The Mellin transform of $f(x)$ is defined in formula (1) below.
(1) $\quad F(s)=\int\limits_0^\infty f(x)\,x^{s-1}\,dx$
Question: For what class of functions $f(x)$ is it true that $F\left(s^*\right)=F(s)^*$ where $F(s)$ is the Mellin transform of $f(x)$ and $^*$ is the complex conjugate?
I know of examples of functions $f(x)$ where $F\left(s^*\right)\ne F(s)^*$ but in these examples $f(x)$ has an imaginary component as well as a real component for $x>0$. I'm wondering whether $f(x)\in\mathbb{R}\ \forall\ x>0\implies F\left(s^*\right)=F(s)^*$ and if not how to define the class of functions $f(x)$ for which $F\left(s^*\right)=F(s)^*$ and what are some counter examples where $f(x)\in\mathbb{R}\ \forall\ x>0$ and $F\left(s^*\right)\ne F(s)^*$.
In this question I'm assuming the Mellin transform $F(s)$ of $f(x)$ exists, so I'm not asking about conditions for the existence of the Mellin transform in general.
$\overline {F(\bar s)}=\overline {\int\limits_0^\infty {f(x)x^{\bar s-1}}dx} =\int\limits_0^\infty \overline {f(x)x^{\bar s-1}}dx=\int\limits_0^\infty \overline {f(x)}x^{s-1}dx=M(\bar f, s)$
Hence $M(f, \bar s)=\overline {M(\bar f, s)}$, so if $f$ is real then yes $M(f, \bar s)=\overline {M(f, s)}$
In particular if $f$ is not real, there is no easy relation between $M(f, \bar s), {M(f, s)}$ since both $\bar f, x^{s}$ have imaginary parts and $f \ne \bar f$
(note $f({\bar x})=\overline {f(x)}$ means $f$ is real on $(0, \infty)$ so the property is true)