Double integral over fractional part of Einstein-like velocity sum $\int_0^1 \int_0^1 \{\frac{u+v}{1-u v}\} \,dudv$ Looking for more interesting and complicated examples of this type of problem I propose this one inspired by Einstein's addition theorem for relativistic velocities
$$i =\int_0^1 \int_0^1 \{\frac{u+v}{1-u v}\} \,dudv = 0.502331527971...\tag{1}$$
Here $\{x\}$ is the fractional part of $x$.
In fact, Einsteins formula has a plus sign in the denominator, and you can start with that exmple for warming up.
The question is the usual one: is there a closed expression for $i$?
Here is the graph of integrand of $i$

I haven't solved the problem yet, but I took a first step substituting integration variables so that one of the new variables is the expression under the fractional part. That is $u\to r, \frac{u+v}{1-u v}\to s$, or inverted  $u\to r, v\to \frac{s-r}{1+r s}$. The Jacobi determinant is $\frac{1+r^2}{(1+r s)^2}$. The range of the new variables follows from $0<v<1$ giving $0<\frac{s-r}{1+r s}<1$ leading to $r<s<\frac{1+r}{1-r}$. And, trivially, $0<r<1$.
Our integral then becomes
$$i_1 = \int_{0}^{1}\,dr \int_{r}^{\frac{1+r}{1-r}} \,ds \frac{1+r^2}{(1+r s)^2} \{s\}\tag{2}$$
Compared to many other cases the $s$-integral has two variable boundaries (instead of, say, (r,$\infty$)). This makes the decomposition into discrete intervals more challenging.   
I stop here for letting us jointly try to solve the problem.
 A: A natural substitution is $u=\tan\theta, v=\tan\varphi$, bringing the given integral into the following form:
$$ I = \iint_{(0,\pi/4)^2}\frac{\{\tan(\theta+\varphi)\}}{\cos^2\theta\cos^2\varphi}\,d\theta\,d\varphi $$
By setting $\theta+\varphi=u$ and $\theta-\varphi=v$ and exploiting symmetry we get
$$ I = 4\int_{0}^{\pi/2}\int_{0}^{\min(u,\pi/2-u)}\frac{\{\tan u\}}{(\cos u+\cos v)^2}   \,dv\,du $$
or
$$ I = \underbrace{4 \int_{0}^{\pi/4}\int_{0}^{u}\frac{\{\tan u\}}{(\cos u+\cos v)^2}   \,dv\,du}_{2-\pi+2\log 2}+4\int_{0}^{\pi/4}\int_{0}^{u}\frac{\{\cot u\}}{(\sin u+\cos v)^2}\,dv\,du $$
or
$$ I = 2-\pi+2\log(2)+4\int_{0}^{1}\int_{0}^{1}\frac{z\left\{\frac{1}{z}\right\}}{(1+z^2 w^2)\left(\frac{z}{\sqrt{z^2+1}}+\frac{1}{\sqrt{z^2 w^2+1}}\right)^2}\,dw\,dz$$
or
$$ I = 2-\pi+2\log(2)+4\int_{1}^{+\infty}\int_{0}^{1}\frac{\left\{z\right\}}{z(z^2+w^2)\left(\frac{1}{\sqrt{1+z^2}}+\frac{z}{\sqrt{w^2+z^2}}\right)^2}\,dw\,dz$$
or
$$ I = 2-\pi+2\log(2)+4\int_{1}^{+\infty}\frac{(1+z^2)(z+(1+z)\log(1+z^2)-(1+z)\log(1+z))}{z^5(z+1)}\{z\}\,dz$$
or
$$ I = 2-\pi+2\log(2)+\small{4\sum_{n=1}^{+\infty}\int_{n}^{n+1}\frac{(1+z^2)(z+(1+z)\log(1+z^2)-(1+z)\log(1+z))}{z^5(z+1)}(z-n)\,dz}$$
or
$$ I = 2-\pi+2\log(2)+4\sum_{n\geq 1}a_n $$
with $a_n$, depending on $n,\arctan(n),\arctan(n+1),\log(n+1),\log(n^2+1)$, is a positive contribution which behaves like $\frac{\log(n+1)}{2n^3}$ as $n\to +\infty$. The last series is a "moral analogue" of $-\frac{1}{2}\zeta'(3)$.
