Integral and polar coordinate transformation I want to calculate 
$$\int \limits_{\mathbb{R^2}} e^{i \ x \cdot \xi} \varphi(\xi) \ \ d\xi$$ using polar coordinate transformation. Where $x,\xi \in \mathbb{R}^2$ , $\varphi \in \mathcal{S}(\mathbb{R})$ (schwartz function) and $x \cdot \xi$ denote the inner product. 
I want to use the fact that $x \cdot \xi = \vert x \vert \vert \xi \vert \cos \theta$ where $\cos \theta= \cos \angle x,\xi$.
$$\int \limits_{\mathbb{R^2}} e^{i \ x \cdot \xi} \varphi(\xi) \ \ d\xi = \int \limits_{0}^{\infty} \int \limits_{0}^{2 \pi} e^{\vert x \vert r \cos\theta} \varphi(?) \ d\theta \ dr$$ 
The idea was to point one of the vectors to the north-pole.
My first question is how to adapt $\varphi$ and the second is does $\varphi$ have to be rotation invariant ? 
Thanks in advance.
 A: For $x \in \mathbb{R}^2$ and $\varphi \in \mathcal{S}(\mathbb{R}^2)$ let
$$ f(x) = \int \limits_{\mathbb{R}^2} \mathrm{e}^{\mathrm{i} x \cdot \xi} \varphi(\xi) \, \mathrm{d} \xi  \, . $$
If we introduce polar coordinates $\xi_1 = r \cos(\theta)$ , $\xi_2 = r \sin(\theta)$ and take into account the Jacobian of this transformation, we obtain
$$ f(x) = \int \limits_0^\infty \int \limits_0^{2\pi} \mathrm{e}^{\mathrm{i} r (x_1 \cos(\theta) + x_2 \sin(\theta))} \,\varphi(r \cos(\theta), r \sin(\theta)) \, r \, \mathrm{d} \theta \,\mathrm{d} r \, .$$
This is the best we can do for functions $\varphi \in \mathcal{S}(\mathbb{R}^2)$ (or $\varphi \in L^1 (\mathbb{R}^2)$) that are not invariant under rotations.
However, if $\varphi$ does have this symmetry, further simplifications are possible. Let $R \in \operatorname{SO}(2)$ be a rotation matrix satisfying $R x = |x| e_1$ . Then the change of variables $\eta = R \xi$ (which has unit Jacobian) yields
$$ f(x) = \int \limits_{\mathbb{R}^2} \mathrm{e}^{\mathrm{i} x \cdot R^{\top} \eta} \varphi(R^\top \eta) \, \mathrm{d} \eta = \int \limits_{\mathbb{R}^2} \mathrm{e}^{\mathrm{i} |x| \eta_1} \varphi(\eta) \, \mathrm{d} \eta \, , $$
where we have used the rotational invariance of $\varphi$ and the inner product. Now the introduction of polar coordinates for $\eta$ leads to
$$ f(x) = \int \limits_0^\infty \int \limits_0^{2\pi} \mathrm{e}^{\mathrm{i}  |x| r \cos(\theta)} \,\varphi(r \cos(\theta), r \sin(\theta)) \, r \, \mathrm{d} \theta \,\mathrm{d} r = \int \limits_0^\infty \int \limits_0^{2\pi} \mathrm{e}^{\mathrm{i}  |x| r \cos(\theta)} \,\psi(r) \, r \, \mathrm{d} \theta \,\mathrm{d} r \, .$$
Here $\psi$ is the function on $[0,\infty)$ satisfying $\varphi(x) = \psi(|x|)$ for $x \in \mathbb{R}^2$. We can integrate with respect to  $\theta$ to find
$$ f(x) = 2 \pi \int \limits_0^\infty \psi(r) \operatorname{J}_0 (r |x|) r \, \mathrm{d} r \,  , $$
where
$$ \operatorname{J}_0 (t) = \frac{1}{2 \pi} \int \limits_0^{2\pi} \mathrm{e}^{\mathrm{i} t \cos(\theta)} \, \mathrm{d} \theta $$
is the zeroth-order Bessel function of the first kind.
