Integrals related to $\int_0^{\pi} \left(\frac{\sin(\alpha u)^\alpha \sin((1-\alpha) u)^{1-\alpha}}{\sin u} \right)^{\rho/\alpha}du$ I meet the following integral when I am reading materials regarding the stable distribution:
$$
\frac{1}{\pi}\int_{0}^{\pi}\left\{%
\frac{\sin^{\alpha}\left(\alpha u\right)\
\sin^{1-\alpha}\,\left(\,{\left[1 - \alpha\right]u}\,\right)\ }{\sin\left(u\right)} \right\}^{\rho/\alpha}\,\mathrm{d}u,
$$
where $\alpha \in \left(0,1\right)$.
$$
\mbox{ It looks the result shall be}\quad
\frac{\Gamma\left(1 - \rho/\alpha\right)}
{\Gamma(1 - \rho/\alpha + \rho)\,\Gamma(1-\rho)},
$$
where $-1 < \Re\left(\rho\right) < \alpha$.
However, I have no idea how to solve it. Wondering someone can help me on this.
Thanks in advance
13/11/2020
I would like to thanks Pisco for the kind and patient help. Now I can understand his proof perfectly. There is a related integral,
$$\int _0^{\infty }\frac{1}{\pi }\int _0^{\pi }\exp\left\{-q x^{-\frac{1-\alpha }{\alpha }}  \left(\frac{(\sin[\alpha
 u])^{\alpha }(\sin[(1-\alpha )u])^{1-\alpha }}{\sin [u]}\right)^{\frac{1}{\alpha }}\right\}du e^{-x}dx=e^{-q^{\alpha }}, \quad \alpha \in (0,1),\ (*)$$
Further, since for the unilateral stable distribution, $S_\alpha$, $\alpha \in (0,1)$, the density function is
$$f_{S_\alpha}(x)=\frac{1}{x}\sum_{k=1}^\infty\frac{(-x^\alpha)^{-k}}{k!\Gamma(-k\alpha)}, \ x>0,$$
and we know the Laplace transform of the above function is $e^{-q^\alpha}$. For example, see "2016A new family of tempered distributions". If we take the Laplace for $f_{S_\alpha}$ and leave the requirement for the exchange of the integral and sum for a while, we have
$$\mathcal{L}_{S_\alpha}(q)=\sum_{k=1}^{\infty}\frac{(-q^\alpha)^k}{k!\Gamma(-k \alpha)}\Gamma(-k\alpha)=e^{-q^\alpha}-1,$$
which is not what we are expecting. Therefore I am also wondering how to find $f_{S_\alpha}$ from $(*)$.
Looking for your help.
 A: The claim seems highly nontrivial, this integral definitely deserves more attention.
It is equivalent to prove
$$\int_0^\pi {{{\left[ {\frac{{\sin^a {{(ax)}}\sin^{1-a} {{((1 - a)x)}}}}{{\sin x}}} \right]}^b}} dx = \pi \frac{{\Gamma (1 - b)}}{{\Gamma (1 - b + ab)\Gamma (1 - ab)}}\qquad 0<a,b<1$$

Let $\alpha>0$, $x>0$, $-\pi/2<\varphi<\pi/2$. Consider
$$I=\Re \int_0^\infty  \exp \left( {itx - {t^\alpha } {e^{i\varphi }}} \right)dt $$
Writing $t$ in polar coordinate $t = re^{i\theta}$, we have
$$\Im(itx - t^\alpha e^{i\varphi }) = rx\cos\theta - r^\alpha  \sin(\alpha\theta+\varphi)$$
so (a portion of) $\{t\in \mathbb{C} \vert \Im(itx - t^\alpha e^{i\varphi})=0 \}$ can be parametrized by $$\tag{*}r(\theta) = {\left( {\frac{{ \sin (\alpha \theta  + \varphi )}}{{x\cos \theta }}} \right)^{1/(1 - \alpha )}}$$
We shall choose $\alpha, \varphi$ so that the term inside the parenthesis is positive when $-\varphi/\alpha < \theta < \pi/2$. We will concentrate on the case $0<\alpha<1$. In this case, $r(\theta)$ travels from $0$ to $\infty$ as $\theta$ increases from $-\varphi/\alpha$ to $\pi/2$, let $$\Gamma = \{ re^{i\theta}| -\varphi/\alpha < \theta < \pi/2, r =r(\theta)\}$$ 
The integrand is decreasing fast enough at infinity to enable the deformation of integration path, giving
$$I=\Re \int_\Gamma  \exp \left( {itx - {t^\alpha }{e^{i\varphi }}} \right)dt = \int_\Gamma \exp( \Re({itx - {t^\alpha } {e^{i\varphi }}})) d(\Re t)$$
the second equality holds because the exponential is real on $\Gamma$. Let $$\mathcal{K}(\alpha,\varphi,\theta) = \frac{{{{( {\sin (\alpha \theta  + \varphi )} )}^{\alpha /(1 - \alpha )}}\cos ((\alpha  - 1)\theta  + \varphi )}}{{{{(\cos \theta )}^{1/(1 - \alpha )}}}}$$
Using parametrization $(*)$, one calculates (details omitted)
$$\begin{aligned}&\Re({itx - {t^\alpha } {e^{i\varphi }}}) = -rx \sin\theta - r^\alpha \cos(\alpha\theta+\varphi) =  - {x^{\alpha /(\alpha  - 1)}}\mathcal{K}(\alpha,\varphi,\theta) \\ &d(\Re t) = d(r\cos\theta) = \frac{\alpha }{{1 - \alpha }}{x^{ - 1/(1 - \alpha )}}\mathcal{K}(\alpha,\varphi,\theta) d\theta \end{aligned}$$
Therefore $$\tag{**}\Re \int_0^\infty  {\exp \left[ {itx - {t^\alpha }{e^{i\varphi }}} \right]dt}  = \frac{\alpha }{{1 - \alpha }}{x^{1/(\alpha  - 1)}}\int_{ - \varphi /\alpha }^{\pi /2} {\exp \left[ { - {x^{\alpha /(\alpha  - 1)}}\mathcal{K}(\alpha,\varphi,\theta)} \right]\mathcal{K}(\alpha,\varphi,\theta)d\theta } $$
The choice of contour $\Gamma$ is adapted from Zolotaryov's One-dimensional stable distributions p. 74-77. $(**)$ is crucial for me to derive the desired integral.

$(**)$ is valid for $x>0$, replace $x$ by $x^c$ ($c>1$), then apply $\int_0^\infty dx$ both sides (easily justified):
$$\Gamma (\frac{1}{c})\Re \int_0^\infty  {{{( - it)}^{ - 1/c}}\exp ( - {t^\alpha }{e^{i\varphi }})dt}  = \Gamma \left( {\frac{{ - 1 + \alpha  + c}}{{\alpha c}}} \right)\int_{ - \varphi /\alpha }^{\pi /2} {{{\mathcal{K}(\alpha,\varphi,\theta)}^{1 - \frac{ - 1 + \alpha  + c}{\alpha c}}}d\theta } $$
Let $\beta = (1-c)/(\alpha c)$ (so $\beta < 0$), then LHS of above displayed equation equals
$$\tag{1}\Gamma (1 + \alpha \beta )\frac{{\Gamma ( - \beta )}}{\alpha }\cos (\frac{\pi }{2}(1 + \alpha \beta ) + \varphi \beta )$$ while RHS is $$\tag{2} \Gamma \left( {1 - \beta + \alpha \beta} \right) \int_{ - \varphi /\alpha }^{\pi /2} {\frac{{{{ {\sin^{\alpha\beta} (\alpha \theta  + \varphi )} }}{{\cos }^{(1 - \alpha )\beta}}((\alpha  - 1)\theta  + \varphi )}}{{{{\cos^\beta \theta }}}}d\theta } $$
We proved the equality $(1) = (2)$ under the following hypotheses:

*

*$0<\alpha<1, \beta<0, \alpha\beta>-1$

*$-\pi/2<\varphi<\pi/2$ such that all exponentials in $(2)$ has positive base throughout range of integration.

In particular, this is true for $\varphi = \pi \alpha/2$,
$$\begin{aligned}&\quad \int_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^{\alpha \beta }}(\alpha \theta  + \frac{\pi }{2}\alpha ){{\cos }^{(1 - \alpha )\beta }}((\alpha  - 1)\theta  + \frac{\pi }{2}\alpha )}}{{{{(\cos \theta )}^\beta }}}d\theta }  = \int_0^\pi  {\frac{{{{\sin }^{\alpha \beta }}\alpha \theta {{\sin }^{(1 - \alpha )\beta }}((1 - \alpha )\theta ))}}{{{{\sin^\beta \theta } }}}d\theta } \\
& = \frac{{\Gamma (1 + \alpha \beta )}}{{\Gamma (1 - \beta  + \alpha \beta )}}\frac{{\Gamma ( - \beta )}}{\alpha }\cos (\frac{\pi }{2} + \pi \alpha \beta )\end{aligned}$$
completing the proof.
