How do you compute this integral? I obtain this using mathematica:
$$\int_{-\pi}^\pi(2+2\cos t)^a\cos(b t)dt=\frac{2\pi\Gamma(1+2a)}{\Gamma(1+a+b)\Gamma(1+a-b)}.$$
This should hold for $\Re(a)>-1/2$.
 A: Let $I$ be the integral. Firstly, we can rewrite the integral with $e^{ibt}$ in place of $\cos bt$, since the rest of the integrand is even, so
$$ I = \int_{-\pi}^{\pi} (2+2\cos t)^a e^{ibt} \, dt . $$
Changing variables, we put $z = e^{it}$, so $-i dz/z = dt $ the integral becomes
$$ -i\int_{\lvert z \rvert=1} \left( \frac{(1+z)^2}{z} \right)^{a} z^{b-1} \, dz = -i\int_{\lvert z \rvert=1} (1+z)^{2a} z^{b-a-1} \, dz , $$
choosing the principal branch of $z \mapsto z^a$. Choosing the principal branch of $z \mapsto z^b$ as well, the integrand is analytic on the disk with the nonpositive real axis removed. We can therefore deform the contour to be the straight line from $-1$ to $-\varepsilon$ below the branch cut, joined to a small anticlockwise circle around the origin, and a straight line from $-\varepsilon$ to $-1$ above the contour.
It remains to calculate the integrals. The small circle has length $2\pi \varepsilon$ and the integrand is asymptotic to $ z^{b-a-1} = \varepsilon^{b-a-1} e^{(b-a-1)it} $, so this part of the integral converges to $0$ as $\varepsilon \to 0$ if $\Re(b-a)>0$.
We put $z = e^{-i\pi} u$ in the lower integral, which gives
$$ ie^{-i\pi} \int_{0}^1 (1-u)^{2a} e^{-i\pi(b-a-1)} u^{b-a-1} \, du = ie^{-i\pi(b-a)} B(1+2a,b-a) , $$
provided that $2a>-1$, the extra $-$ being because the contour is traversed in the wrong direction. For the upper integral, put $z = e^{i\pi} u$, which gives
$$ -ie^{i\pi} \int_{0}^1 (1-u)^{2a} e^{i\pi(b-a-1)} u^{b-a-1} \, du = -ie^{i\pi(b-a)} B(1+2a,b-a) , $$
and so the sum is
$$ I = ie^{-i\pi(b-a)} B(1+2a,b-a) - ie^{i\pi(b-a)} B(1+2a,b-a) = 2\sin (\pi(b-a)) B(1+2a,b-a) . $$
To get this into the form Mathematica gives, we need to do two things: extend the range, and get rid of the sine. For the second, we have
$$ B(1+2a,b-a) = \frac{\Gamma(1+2a)\Gamma(b-a)}{\Gamma(1+a+b)} , $$
and then applying
$$ \Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin \pi s} $$
gives
$$ I = \frac{2\pi \Gamma(1+2a)}{\Gamma(1+a+b)\Gamma(1+a-b)} \quad \Re(b-a)>0 , \Re(a)>-1/2 $$
as hoped. Lastly, we need to extend the range of validity to remove the condition that $\Re(b-a)>0$. We can do this by using analytic continuation: the original integral is a analytic function of $b$, and $1/\Gamma(s)$ is an analytic function of $s$, so in fact the right-hand side is also an analytic function of $b$. Since these analytic functions on the region $\Re(b-a)>0$, and both are analytic on the larger region $b \in \mathbb{C}$, the identity theorem implies that they are equal on the larger region. Hence
$$ I = \frac{2\pi \Gamma(1+2a)}{\Gamma(1+a+b)\Gamma(1+a-b)} \quad \Re(a)>-1/2 $$
as required.
