ on the second integral, yields $I = {\,\pi^2/18} + {\,\pi^2/9}$. Therefore, $\sum_{n=1}^\infty{1/n^2} = {\,\pi^2/18} + {\,\pi^2/9} = {\,\pi^2/6}$. $\Box$

The other proof, in which we are more interested here, was given by Beukers, Calabi, and Kolk (1993) [@Calabi]. 

Similarly to Apostol’s proof, Beukers, Calabi, and Kolk (BCK) start by showing that $$\begin{aligned}
K &:=& \int_0^1\!\!\!{\int_0^1{\frac{1}{1-x^2 y^2} \, dx} \, dy} = \int_0^1\!\!\!{\int_0^1{\sum_{n=0}^\infty{x^{2n} y^{2n}} \, dx} \, dy} \nonumber \\
&=& \sum_{n=0}^\infty{\int_0^1\!\!\!{\int_0^1{x^{2n} y^{2n}} \, dx} \, dy} = \sum_{n=0}^\infty{\frac{1}{(2n+1)^2}} \, . 
\label{eq:intK}\end{aligned}$$ Again, as each integrand is nonnegative the absolute convergence allows the interchange of the sum and the integral.  Since $\,\sum_{n=1}^\infty{{\,1/(2n)^2}} = \frac14 \, \sum_{n=1}^\infty{{\,1/n^2}}$, then $\:\sum_{n=0}^\infty{\frac{1}{(2n+1)^2}} = \frac34 \, \sum_{n=1}^\infty{\frac{1}{n^2}}$. They then evaluate the unit square integral that defines $K$ by applying the following non-trivial, trigonometric change of variables [@Calabi]: $$x = \frac{\sin{u}}{\cos{v}} \; , \quad y = \frac{\sin{v}}{\cos{u}} \: .$$ For this change of variables, one easily finds that $|J|= 1-\tan^2{u}\,\tan^2{v}=1-x^2 y^2$, which reduces the integral for $K$ to $\, {\int \! \! \!  \int}_{\!\!  T} \: 1 \; du \, dv \,$ where $T$ is the domain in the $u v$-plane defined by $\{(u,v): \, u \ge 0$, $v \ge 0$, $u+v \le {\,\pi/2} \}$ [@Kalman; @bk:TheBook; @Calabi].  This double integral exactly evaluates to ${\,\pi^2/8\,}$, i.e. the area of the triangle corresponding to $\,T$, which implies Lemma \[lem:z2\].  $\Box$

Here in this work, I introduce a hyperbolic version of BCK change of variables for double integrals over the unit square.  By applying this new change of variables to $\int_0^1\!\!{\int_0^1{1/(1-x^2 y^2) \, dx} \, dy}$, I derive exact closed-form expressions for some definite integrals.  From one of these integrals, I deduce a closed-form expression for $\,\mathrm{Li}_2{(\sqrt{2}-1)} + \mathrm{Li}_2{\left(1-{\,\sqrt{2}/2}\right)}$.

A hyperbolic version of BCK change of variables
===============================================

Let us introduce the following *hyperbolic version* of BCK change of variables for double integrals over the unit square: $$x = \frac{\sinh{u}}{\cosh{v}} \quad \text{and} \quad y = \frac{\sinh{v}}{\cosh{u}} \, . 
\label{eq:hyper}$$ The corresponding Jacobi determinant is: $$J = \left|
  \begin{array}{cc}
    \dfrac{\cosh{u}}{\cosh{v}} & -\dfrac{\sinh{u} \, \sinh{v}}{\cosh^2{v}} \\
    \\
    -\dfrac{\sinh{u} \, \sinh{v}}{\cosh^2{u}} & \dfrac{\cosh{v}}{\cosh{u}} \\
  \end{array}
\right| = 1 -\tanh^2{u}\,\tanh^2{v} . 
\label{eq:detJ}$$ As $0 \le x \, y \le 1$ in the unit square, then $1 - x^2\,y^2 \ge 0$, then $|J|$ is also equal to $1 -\tanh^2{u}\,\tanh^2{v}$. Therefore $$\begin{aligned}
\int_0^1\!\!\!{\int_0^1{F(x,y) \, dx} \, dy} &=& {\int \! \! \!  \int}_{\!\!  S} \, F(x(u,v),y(u,v)) \: \left|J(u,v)\right| \: du \, dv \, \nonumber \\
&=& {\int \! \! \!  \int}_{\!\!  S} \, \tilde{F}(u,v) \: \left(1 -\tanh^2{u}\,\tanh^2{v}\right) \, du \, dv \, ,
\label{HypChange}\end{aligned}$$ where $S$ is the cusped hyperbolic ‘quadrilateral’ on the right-hand side of Fig.  \[fig:mapeamento\].  We are assuming that this change of variables is a $C^{1}$ diffeomorphism, which is reasonable since the functions $x=x(u,v)$ and $y=y(u,v)$ given in Eq.  , as well as $$u = u(x,y) = \cosh^{-1}{\sqrt{\frac{1+x^2}{1-x^2 \, y^2}}}$$ and $$v = v(x,y) = \cosh^{-1}{\sqrt{\frac{1+y^2}{1-x^2 \, y^2}}} \, ,$$ which define the *inverse* change,[^3] are continuously differentiable in all points of their domains (see the hachured regions in Fig.  \[fig:mapeamento\]), the only exception being point $(x,y) = (1,1)$. As indicated there in Fig.  \[fig:mapeamento\], $S$ is bounded by the axes $u$ and $v$, the curve $v = f(u) = \sinh^{-1}{(\cosh{u})}$, above the diagonal $v=u$, and the curve $v = g(u) = \cosh^{-1}{(\sinh{u})}$, below the diagonal.  The parameter $\,\alpha\,$ can be readily evaluated by noting that it is the abscissa of the point where $v = g(u)$ intersects the $u$-axis.  There in that point, one has $\,\cosh^{-1}{\!\left(\sinh{\alpha}\right)} = 0$, thus $\sinh{\alpha} = 1$. Since $\sinh^{-1}{x} = \ln{\left(x+\sqrt{x^2+1}\,\right)}$, $\forall \, x \in \mathbb{R}$, then $$\alpha = \sinh^{-1}{(1)} = \ln\!{\left( 1+\sqrt{2}\, \right)} . 
\label{eq:alfa}$$

New definite integrals and a dilogarithm identity
=================================================

Let us now apply our hyperbolic change of variables to the unit square integral $\,K$, as defined in Eq.  . 

\[teo1\] $$\int_0^{\,\infty}\!{\left[\sinh^{-1}{\left(\cosh{u}\right)} -u\right] \, du} = \frac{\pi^2}{16} \, .$$

From the BCK proof, one knows that $\int_0^1\!\!{\int_0^1{{\,1/(1-x^2 y^2)} \, dx} \, dy} = {\,\pi^2/8}$. By combining this result with , one has $$\begin{aligned}
\frac{\pi^2}{8}  &=& {\int \! \! \!  \int}_{\!\!  S} \, \frac{1}{1 -\tanh^2{u}\,\tanh^2{v}} \, \left(1 -\tanh^2{u}\,\tanh^2{v}\right) \: du \, dv \nonumber \\
&=& {\int \! \! \!  \int}_{\!\!  S} \: 1 \: du \, dv \, = \, \text{area of~} S .\end{aligned}$$ The area of the domain $S$ can be equalized to a single definite integral by noting that $S$ is symmetric with respect to the diagonal $v=u$ (see Fig.  \[fig:mapeamento\]).  Then, by taking only the upper half of $S$ into account, one has $$\text{area of~} S = 2 \int_0^{\,\infty}{\!\!\!{\int_u^{f(u)} \: 1 \: dv} \, du} = 2 \int_0^{\,\infty}\!{\left[\sinh^{-1}{\left(\cosh{u}\right)} -u\right] \, du} \, ,$$ which leads to the desired result.  $\Box$

\[teo2\] $$\int_\alpha^{\,\infty}\!{\left[u -\cosh^{-1}{\left(\sinh{u}\right)}\right] \, du} = \frac{\pi^2}{16} -\frac12\,\ln^2\!{\left(1+\sqrt{2}\,\right)} \, .$$

Similarly to the previous proof, by taking the lower half of $S$ into account and noting that it is bounded by the $u$-axis, the diagonal $v=u$, and the curve $v = g(u)$, as seen in Fig.  \[fig:mapeamento\], one has, from Eq.  , that: $$\text{area of~} S = 2 \, \left[\frac{\alpha^2}{2} + \int_\alpha^{\,\infty}{\!\!\!{\int_{g(u)}^u \: 1 \: dv} \, du} \right] = \alpha^2 + 2 \, \int_\alpha^{\,\infty}\!{\left[u - \cosh^{-1}{\left(\sinh{u}\right)} \right] \, du} \, .$$ Again, since the area of $S$ is ${\,\pi^2/8}$, one finds $$\int_\alpha^{\,\infty}\!{\left[u - \cosh^{-1}{\left(\sinh{u}\right)} \right] \, du} = \frac12 \left(\frac{\pi^2}{8} -\alpha^2\right) = \frac{\pi^2}{16} -\frac{\alpha^2}{2} \, .$$

$\Box$

\[teo3\] $$\int_{\,\alpha/2}^{\,\infty}{\,\ln{\left(\tanh{z}\right)} \: dz} = \frac14 \, \ln^2{\!\left(1+\sqrt{2}\,\right)} -\frac{\pi^2}{16} \, .$$

From Theorem \[teo1\], we know that the area of $S$ evaluates to $$2 \int_0^{\,\infty}{\!\!{\int_u^{\,f(u)} 1 \: dv} \: du} = \frac{\pi^2}{8} \, ,$$ where $f(u)=\sinh^{-1}{(\cosh{u})}$. Let us rotate the coordinate axes by the angle $\pi/4$ rad, as in Apostol’s proof.  Since “pure” rotations about the origin always yield an unitary Jacobi determinant, one has: $$\begin{aligned}
\frac{\pi^2}{8} &=& 2 \int_0^{\,\alpha/\sqrt{2}}{\!\!\int_0^X{ 1 \: dY} \, dX} +2 \int_{\alpha/\sqrt{2}}^\infty{\,\int_0^{\,h(X)}{1 \; dY}\,dX} \nonumber \\
&=& 2 \int_0^{\,\alpha/\sqrt{2}}{X \, dX} +2 \int_{\alpha/\sqrt{2}}^\infty{\,h(X) \, dX} \nonumber \\
&=& \frac{\alpha^2}{2} + 2 \int_{\alpha/\sqrt{2}}^\infty{\,h(X) \: dX} ,
\label{eq:aux}\end{aligned}$$ where $$u = \frac{X-Y}{\sqrt{2}} \; , \quad v = \frac{X+Y}{\sqrt{2}} \, ,$$ and $h(X)$ is determined by changing the variables $u$ and $v$ on the equation $v=f(u)$ to the new variables $X$ and $Y$, as follows.  Since $v = f(u) = \sinh^{-1}(\cosh{u})$, then $\sinh{v} = \cosh{u}$, so $$\sinh{\left( \frac{X+Y}{\sqrt{2}} \right)} = \cosh{\left( \frac{X-Y}{\sqrt{2}} \right)} \, .$$ By making use of the addition and subtraction formulas for hyperbolic functions, as well as of the substitutions $X' := X/\sqrt{2}$ and $Y' := Y/\sqrt{2}$, the above equality expands to $$\sinh{X'} \: \cosh{Y'} + \sinh{Y'} \: \cosh{X'} = %\nonumber \\
\cosh{X'} \: \cosh{Y'} - \sinh{X'}  \: \sinh{Y'} \, .$$ By dividing both sides by $\cosh{X'} \: \cosh{Y'}$, one finds $$\tanh{X'} + \tanh{Y'} = 1 -\tanh{X'} \cdot \tanh{Y'} \, ,$$ which can be written as $$\tanh{Y'} = \frac{1 -\tanh{X'} }{1 +\tanh{X'} } \, . 
\label{eq:dem1}$$ Now, for simplicity, let us define $\, t := \tanh{X'}$. After some algebra, Eq.   becomes $$Y' = \frac12 \: \ln{\left( \frac{1 + \dfrac{1-t}{1+t}}{1 -\dfrac{1-t}{1+t}} \right)} = \frac12 \: \ln{\left(\frac{1}{t}\right)} = -\frac12 \: \ln{t} \, ,$$ which means that $$Y = h(X) = -\frac{\sqrt{2}}{2} \; \ln{\!\left[ \tanh{\!\left(\frac{X}{\sqrt{2}}\right)} \right]} .$$ By putting this function on the last integral of Eq.  , one has $$\frac{\pi^2}{8} = \frac{\alpha^2}{2} - \sqrt{2} \, \int_{\alpha/\sqrt{2}}^\infty{\:\ln{\!\left[\tanh{\left(\frac{X}{\sqrt{2}}\right)}\right]} \, dX} . 
\label{eq:aux2}$$ The simple substitution $\,z = {\,X/\sqrt{2}}\,$ promptly reduces this equation to $$\int_{\alpha/2}^{\,\infty}{\ln{\left(\tanh{z}\right)} \: dz} = \, \frac{\alpha^2}{4} -\frac{\pi^2}{16} \, . 
\label{eq:aux3}$$

$\Box$

After this proof, we make the substitution $\,t = \tanh{z}\,$ in the integral in Eq.  , above.  When the result is expressed in terms of the dilogarithm function, an interesting identity arises. 

\[teo4\] Let us define $\,\mathrm{Li}_2{(z)} := \sum_{k=1}^\infty{{\,z^k/k^2}}$, which is valid for all complex $z$ with $|z| \le 1$ *[@Lewin]*.   Then $$\mathrm{Li}_2{\left(\sqrt{2}-1\right)} +\mathrm{Li}_2{\left( 1 -\frac{1}{\sqrt{2}} \right)} = \frac{\pi^2}{8} -\frac{\ln^2{\!\left(1+\sqrt{2}\,\right)}}{2} -\frac18 \, \ln^2{2} \, . 
\label{eq:fim}$$

By applying the substitution $\,t = \tanh{z}\,$ ($dt = \mathrm{sech}^2{z}\,dz$) to the definite integral of Theorem \[teo3\], in the previous section, one finds $$\frac{\alpha^2}{4} -\frac{\pi^2}{16} = \int_{\tanh{\!({\alpha/2})}}^1{\!\frac{\ln{t}}{1-t^2} \; dt} %\nonumber \\
= \int_{\sqrt{2}-1}^1{\frac{\ln{t}}{1-t^2} \; dt} \, ,
\label{eq:ln}$$ since $$\tanh{\!\left(\frac{\alpha}{2}\right)} = \tanh{\!\left(\ln{\sqrt{1+\sqrt{2}}}\right)} = \frac{\sqrt{1+\sqrt{2}} -\dfrac{1}{\sqrt{1+\sqrt{2}}}}{\sqrt{1+\sqrt{2}} +\dfrac{1}{\sqrt{1+\sqrt{2}}}} = \frac{\sqrt{2}}{2+\sqrt{2}} = \sqrt{2}-1 .$$ By expanding the integrand in Eq.   in partial fractions, one has $$\frac{\alpha^2}{4} -\frac{\pi^2}{16}
%&=& \frac12 \, \int_{\sqrt{2}-1}^1{\frac{\ln{t}}{1+t} \: dt} + \frac12 \, \int_{\sqrt{2}-1}^1{\frac{\ln{t}}{1-t} \: dt} \nonumber \\
= \frac12 \, \int_{\sqrt{2}-1}^1{\frac{\ln{t}}{1+t} \: dt} -\frac12 \, \int_1^{\sqrt{2}-1}{\!\frac{\ln{t}}{1-t} \: dt} \, . 
\label{eq:2lns}$$ These integrals can be reduced to special values of the dilogarithm function if one adopts the more general integral definition for this function [@Lewin], namely $$\mathrm{Li}_2{(z)} := -\int_0^{z}{\frac{\ln{(1-s)}}{s} \; ds} \, ,$$ valid for $z \in \mathbb{C} \backslash [1,\infty)$.[^4] Note that, by expanding the logarithm in powers of $z$ in this integral definition, one finds $\mathrm{Li}_2{(z)} = \sum_{k=1}^\infty{{\,z^k/k^2}}$, which agrees with the basic definition stated in the theorem for all complex $z$ with $|z| \le 1$ [@Lewin].  By taking into account this summation form, it is clear that $\mathrm{Li}_2{(1)} = {\,\pi^2/6}$, which implies that $$\mathrm{Li}_2{(z)}  = \frac{\pi^2}{6} -\,\int_1^{\,z}{\frac{\ln{(1-s)}}{s} \: ds} \, . 
\label{eq:defLi2}$$ From this integral form, it is easy to deduce that $$\int_1^{\,z}{\frac{\ln{t}}{1-t} \: dt} = \mathrm{Li}_2{(1-z)}$$ and $$\int_z^{\,1}{\frac{\ln{t}}{1+t} \: dt} = -\mathrm{Li}_2{(-z)} -\ln{z}\,\ln{(z+1)} -\frac{\pi^2}{12} \, .$$

By putting $z=\sqrt{2}-1$ in these integrals and then substituting the resulting dilogarithms in Eq.  , one finds $$\begin{aligned}
\frac{\alpha^2}{4} -\frac{\pi^2}{16} &=& \frac12 \, \left[ -\mathrm{Li}_2{\left(1-\sqrt{2}\right)} -\ln{\!\left(\sqrt{2}-1\right)}\,\ln{\sqrt{2}} -\frac{\pi^2}{12} \right] -\frac12 \, \mathrm{Li}_2{\left(2-\sqrt{2}\right)} \nonumber \\
&=& -\frac{\mathrm{Li}_2{\left(1-\sqrt{2}\,\right)}}{2} -\frac{\mathrm{Li}_2{\left(2-\sqrt{2}\,\right)}}{2} -\frac14\,\ln{\!\left(\sqrt{2}-1\right)}\,\ln{2} -\frac{\pi^2}{24} \, . 
\label{eq:auxLi2}\end{aligned}$$ Now, by applying the Euler reflection formula for $\mathrm{Li}_2{(z)}$ about $z=\frac12$, which reads (see Eq.  (1.11) in Ref.  [@Lewin]) $$\mathrm{Li}_2{(1-z)} = -\mathrm{Li}_2{(z)} -\ln{z}\,\ln{(1-z)} +\frac{\pi^2}{6} \, ,$$ to $\mathrm{Li}_2{\left(2-\sqrt{2}\,\right)}$ in Eq.  , one finds $$\mathrm{Li}_2{\left(\sqrt{2}-1\right)} - \mathrm{Li}_2{\left(1 -\sqrt{2}\,\right)} = \frac{\alpha^2}{2} -\frac{\pi^2}{8} + \frac{\pi^2}{4} -\ln^2{\!\left(\sqrt{2}-1\right)} . 
\label{eq:tmp}$$ By applying the Landen’s formula, namely (see Eq.  (1.12) in Ref.  [@Lewin]) $$\mathrm{Li}_2{(z)} = -\mathrm{Li}_2{\left( \frac{z}{z-1} \right)} -\frac12 \, \ln^2{(1-z)} \, ,$$ which is valid for all complex $z$, $z \not \in [1,\infty)$, to $\,\mathrm{Li}_2{\left(1 -\sqrt{2}\,\right)}$, one has $$\mathrm{Li}_2{\left(1 -\sqrt{2}\,\right)} = -\mathrm{Li}_2{\left( 1 -\frac{1}{\sqrt{2}} \right)} -\frac18 \, \ln^2{2} \, .$$ Finally, by substituting this result on Eq.  , one finds $$\mathrm{Li}_2{\left(\sqrt{2}-1\right)} +\mathrm{Li}_2{\left( 1 -\frac{1}{\sqrt{2}} \right)} = \frac{\pi^2}{8} -\frac{\alpha^2}{2} -\frac18 \, \ln^2{2} \, . 
\label{eq:fim2}$$

$\Box$

It should be mentioned that neither Mathematica (release 7) nor Maple (release 13) are able to generate any closed-form expression for this sum.  Indeed, this identity is not found in standard bibliographic sources for dilogarithms [@Lewin; @Zagier], though it resembles the identity $$\mathcal{L}{\left(\sqrt{2}-1\right)} +\mathcal{L}{\left( 1 -\frac{1}{\sqrt{2}} \right)} = \frac34 \: ,
\label{eq:rogers}$$ where $\:\mathcal{L}{(x)} := \dfrac{6}{\pi^2} \left[ \mathrm{Li}_2{(x)} +\frac12 \, \ln{x} \: \ln{(1-x)} \right]\,$ is the *normalized Roger’s dilogarithm*, as found by Bytsko (1999) [@Bytsko]. 

It seems plausible to extend the results obtained here in this paper to other known unit square integrals, such as $\,\int_0^1\!\!{\int_0^1{\ln{(x y)}/(1-x^2 y^2) \, dx} \, dy} = -\frac74 \, \zeta{(3)}\,$, $\,\int_0^1\!\!{\int_0^1{\ln{x}/(1-x^2 y^2) \, dx} \, dy} = -\frac78 \, \zeta{(3)}\,$, $\,\int_0^1\!\!{\int_0^1{\ln{(x y)}/(1-x y) \, dx} \, dy} = -2 \, \zeta{(3)}\,$, $\,\int_0^1\!\!{\int_0^1{\ln{(1-x y)}/(1-x y) \, dx} \, dy} = -\zeta{(3)}\,$, and many others (see, e.g., Ref.  [@Wolfram]).  Similarly, it should also be interesting to include a third dimension in the hyperbolic change of variables in order to investigate some triple integrals over the unit cube, such as $$\int_0^1\!\!{\int_0^1\!\!{\int_0^1{\frac{1}{1 -x y z} \: dx} \, dy} \, dz} = \zeta{(3)} \quad \text{and} \; \int_0^1\!\!{\int_0^1\!\!{\int_0^1{\frac{1}{1 -x^2 y^2 z^2} \: dx} \, dy} \, dz} = \frac78 \, \zeta{(3)} \, .$$ Extensive work in both directions is already ongoing by the author. 

Acknowledgments {#acknowledgments .unnumbered}
===============

The author acknowledges a postdoctoral fellowship from CNPq (Brazilian agency) during the course of this work. 

[99]{}

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D. Kalman, Six ways to sum a series, Coll.  Math.  J. **24** (1993) 402–421. 

R. Chapman, Evaluating $\zeta{(2)}$. Available at:

     http://secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf 

T. M. Apostol, A proof that Euler missed: evaluating $\zeta{(2)}$ the easy way, Math.  Intelligencer **5**:3 (1983) 59–60. 

W. Rudin, *Principles of Mathematical Analysis*, 3rd ed.,  McGraw-Hill, New York, 1976. 

M. Aigner and G. M. Ziegler, *Proofs from THE BOOK*, 4th ed.,  Springer, New York, 2010.  Chap.  8. 

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L. Lewin, *Polylogarithms and Associated Functions*, Elsevier, New York, 1981. 

D. Zagier, *The Dilogarithm Function*.  In: *Frontiers in Number Theory, Physics, and Geometry II*, Ed.  by P. Cartier, P. Moussa, B. Julia, and P. Vanhove.  Springer, New York, 2007.  Pages 3–61. 

A. G. Bytsko, Two-term dilogarithm identities related to conformal field theory, Lett.  Math.  Phys. ** 50** (1999) 213–228. 

E. W. Weisstein, Unit square integrals.  From *MathWorld*, a Wolfram Web resource, available at:

     http://mathworld.wolfram.com/UnitSquareIntegral.html 

Figures {#figures .unnumbered}
=======

[^1]: Note that ${\,1/(1-x y)}$ is a positive real number at every point in the unit square, the only exception being the point $(1,1)$, where it is undefined.  Of course, usual methods for treating improper integrals apply to this case. 

[^2]: For this change of variables, it is easy to see that $\left|J\right| = 1$, $J$ being the Jacobi determinant $\frac{d(x,y)}{d(u,v)} = \left|
      \begin{array}{cc}
        {\partial x/\partial u} & {\partial x/\partial v} \\
        {\partial y/\partial u} & {\partial y/\partial v} \\
      \end{array} \right|$.

[^3]: Note that the real function $\sinh^{-1}{x}$ is the inverse of the *bijective* function $\sinh{x}$ for all $x \in \mathbb{R}$, as usual.  However, since the function $\cosh{x}$ is not bijective over the entire real domain, we can reduce its domain to the nonnegative reals and then *define* its inverse as $\cosh^{-1}{x} := \ln{\left(x+\sqrt{x^2-1}\,\right)}$, $\forall \, x \ge 1$.

[^4]: We are considering here the principal branch of the dilogarithm function $\mathrm{Li}_2{(z)}$, defined by taking the principal branch of $\ln{z}$, for which it has a cut along the negative real axis, with $|\arg{(z)}| < \pi$. This defines the principal branch of $\mathrm{Li}_2{(z)}$ as a single-valued function in the complex plane cut along the real axis, from $1$ to $+\infty$.
