Why does Integrate not evaluate the integral?

Question

I am following a caluclation in this paper, equation (7). There, it is stated that the integral $$\int_0^{2\pi}d\phi e^{-ik_rr\cos(v_{\phi}-\phi)}=J_0(k_rr)$$results in the zeroth-order Bessel function. Why, then, does Mathematica not provide the same result?

Assuming[r \[Element] Reals && vr \[Element] Reals && v\[Phi] \[Element] Reals
 , Integrate[E^(-I r Subscript[v, r] Cos[v\[Phi] - \[Phi]]), {\[Phi], 0, 
    2 \[Pi]}] // FullSimplify]

As an output, I simply get the integral in LateX form as if I would have written it down by hand.

Edit: Even weirder, if I leave out the $v_{\phi}$ in the integrand, I get the result, even though both should be the same as they are related by a simple change of variables: $v_{\phi} - \phi = \alpha$

Answer 1

As @Artes' answer begins, "As a rule-based, term-rewriting system...." Mathematica has a missing rule here, it seems.

If we reduce the number of parameters via substitution, which can be done without losing the original integral, Mathematica succeeds. (Here I used the periodicity of cosine to translate the interval.)

Integrate[E^(-I rr Cos[-tt]), {tt, 0, 2 π}, 
  Assumptions -> rr ∈ Reals] /. {rr -> r*Subscript[v, r]}
(*  2 π BesselJ[0, Abs[r Subscript[v, r]]]  *)

Not translating the interval fails (an appropriate rule seems to be missing):

Integrate[E^(-I rr Cos[-tt]), {tt, 0 - vϕ, 2 π - vϕ}, 
 Assumptions -> rr ∈ Reals && vϕ ∈ Reals]
(* ... input returned ... *)

A success:

Integrate[E^(-I rr Cos[-tt]), {tt, 0 - Pi/2, 2 π - Pi/2}, 
 Assumptions -> rr ∈ Reals]
(*  2 π BesselJ[0, Abs[rr]]  *)

More failures. One might or might not excuse the interval that depended on the parameter vϕ, but given the success with translation by Pi/2, I wonder at the failures with translation by other values.

Integrate[E^(-I rr Cos[-tt]), {tt, 0 - 1, 2 π - 1}, 
 Assumptions -> rr ∈ Reals]

Integrate[E^(-I rr Cos[-tt]), {tt, 0 - Pi/3, 2 π - Pi/3}, 
 Assumptions -> rr ∈ Reals]

Aside & caveat: Note that symbolically, r and Subscript[v, r], which has the form f[v, r], both depend on r as expressions. This is probably not what was meant.