Mellin transform of $f(1-x)$

Question

The Mellin transform of f(x) is defined as follows:
$$M[f(x);s]=\int_0^\infty x^{s-1}f(x) dx\ = F(s)$$ This integral transform obeys several nice (and easy to prove) properties, such as: $$M[x^n f(x);s] = F(s+n)$$ $$M[f(x^p);s] = \frac{1}{p}F(s/p) $$ Does there exist a similar property for the Mellin transform of $f(1-x)$ $$ M[f(1-x);s] = \int_0^\infty x^{s-1}f(1-x) dx\ $$ in terms of $F(s)$?

Answer 1

No. For one thing, most of the values of $f(x)$ in the integral for $F(s)$ are completely different from the values of $f(1-x)$ in $M(f(1-x))$: the former takes values from $f([0,\infty))$, while the latter uses $f((-\infty,1])$.