# Mellin transform

In mathematics, the Mellin transform is an integral transform that may be regarded as the multiplicative version of the two-sided Laplace transform. This integral transform is closely connected to the theory of Dirichlet series, and is often used in number theory and the theory of asymptotic expansions; it is closely related to the Laplace transform and the Fourier transform, and the theory of the gamma function and allied special functions.

The Mellin transform of a function f is

$\left\{{\mathcal {M}}f\right\}(s)=\varphi (s)=\int _{0}^{\infty }x^{s}f(x){\frac {dx}{x}}.$ The inverse transform is

$\left\{{\mathcal {M}}^{-1}\varphi \right\}(x)=f(x)={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }x^{-s}\varphi (s)ds.$ The notation implies this is a path integral taken over a vertical line in the complex plane. Conditions under which this inversion is valid are given in the Mellin inversion theorem.

The transform is named after the Finnish mathematician Robert Hjalmar Mellin (1854 - 1933).

## Relationship to other transforms

The two-sided Laplace transform may be defined in terms of the Mellin transform by

$\left\{{\mathcal {B}}f\right\}(s)=\left\{{\mathcal {M}}f(-\ln x)\right\}(s)$ and conversely we can get the Mellin transform from the two-sided Laplace transform by

$\left\{{\mathcal {M}}f\right\}(s)=\left\{{\mathcal {B}}f(e^{-x})\right\}(s)$ The Mellin transform may be thought of as integrating using a kernel xs with respect to the multiplicative Haar measure, ${\frac {dx}{x}}$ , which is invariant under dilation $x\mapsto ax$ , so that ${\frac {d(ax)}{ax}}={\frac {dx}{x}}$ ; the two-sided Laplace transform integrates with respect to the additive Haar measure $dx$ , which is translation invariant, so that $d(x+a)=dx$ .

We also may define the Fourier transform in terms of the Mellin transform and vice-versa; if we define the two-sided Laplace transform as above, then

$\left\{{\mathcal {F}}f\right\}(s)=\left\{{\mathcal {B}}f\right\}(is)=\left\{{\mathcal {M}}f(-\ln x)\right\}(is)$ We may also reverse the process and obtain

$\left\{{\mathcal {M}}f\right\}(s)=\left\{{\mathcal {B}}f(e^{-x})\right\}(s)=\left\{{\mathcal {F}}f(e^{-x})\right\}(-is)$ 