<s>[INST] 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.
 [/INST] 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 $I_{\, 2}$.
</s>