Prove $\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \text{d}\varphi\text{d}\theta = \pi\ln2$ Month ago I encounter a nice result numerically checked by Mathematica
$$
\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \mathrm{d}\varphi\mathrm{d}\theta = \pi\ln2
$$
where the integrated function is actually well-defined even around its singularity $\theta=\varphi=0$.
At my first sight, I thought it might be a trivial conclusion derived from a kind of typical integral like
$$
\int_{0}^{\pi} \frac{\cos n\theta}{\cos\theta-\cos\varphi} \mathrm{d}\theta = \pi\frac{\sin n\varphi}{\sin\varphi}
$$
just using a proper series expansion. However, when I review it in detail, the result over $(0,\pi/2)$ will be awkwardly complicated. I realize this double integrals may not be done directly, or I may lack some essential insight to solve it.
So I question it here for some further discussion, and thanks in advance for any suggestion.
 A: Here is a solution based on @G Cab's remarkable observation.

Reduction. Let $I$ denote the integral, and apply the substitution $(\theta,\varphi)\mapsto(2\theta,2\varphi)$ to write
$$ I = 4 \int_{0}^{\pi/4} \int_{0}^{\pi/4} \frac{2\theta\cot(2\theta) - 2\varphi \cot(2\varphi)}{\cos(2\theta)-\cos(2\varphi)} \, \mathrm{d}\theta\mathrm{d}\varphi. $$
If we define the $1$-form $\omega$ by
$$ \omega = \left( \log\frac{\cot\theta}{\cot\varphi} \right) \biggl( \frac{\mathrm{d}( (\theta-\varphi)^2 )}{2\sin^2(\theta-\varphi)} - \frac{\mathrm{d}( (\theta+\varphi)^2 )}{2\sin^2(\theta+\varphi)} \biggr), $$
then $\omega$ is smooth in the region $(0, \pi/2)^2$ via continuation along the diagonal $\varphi = \theta$. Moreover, @G Cab's computation for the differential reduces to1)
$$ \mathrm{d}\omega
= 4 \left( \frac{2\theta\cot(2\theta) - 2\varphi \cot(2\varphi)}{\cos(2\theta)-\cos(2\varphi)} \right) \, \mathrm{d}\theta \wedge \mathrm{d}\varphi. $$
Now let $\epsilon > 0$ be small and consider the triangular region
$$ \mathcal{T}_{\epsilon} = \bigl\{ (\theta, \varphi) : \epsilon \leq \varphi \leq \theta \leq \tfrac{\pi}{4} \bigr\}. $$
Its boundary $\partial\mathcal{T}_{\epsilon}$ consists of three oriented line segments

Then by the symmetry and the Stokes' theorem,
\begin{align*}
I
&= 2 \lim_{\epsilon \to 0^+} \int_{\mathcal{T}_{\epsilon}} \mathrm{d}\omega
= 2 \lim_{\epsilon \to 0^+} \int_{\partial \mathcal{T}_{\epsilon}} \omega
= 2 (I_{\rightarrow} + I_{\uparrow} + I_{\swarrow}),
\end{align*}
where $\rightarrow$, $\uparrow$, and $\swarrow$ denote the oriented bottom/height/hypothenus of $\partial\mathcal{T}_{\epsilon}$, respectively, and
$$ I_{\bullet} =  \lim_{\epsilon \to 0^+} \int_{\bullet} \omega $$
for each $\bullet \in \{ \rightarrow, \uparrow, \swarrow \}$.

Integral Computations. We will evaluate each $I_{\bullet}$:
1. It is not hard to check that $\omega = 0$ along $\varphi = \theta$, and so, we get $I_{\swarrow} = 0$.
2. For $I_{\uparrow}$, we note that
\begin{align*}
I_{\uparrow}
&= \int_{\varphi = 0}^{\varphi = \pi/4} \omega|_{\theta=\pi/4} \\
&= \int_{\varphi = 0}^{\varphi = \pi/4} \left( - \log\cot\varphi \right) \biggl( \frac{\mathrm{d}( (\pi/4-\varphi)^2 )}{2\sin^2(\pi/4-\varphi)} - \frac{\mathrm{d}( (\pi/4+\varphi)^2 )}{2\sin^2(\pi/4+\varphi)} \biggr) \\
&= 2 \int_{0}^{\pi/4} \left( \log\cot\varphi \right) \biggl( \frac{\pi/4-\varphi}{1-\sin(2\varphi)} + \frac{\pi/4+\varphi}{1+\sin(2\varphi)} \biggr) \, \mathrm{d}\varphi \\
&= 2 \int_{-\pi/4}^{\pi/4} \frac{\pi/4+\varphi}{1+\sin(2\varphi)} \log \left| \cot\varphi \right|  \, \mathrm{d}\varphi \\
&= \frac{1}{2} \int_{0}^{\pi} \frac{x}{1-\cos x} \log \left| \cot \left(\frac{x}{2} - \frac{\pi}{4}\right) \right|  \, \mathrm{d}x,
\end{align*}
where the last step is the result of the substitution $\varphi=\tfrac{x}{2}-\tfrac{\pi}{4}$. Then by noting that
$$ \log \left| \cot \left(\frac{\pi}{4} - \frac{x}{2} \right) \right|
= \operatorname{Re}\log\left(\frac{1-ie^{ix}}{1+ie^{ix}}\right)
= i (\arctan(e^{-ix}) - \arctan(e^{ix})), $$
we get
\begin{align*}
I_{\uparrow}
&= \frac{1}{2}\int_{0}^{\pi} \frac{ix}{1-\cos x} (\arctan(e^{-ix}) - \arctan(e^{ix}))  \, \mathrm{d}x \\
&= \frac{1}{2}\int_{-\pi}^{\pi} \frac{ix}{1-\cos x} \left(\frac{\pi}{4} - \arctan(e^{ix})\right)  \, \mathrm{d}x \\
&= i \int_{\mathcal{C}} \frac{(\log z )(\frac{\pi}{4} - \arctan z)}{(z-1)^2}  \, \mathrm{d}z,
\end{align*}
where the substitution $z = e^{ix}$ is applied to the last step and $\mathcal{C}$ is the path parametrized by $e^{ix}$ for $-\pi < x < \pi$. Now deforming the contour $\mathcal{C}$ so as to enclose the principal branch cut for $\log$,
\begin{align*}
I_{\uparrow}
&= 2\pi \int_{0}^{1} \frac{\frac{\pi}{4} + \arctan t}{(t+1)^2}  \, \mathrm{d}t
= 2\pi \int_{0}^{1} \frac{1}{(t^2+1)(t+1)} \, \mathrm{d}t
= \frac{\pi^2}{4} + \frac{\pi}{2} \log 2.
\end{align*}
3. For the integral along the bottom,
\begin{align*}
\int_{\rightarrow} \omega
&= \int_{\theta = \epsilon}^{\theta = \pi/4} \omega|_{\varphi=\epsilon} \\
&= \int_{\theta = \epsilon}^{\theta = \pi/4} \log\left( \frac{\cot\theta}{\cot\epsilon} \right) \biggl( \frac{\mathrm{d}( (\theta-\epsilon)^2 )}{2\sin^2(\theta-\epsilon)} - \frac{\mathrm{d}( (\theta+\epsilon)^2 )}{2\sin^2(\theta+\epsilon)} \biggr) \\
&= \int_{\epsilon}^{\pi/4} \log\left( \frac{\cot\theta}{\cot\epsilon} \right) \biggl( \frac{\theta-\epsilon}{\sin^2(\theta-\epsilon)} - \frac{\theta+\epsilon}{2\sin^2(\theta+\epsilon)} \biggr) \,  \mathrm{d}\theta.
\end{align*}
Substituting $\theta = \epsilon x$,
\begin{align*}
\int_{\rightarrow} \omega
&= \int_{1}^{\pi/4\epsilon} \log\left( \frac{\cot(\epsilon x)}{\cot\epsilon} \right) \biggl( \frac{\epsilon^2 (x-1)}{\sin^2(\epsilon(x-1))} - \frac{\epsilon^2 (x+1)}{2\sin^2(\epsilon(x+1))} \biggr) \, \mathrm{d}x.
\end{align*}
It is not hard (although not entirely trivial) to show that the above integral converges to
\begin{align*}
I_{\rightarrow}
= \lim_{\epsilon \to 0^+} \int_{\rightarrow} \omega
= \int_{1}^{\infty} (-\log x) \biggl( \frac{1}{x-1} - \frac{1}{x+1} \biggr) \, \mathrm{d}x
= -\frac{\pi^2}{4}.
\end{align*}

Conclusion. Combining all these observations altogether, we have
$$ \bbox[10px,#ffd]{ I = 2 (I_{\rightarrow} + I_{\uparrow} + I_{\swarrow}) = \pi \log 2 } $$
as desired.

${}^{1)}$ We include the sketch of computation for $\mathrm{d}\omega$, which is essentially @G Cab's computation in backward direction. Substituting $u=\theta-\phi$ and $v=\theta+\phi$,
$$ \omega = \left( \log \left( \frac{\sin v - \sin u}{\sin v + \sin u} \right) \right)\biggl( \frac{u \, \mathrm{d}u}{\sin^2 u} - \frac{v \, \mathrm{d}v}{\sin^2 v} \biggr) . $$
Taking $\mathrm{d}$,
\begin{align*}
\mathrm{d}\omega
&= \mathrm{d}\left( \frac{u}{\sin^2 u} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \wedge \mathrm{d}u \\
&\qquad - \mathrm{d}\left( \frac{v}{\sin^2 v} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \wedge \mathrm{d}v \\
&= \frac{u}{\sin^2 u} \left( \frac{\partial}{\partial v} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \mathrm{d}v  \wedge \mathrm{d}u\\
&\qquad - \frac{v}{\sin^2 v} \left( \frac{\partial}{\partial u} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \mathrm{d}u \wedge \mathrm{d}v \\
&= 4 \left( \frac{v \cos u \csc v - u \cos v \csc u}{\cos(2u) - \cos (2v)} \right) \, \mathrm{d}u \wedge \mathrm{d}v.
\end{align*}
By noting that $\mathrm{d}u \wedge \mathrm{d}v = 2 \, \mathrm{d}\theta \wedge \mathrm{d} \varphi$, the question boils down to showing
$$ 2 \left( \frac{v \cos u \csc v - u \cos v \csc u}{\cos(2u) - \cos (2v)} \right) = \frac{(v+u)\cot(v+u) - (v-u)\cot(v-u)}{\cos(v+u) - \cos(v-u)}. $$
This can be verified by expanding the right-hand side.
A: The following illustrates an approach that looks promising, but
that I could not yet complete: wish you and the community here might help
to finalize it.
Starting from the integrand
$$
f(\theta ,\varphi )
 = {{\theta \cot \theta  - \varphi \cot \varphi } \over {\cos \theta  - \cos \varphi }}
 = {{\theta {{\cos \theta } \over {\sin \theta }} - \varphi {{\cos \varphi } \over {\sin \varphi }}}
 \over {\cos \theta  - \cos \varphi }}
 = f(\varphi ,\theta )
$$
it's structure suggests the idea that Green's Theorem might be applied.
Formulated as above, it is difficult to find the needed potential function.
But applying the transformation
$$
\left\{ \matrix{
  \theta  = x + y \hfill \cr 
  \varphi  = y - x \hfill \cr}  \right.\quad  \Leftrightarrow \quad \left\{ \matrix{
  x = {{\theta  - \varphi } \over 2} \hfill \cr 
  y = {{\theta  + \varphi } \over 2} \hfill \cr}  \right.
$$
we get
$$
\eqalign{
  & g(x,y) = f(x + y,\;y - x) =   \cr 
  &  = {{\left( {y - x} \right){{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}}
 - \left( {y + x} \right){{\cos \left( {y + x} \right)}
 \over {\sin \left( {y + x} \right)}}}
 \over {\cos \left( {y - x} \right) - \cos \left( {y + x} \right)}} =   \cr 
  &  = {{y\left( {{{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}}
 - {{\cos \left( {y + x} \right)} \over {\sin \left( {y + x} \right)}}} \right)
 - x\left( {{{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}}
 + {{\cos \left( {y + x} \right)} \over {\sin \left( {y + x} \right)}}} \right)}
 \over {2\sin x\sin y}} =   \cr 
  &  = {{y\sin \left( {2x} \right) - x\sin \left( {2y} \right)}
 \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}} =   \cr 
  &  = y{{\sin \left( {2x} \right)} \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}}
 - x{{\sin \left( {2y} \right)} \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}} =   \cr 
  &  = y{{\cos x} \over {\sin y\left( {\sin ^2 y\cos ^2 x - \cos ^2 y\sin ^2 x} \right)}} - x{{\cos y}
 \over {\sin x\left( {\sin ^2 y\cos ^2 x - \cos ^2 y\sin ^2 x} \right)}} =   \cr 
  &  = y{{\cos x} \over {\sin y\left( {\cos ^2 x - \cos ^2 y} \right)}}
 - x{{\cos y} \over {\sin x\left( {\cos ^2 x - \cos ^2 y} \right)}} =   \cr 
  &  = g( - x,y) \cr} 
$$
Putting:
$$
\eqalign{
  & M(x,y) =  - {y \over {2\sin ^2 y}}\ln \left( {{{ - \sin x + \sin y}
 \over {\sin x + \sin y}}} \right)  \cr 
  & L(x,y) = {x \over {2\sin ^2 x}}\ln \left( {{{ - \sin x + \sin y}
 \over {\sin x + \sin y}}} \right) \cr} 
$$
then we have the expression required for applying the Green's theorem
$$
g(x,y) = {\partial  \over {\partial x}}M(x,y) - {\partial  \over {\partial y}}L(x,y)
$$
With the help of this sketch

we have
$$
\eqalign{
  & I = \int_0^{\pi /2} {\int_0^{\pi /2} {f(\theta ,\varphi )\,d\varphi \,d\theta } }  =   \cr 
  &  = \int\!\!\!\int\limits_{square} {f(\varphi ,\theta )\,d\varphi \,d\theta }
  = 2\int\!\!\!\int\limits_{triang} {f(\varphi ,\theta )\,d\varphi \,d\theta }  =   \cr 
  &  = 4\int\!\!\!\int\limits_{triang} {g(x,y)\,dx\,dy}
  = 4\int\!\!\!\int\limits_{triang} {\left( {{\partial  \over {\partial x}}M
 - {\partial  \over {\partial y}}L} \right)\,dx\,dy}  =   \cr 
  &  = 4\oint\limits_{triang} {Ldx + Mdy}  \cr} 
$$
that is
$$
\eqalign{
  & {I \over 4} = \oint\limits_{triang} {Ldx + Mdy}  =   \cr 
  &  = \left( \matrix{
  \int_{y = 0}^{\pi /4} {\left( {L\left( {y,y} \right) + M(y,y)} \right)dy}  +  \hfill \cr 
   + \int_{y = \pi /4}^{\pi /2} {\left( { - L\left( {\pi /2 - y,y} \right) + M(\pi /2 - y,y)} \right)dy}
  +  \hfill \cr 
   - \int_{y = 0}^{\pi /2} {M(0,y)dy}  \hfill \cr}  \right) =   \cr 
  &  = I_{\,1}  + I_2  + I_3  \cr
} 
$$
Now
$$
I_3  =  - \int_{y = 0}^{\pi /2} {M(0,y)dy}
  = \int_{y = 0}^{\pi /2} {{y \over {2\sin ^2 y}}\ln \left( {{{\sin y} \over {\sin y}}} \right)dy}
  = 0
$$
$I_{\,1}$ instead,  is not null as it might appear.
If the path is taken to approach $y=x$ while maintaining positive the difference $y-x$ then a numerical
computation gives a value of $\approx -1.233688 \ldots$, which together with the computation of $I_{\, 2}$ returns the
expected value for $I$.
I am blocked at this point:  analytically establish and justify the path for $I_{\, 1}$ and find a close expression for it and for $I_{\, 2}$.
