Why does this integral not depend on the parameter? While working with stable distributions, we for $\alpha \in (1,2) $ have encountered the following integral:
$$\lim_{\varepsilon \to 0} \int_{(\varepsilon, \pi/2 - \varepsilon) \cup (\pi/2 + \varepsilon, \pi-\varepsilon)} \frac{\log \left(|\cos \theta|^\alpha + |\sin \theta |^\alpha + |\cos \theta + \sin \theta|^\alpha \right)}{\alpha \cos \theta \sin \theta} d\theta .$$
Mathematica suggests that it is independent of $ \alpha $, and we have other resons to believe that this is also the truth, but we do not know neither how to formally solve the integral now how to prove this independence. Is there any simple way to do either of these things which we (clearly) haven't thought of?
 A: A partial answer, designed for simplifying further attempts. Now an answer, with some detours not really needed but left here for documentation purposes.
$\cos\left(\theta+\frac{\pi}{2}\right)=-\sin\theta$ and $\sin\left(\theta+\frac{\pi}{2}\right)=\cos\theta$, hence the given integral can be written as
$$ \int_{0}^{\pi/2}\log\left(\frac{\sin^\alpha\theta+\cos^\alpha\theta+|\cos\theta+\sin\theta|^{\alpha}}{\sin^\alpha\theta+\cos^\alpha\theta+|\cos\theta-\sin\theta|^{\alpha}}\right)\frac{d\theta}{\alpha\sin\theta\cos\theta}$$
in the improper Riemann sense. By substituting $\theta=\arctan u$ we are left with
$$ \int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+|1+u|^{\alpha}}{1+u^\alpha+|1-u|^\alpha}\right)\frac{du}{\alpha u}=2\int_{0}^{1}\log\left(\frac{1+u^\alpha+(1+u)^{\alpha}}{1+u^\alpha+(1-u)^\alpha}\right)\frac{du}{\alpha u}$$
or
$$\frac{2}{\alpha}\int_{0}^{+\infty}\log\left(\frac{2\cosh\frac{\alpha x}{2}+\left(2\cosh\frac{x}{2}\right)^\alpha}{2\cosh\frac{\alpha x}{2}+\left(2\sinh\frac{x}{2}\right)^{\alpha}}\right)\,dx=\frac{4}{\alpha}\int_{0}^{+\infty}\log\left(\frac{2\cosh\alpha x+\left(2\cosh x\right)^\alpha}{2\cosh\alpha x+\left(2\sinh x\right)^{\alpha}}\right)\,dx.$$
We already know some integrals that, due to the substitution $u\mapsto\frac{1}{u}$, do not really depend on their parameter $\beta>0$:
$$ \int_{0}^{+\infty}\frac{du}{(1+u^2)(1+u^\beta)}=\frac{\pi}{4},\qquad \int_{0}^{+\infty}\frac{\log(u)\,du}{(1-u^2)(1+u^\beta)}=-\frac{\pi^2}{8} $$
$$\int_{0}^{+\infty}\frac{du}{(1+u+u^2)(1+u^\beta)}=\frac{\pi}{3\sqrt{3}},\qquad \int_{0}^{+\infty}\frac{g\left(\frac{u-1}{u+1}\right)\,du}{(1-u^2)(1+u^\beta)}\;\text{ with }g\text{ odd}$$
hence it is not unlikely that by choosing a suitable $g$, applying $\int_{0}^{\alpha}(\ldots)d\beta$ and performing a substitution we can prove that
$$\frac{\partial}{\partial\alpha}\int_{0}^{1}\log\left(\frac{1+u^\alpha+(1+u)^{\alpha}}{1+u^\alpha+(1-u)^\alpha}\right)\frac{du}{u}=\text{ const.}$$
as wanted. Indeed
$$\begin{eqnarray*} \int_{0}^{1}\log\left(\frac{1+u^\alpha+(1-u)^\alpha}{2}\right)\frac{du}{u}&=&\int_{1}^{+\infty}\log\left(\frac{1+\frac{1}{u^\alpha}+\left(1-\frac{1}{u}\right)^\alpha}{2}\right)\frac{du}{u}\\&=&\int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2(u+1)^\alpha}\right)\frac{du}{1+u}\end{eqnarray*}$$
but the LHS, by symmetry, is also 
$$\begin{eqnarray*} \int_{0}^{1}\log\left(\frac{1+u^\alpha+(1-u)^\alpha}{2}\right)\frac{du}{1-u}&=&\int_{1}^{+\infty}\log\left(\frac{1+\frac{1}{u^\alpha}+\left(1-\frac{1}{u}\right)^\alpha}{2}\right)\frac{du}{u(u-1)}\\&=&\int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2(u+1)^\alpha}\right)\frac{du}{u(1+u)}\end{eqnarray*}$$
while
$$\begin{eqnarray*} \int_{0}^{1}\log\left(\frac{1+u^\alpha+(1+u)^\alpha}{2}\right)\frac{du}{u}&=&\int_{1}^{+\infty}\log\left(\frac{1+\frac{1}{u^\alpha}+\left(1+\frac{1}{u}\right)^\alpha}{2}\right)\frac{du}{u}\\&=&\int_{1}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2 u^\alpha}\right)\frac{du}{u}\end{eqnarray*}.$$
Hence
$$
\begin{eqnarray*} && 2\int_{0}^{1}\log\left(\frac{1+u^\alpha+(1+u)^{\alpha}}{1+u^\alpha+(1-u)^\alpha}\right)\frac{du}{\alpha u}\\
&=&
\int_{0}^{1}\log\left(\frac{1+u^\alpha+(1+u)^\alpha}{2}\right)\frac{du}{\alpha u}
+
\int_{1}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2 u^\alpha}\right)\frac{du}{\alpha u}
\\&\qquad-&
\int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2(u+1)^\alpha}\right)\frac{du}{\alpha (1+u)}
-
\int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2(u+1)^\alpha}\right)\frac{du}{\alpha u(1+u)}
\\&=&
\int_{0}^{\infty}\log\left(\frac{1+u^\alpha+(1+u)^\alpha}{2\max(1,u)^\alpha}\right)\frac{du}{\alpha  u}
 - 
\int_{0}^{+\infty}\log\left(\frac{1+u^\alpha+(u+1)^\alpha}{2(u+1)^\alpha}\right)\frac{du}{\alpha u}
\\&=&
\int_{0}^{\infty}\log\left(\frac{(u+1)^\alpha}{\max(1,u)^\alpha}\right)\frac{du}{\alpha  u}
\\&=&
\int_{0}^{\infty}\log\left(\frac{u+1}{\max(1,u)}\right)\frac{du}{u}
\\&=&
2\int_{0}^{1} \frac{\log\left( u+1 \right)}{u} du.
\end{eqnarray*}
$$
The last expression is clearly independent of $\alpha$, and in particular, for the original integral we get

$$
\int_{0}^\pi \frac{\log(|\cos \theta|^\alpha+|\sin\theta|^\alpha+|\cos \theta + \sin \theta|^\alpha)}{\alpha \cos \theta \sin \theta} d\theta
= 
\int_{0}^{1} \frac{\log\left( u+1 \right)}{u} du
= \frac{\pi^2}{12}.
$$

The case $\alpha=3$ is not instantly recognized by (my version of) Mathematica. It turns out to be a sort of Ahmed-like integral.
