cv.complex variables - Getting a differential equation for a function from a functional equation of its Mellin transform

If $f$ is a locally integrable function then its Mellin transform
$mathcal{M}[f]$ is defined by
$$ mathcal{M}[f] (s) = int_0^{infty} x^{s - 1} f (x) dx . $$
This integral usually converges in a strip $alpha < Re ; s < beta$ and
defines an analytic function. For our purposes we can assume that
$mathcal{M}[f]$ converges in the right half-plane.

Let us denote $F (s) =mathcal{M}[f] (s)$. Provided that the corresponding
Mellin transforms exist, the basic general theory tells us that, for instance,
$$
mathcal{M} left[ frac{d}{d x} f (x) right] = - (s - 1) F
(s - 1),quad
mathcal{M} left[ x^{mu} f (x) right] = F (s + mu) .
$$
This allows us to translate a differential equation for $f (x)$ into a
functional equation for its Mellin transform $F (s)$.

Example:
For instance, the function $f (x) = e^{- x}$ satisfies the differential
equation
$$ f' (x) + f (x) = 0 $$
which translates to the functional equation
$$ - (s - 1) F (s - 1) + F (s) = 0 $$
for its Mellin transform. Of course, the Mellin transform of $e^{- x}$ is
nothing but the gamma function $Gamma (s)$ which is well-known for
satisfying exactly this functional equation.

Now, let us assume that we are given a function $f (x)$ and its Mellin
transform $F (s)$. Further, suppose that we know that $F (s)$, just as the
gamma function, can be analytically extended to the whole complex plane with
poles at certain nonpositive integers. We also know that $F (s)$ satisfies a
functional equation which we would like to translate back into a differential
equation for $f (x)$. Formally, we obtain, say, a third order differential
equation with polynomial coefficients. Can we conclude that $f (x)$ solves
this DE?

The issue is that in our case the derivatives of $f (x)$ develop singularities
in the domain and are no longer integrable. So Mellin transforms can't be
defined in the usual way for them (and so we can't just use Mellin inversion).

What I am looking for is conditions under which we can still conclude that the
functional equation for $F (s)$ translates into a differential equation for $f
(x)$. Preferably, these should be conditions on $F (s)$ and not on $f (x)$. If
it helps, we can assume $f (x)$ to be compactly supported.

Any help or references are greatly appreciated!