Multidimensional Mellin transform

Remarks

For $n = 1$ this is the Mellin transform. Absolute convergence at $s$ depends only on $(\operatorname{Re} s_1, \dots, \operatorname{Re} s_n)$, and the set of such real parts is convex.

Example: $f(x) = e^{-x_1 - \cdots - x_n}$. The integral splits into a product, and $(\mathcal{M} f)(s) = \Gamma(s_1) \cdots \Gamma(s_n)$ for $\operatorname{Re} s_j > 0$, by the gamma function integral.

Non-example: $f = 1$ has no Mellin transform. For every $s$, the factor $\int_0^\infty x_1^{\operatorname{Re} s_1 - 1}\, dx_1$ diverges, at $0$ or at $\infty$.