# 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 $f(x)$?

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.

• $dx$ is real measure so conjugation goes through Feb 16 at 21:12
• @Conrad Are you saying $f\left(x^*\right)=f(x)^*\implies F\left(s^*\right)=F(s)^*$? Feb 16 at 22:30

$$\overline {F(\bar s)}=\overline {\int\limits_0^\infty {f(x)x^{\bar s-1}}dx} =\int\limits_0^\infty \overline {f(x)x^{\bar s-1}}dx=\int\limits_0^\infty \overline {f(x)}x^{s-1}dx=M(\bar f, s)$$
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)