Multi variable Mellin transformation and its inverse transformation

First of all, I know Fourier transformation and its inverse transformation for multivariable. i.e., The multivariable Fourier transformation of $f(\vec{x})$ and inverse Fourier transformation of $g(\vec{\zeta})$ is given by \begin{align} &\mathcal{F} (f) := \int e^{-2 \pi i \vec{x} \cdot \vec{\zeta}} f(\vec{x}) d\vec{x} \\ &\mathcal{F}^{-1} (g):= \int e^{2\pi i \vec{x} \cdot \vec{\zeta} } g(\vec{\zeta}) d \vec{\zeta} \end{align}

Now I want to know the Multi-variable extension of Mellin transformation and its inverse transformation.

For the single variable case, I know the following definitions from Fourier and its inverse transformation.

Let $f(t)$ be a function defined on the positive real axis $0<t<\infty$. The Mellin transformation $\mathcal{M}$ is the operation mapping the function $f$ into the function $F$ defined on the complex by the relation \begin{align} \mathcal{M}[f;s]= F(s) := \int_0^{\infty} f(t) t^{s-1} dt \end{align} In general the integral does exists only for complex values $s=a+jb$ such that $a_1<a<a_2$, where $a_1, a_2$ depend on the function $f(t)$ to transform. Denote this by $S(a_1,a_2)$ and call the strip.

And the Inverse Melline transformation is given by \begin{align} f(t) = \frac{1}{2\pi j} \int_{a-j\infty}^{a+j\infty} F(s) t^{-s} ds \end{align} where the integration is along a vertical line through $\operatorname{Re}[s]=a$.

Solution 1:

Consider the multivariable Fourier transformation in more detail. i.e., Fourier transformation of $f(x_1, \cdots, x_n)$ into $\mathcal{F}(f)(\zeta_1, \cdots, \zeta_n)$ is given by

\begin{align} \mathcal{F}(f) = \int e^{-2 \pi i (\sum_i x_i \zeta_i)} f(x_1, \cdots, x_n) dx_1 \cdots dx_n \end{align}

From \begin{align} \mathcal{M}[f(x);s=a+2 \pi ib] = \mathcal{F}[f(e^{-x})e^{-ax};b] \end{align} with the comment of @reuns, extend $s$ to $\vec{s}$ \begin{align} \mathcal{M}[f(x_1, \cdots, x_n);\vec{s}] = \mathcal{F}[f(e^{-x_1}, e^{-x_2}, \cdots, e^{-x_n}) ; \vec{b}] = \int_{-\infty}^{\infty} f(e^{-x_1}, \cdots, e^{-x_n}) e^{-\vec{s} \cdot \vec{x}} dx_1 \cdots dx_n \end{align} Now back to $t$, for one variatble case using $t=e^{-x}, dt = - e^{-x}$, I have $\int_{0}^{\infty} f(t) t^{s-1} dt$, hence \begin{align} \mathcal{M}[f(x_1, \cdots, x_n);\vec{s}] = \int_0^{\infty} f(t_1, \cdots, t_n) t_1^{s_1 -1} \cdots t_n^{s_n-1} dt_1 \cdots dt_n \end{align} and similar arguement can be applied to inverse Mellin as well. i.e., the inverse Mellin transformation can be achieved via \begin{align} f(t_1, \cdots, t_n) = (\frac{1}{2 \pi i})^n \int_{a_n - i \infty}^{a_n + i\infty} \cdots \int_{a_1 - i \infty}^{a_1 + i\infty} F(s_1, \cdots, s_n) t_1^{-s_1} \cdots t_n^{-s_n} ds_1 \cdots ds_n \end{align} where the integration is along a vertical line through $\operatorname{Re}[s_i]=a_i$.