Multidimensional angular integrals with complex exponentials I am doing some multidimensional integrations, the angular parts of which are one of the following types:
\begin{align}
\int_0^{2\pi} d\phi \int_0^{2\pi} d\phi'\, \frac{e^{-im'\phi'} e^{im\phi}}{a + b \cos(\phi - \phi')}
\end{align}
or
\begin{align}
\int_0^{2\pi} d\phi \int_0^{2\pi} d\phi'\, \frac{e^{-im'\phi'} e^{im\phi}}{[a + b \cos(\phi - \phi')]^2}.
\end{align}
For $m, m' = 0$ the first integral simplifies to 
\begin{align}
\int_0^{2\pi} d\phi \int_0^{2\pi} d\phi' \frac{1}{a + b\cos (\phi - \phi')}
\end{align}
which I calculate to be 
\begin{align}
\frac{4\pi^2}{\sqrt{a^2-b^2}}
\end{align}
However I am not sure how to do the integrals when $m, m' \ne 0$. I suspect that they will be zero but I am unable to prove that. 
I was thinking that maybe I can substitute $\phi - \phi'$ with $z$, and then integrate over $\phi$ and $z$. For example if I do so in the first integral then that becomes
\begin{align}
\int_0^{2\pi} d\phi \,e^{-i(m - m')\phi} \int_\phi^{\phi-2\pi} dz \frac{e^{-im'z}}{a + b \cos z}. 
\end{align}
I am unable to progress from there.
 A: I'm going to assume that $m, m' \in \mathbb Z$ and $0 < |b| < a$. Due to periodicity, the inner integral becomes
$$\int_0^{2 \pi} \frac {e^{-i m' \phi'}} {a + b \cos (\phi - \phi')} d\phi' =
e^{-i m' \phi} \int_0^{2 \pi} \frac 1 {\sqrt {a^2 - b^2}}
 \frac {1 - r^2} {1 - 2 r \cos \phi' + r^2} e^{-i m' \phi'} d\phi',$$
where $r = (\sqrt {a^2 - b^2} - a)/b$. This, up to a constant factor, is the $n$th Fourier series coefficient of a Poisson kernel, which is known and equals $r^{|n|}$. Then
$$\int_0^{2 \pi} \int_0^{2 \pi}
 \frac {e^{i (m \phi - m' \phi')}} {a + b \cos (\phi - \phi')} d\phi' d\phi =
\frac {2 \pi r^{|m'|}} {\sqrt {a^2 - b^2}}
 \int_0^{2 \pi} e^{i (m - m') \phi} d\phi =
\frac {4 \pi^2 r^{|m|}} {\sqrt {a^2 - b^2}} \delta_{m, m'}.$$
The $n$th Fourier coefficient for the second integrand is the convolution of $r^{|n|}$ with itself:
$$\sum_{k \in \mathbb Z} r^{|k|} r^{|n - k|} =
\left( |n| + \frac {1 + r^2} {1 - r^2} \right) r^{|n|}, \\
\int_0^{2 \pi} \int_0^{2 \pi} \frac
 {e^{i (m \phi - m' \phi')}}
 {(a + b \cos (\phi - \phi'))^2} d\phi' d\phi =
\frac {4 \pi^2 r^{|m|}} {a^2 - b^2}
 \left( |m| + \frac {1 + r^2} {1 - r^2} \right) \delta_{m, m'}.$$
