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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?

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  • $\begingroup$ You could try taking a derivative with respect to $\alpha$. If it is truly independent of $\alpha$, then the derivative should be zero. $\endgroup$ – Cameron Williams Mar 19 '18 at 16:34
  • $\begingroup$ That was the first thing we tried, but it does not become simpler. If anything, it makes the integral worse, and in particular, I cannot show it is zero $\endgroup$ – malin Mar 19 '18 at 19:58
  • $\begingroup$ Do you mean the integral is independent or just the limit. $\endgroup$ – Keith McClary Mar 20 '18 at 0:34
  • $\begingroup$ Is it related to the Poisson kernel or the topological degree of some curve? $\endgroup$ – Jack D'Aurizio Mar 20 '18 at 1:02
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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.

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  • $\begingroup$ @DavidK: I mean that Mathematica does not immediately simplify the integral for $\alpha=3$ as $\frac{\pi^2}{4}$, and, machine intervention or not, the highlighted identity is a sort of Ahmed integral. $\endgroup$ – Jack D'Aurizio Mar 20 '18 at 14:58
  • $\begingroup$ @DavidK: but "X is a sort of Blah-integral" is not adversative to "Mathematica cannot manage X". I am splitting the sentence to avoid any ambiguity. $\endgroup$ – Jack D'Aurizio Mar 20 '18 at 18:25
  • $\begingroup$ "Adversative" is too strong a requirement for using "but." The sentence as originally written did not make sense to me, but it is OK now that it is split into two sentences. $\endgroup$ – David K Mar 20 '18 at 18:37

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