Let a function in spherical coordinates
$$\vec F(\vec r) = \int{ d^3\vec r\,' \vec j(\vec r\,') } \,e^{-ik\hat r \cdot \vec r \,'}$$ Where $\vec j$ is a vector function. So $\vec F$ only depends on the angular variables, ie $\vec F(\vec r) = \vec F(\theta, \phi)$. Someone says its curl is zero, that is $\nabla \times \vec F = 0$.
How is it derived? Does this result depend on how $\vec F(\theta, \phi)$ is defined (an exponential dependency in the above one), or is it generally true for all functions with only angular dependencies (that is, with no radial dependency)?
A vector field independent of $r$ can have nonzero curl. For example, $$\vec F(x,y,z) = \left(\frac{-y}{\sqrt{x^2+y^2}}, \frac{x}{\sqrt{x^2+y^2}},0\right)$$ has nonzero curl: its $z$-component is $1/\sqrt{x^2+y^2}$.
So, the integral form matters.
Actually, I don't think the result is true as stated: my computations do not produce zero curl for this integral, unless something is assumed about $\vec j$.
Here is a similar integral which does have zero curl: $$\vec F(\vec r) = \int d^3 \vec v\, \varphi(\vec v) \,g(\vec r\cdot \vec v) \vec v $$ where $\varphi$ and $g$ are arbitrary smooth scalar-valued functions.
Indeed, let $G$ be an antiderivative of $g$, that is $G'=g$ (recall these are functions of a scalar argument). Then the chain rule implies $$\vec F(\vec r) = \nabla \left ( \int d^3 \vec v\, \varphi(\vec v) \,G(\vec r\cdot \vec v) \right ) $$ which means $\vec F$ is conservative. Hence it has zero curl.