TITLE: Calculate a double integral
QUESTION [5 upvotes]: Why does the following identity hold?
$$\int_0^{2\pi}\int_0^{2\pi}\arcsin(k\cdot \cos(\theta-\phi))e^{i(\theta-\phi)} \, d\theta \, d\phi=2\pi\int_0^{2\pi}\arcsin(k\cdot \cos x)e^{ix} \, dx$$
where $k\in[-1,1]$ is a constant.
I'm trying to use substitution $x=\theta-\phi$ but it still needs some tricks. Thanks!

REPLY [2 votes]: Substitution $x = \theta - \phi$, then 
$$ \frac{dx}{d\theta} = 1$$
$\theta = 0$ gives $x = -\phi$, $\theta = 2\pi$ gives $x = 2\pi - \phi$
$$\int_0^{2\pi}\int_0^{2\pi}\arcsin(k\cdot \cos(\theta-\phi))e^{i(\theta-\phi)} \, d\theta \, d\phi=$$
$$\int_0^{2\pi}\int_{-\phi}^{2\pi - \phi}\arcsin(k\cdot \cos(x))e^{i(x)} \, dx \, d\phi=$$
$$\int_0^{2\pi}\int_0^{2\pi}\arcsin(k\cdot \cos(x))e^{i(x)} \, dx \, d\phi=$$
$$2\pi \int_0^{2\pi}\arcsin(k\cdot \cos(x))e^{i(x)} \, dx=$$
Where we used that the integrand is a function that is $2\pi$ periodic, to convert the boundary terms in the third integral. In the last step we used that the inner integral does not depend on $\phi$.