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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.