LogSine Generating Fn $ \int_0^\pi \big(2\sin\frac{\theta}{2}\big)^x e^{\theta y} d\theta$ This is related to generating functions for Ls (Log Sine Integrals.)   I am trying to calculate
$$
\int_{0}^{\pi}\left[2\sin\left(\theta \over 2\right)\right]^{x}
{\rm e}^{\theta y}\,{\rm d}\theta.
$$
This does have a closed form (it is quite pretty).  Note $x,y\in \mathbb{R}$.  Thanks.  I am not really sure how to calculate this one, because of the term
$$
\left[2\sin\left(\theta \over 2\right)\right]^{x}.
$$
Thanks.
 A: Since this was not answered, I will go ahead and give my two cents.
Begin with the known result (The proof of which I am sure can be found online under Cauchy's Beta Integral):
$$\int_{-\infty}^{\infty}\frac{1}{(u+ix)^{a}(v-ix)^{b}}dx=\frac{2\pi \Gamma(a+b-1)}{(u+v)^{a+b-1}\Gamma(a)\Gamma(b)}$$
Let $u=v=1$ and make the sub $x=\cot(t), \;\ dx=-\csc^{2}(t)dt$.
$$\int_{0}^{\pi}\frac{1}{\left(1+\frac{\cos(x)}{\sin(x)}i\right)^{a}\left(1-\frac{\cos(x)}{\sin(x)}i\right)^{b}}\cdot \frac{1}{\sin^{2}(x)}dx=\frac{2\pi \Gamma(a+b-1)}{2^{a+b-1}\Gamma(a)\Gamma(b)}$$
rewrite the left side:
$$\int_{0}^{\pi}\frac{\sin^{a+b-2}(x)}{(\sin(x)+i\cos(x))^{a}(\sin(x)-i\cos(x))^{b}}dx=\frac{2\pi \Gamma(a+b-1)}{2^{a+b-1}\Gamma(a)\Gamma(b)}$$
But, $\displaystyle \sin(x)+i\cos(x)=ie^{-ix}, \;\ \sin(x)-i\cos(x)=-ie^{ix}$
so we may write:
$$e^{-\pi (a-b)i/2}\int_{0}^{\pi}\sin^{a+b-2}(x)e^{-i(b-a)x}dx=\frac{2\pi \Gamma(a+b-1)}{2^{a+b-1}\Gamma(a)\Gamma(b)}$$
Let $\displaystyle y=a-b, \;\ x=a+b-2$
$$e^{-\frac{\pi y i}{2}}\int_{0}^{\pi}\sin^{x}(\theta)e^{iy\theta}d\theta=\frac{2\pi \Gamma(x+1)}{2^{x+1}\Gamma\left(1+\frac{x+y}{2}\right)\Gamma\left(1+\frac{x-y}{2}\right)}$$
$$\int_{0}^{\pi}\sin^{x}(\theta)e^{iy\theta}d\theta=\frac{\pi \Gamma(x+1)e^{\frac{\pi y i}{2}}}{2^{x}\Gamma\left(1+\frac{x+y}{2}\right)\Gamma\left(1+\frac{x-y}{2}\right)}$$
Is this similar to the result you were shooting for?. This is a slight variation on your integral.
Here is a link I just found that uses the Beta to derive one of the Cauchy Beta Integrals with cos. With some adjustments one may probably use this method to derive this one. 
http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R%28x,cos%29
I made some adjustments to one of these methods to derive this. The form I derived is the sin version of the Beta integral I have commonly encountered
