📂Fourier Analysis

Mellin Transformation

Definition

For a given function $f : [0, \infty) \to \mathbb{C}$, if the following integral exists, it is called the Mellin transform of $f$, and such integral transformations are denoted by $\mathcal{M}$.

$$\mathcal{M}f (s) = \int_{0}^{\infty} x^{s-1}f(x)dx = \phi (s),\quad s \in \mathbb{C}$$

The inverse Mellin transform is as follows.

$$\mathcal{M}^{-1}\phi (x) = \dfrac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty} x^{-s}\phi (s) ds$$

Explanation

It is a kind of integral transformation. The inverse Mellin transform is independent of the value of constant $c$. Mellin transform is used in computer science, number theory, mathematical statistics, quantum mechanics, tomography, etc., and is related to the Laplace transform, Fourier transform, gamma function, and others.

Relation to Fourier Transform

The Fourier transform can be appropriately modified to become a Mellin transform. In other words, $\mathcal{M}f = \mathcal{F}[f \circ \exp]$.

$$\begin{align*} \mathcal{F}[f(e^x)] (\xi) &= \int_{-\infty}^{\infty}f(e^x)e^{-i\xi x} dx \\ &= \int_{0}^{\infty} f(e^x)(e^x)^{-i\xi}\dfrac{1}{e^x}d(e^x) \\ &= \int_{0}^{\infty} f(t)t^{-i\xi}\dfrac{1}{t}dt \\ &= \int_{0}^{\infty} f(t)t^s\dfrac{1}{t}dt \\ &= \int_{0}^{\infty} f(t)t^{s-1}dt \\ &= \mathcal{M}[f(x)] (s) \end{align*}$$

The second equality holds because of $\frac{d e^x}{dx}=e^x \implies dx=\frac{1}{e^x}de^x$. The fourth equality holds if we set it to $s=-i \xi$.

Relationship with the Gamma Function

Let’s say $f(x)=e^{-x}$. Then, the Mellin transform of $f$ is equivalent to the gamma function.

$$\mathcal{M}f(s) = \int_{0}^{\infty} x^{s-1}e^{-x}dx=\Gamma (s)$$

Convolution

A convolution is a function that makes the integral transformation of a product equal to the product of the integral transformations. The convolution of the Mellin transform is as follows.

$$(f \times g) (y) = \int _{0}^{\infty} f(x)g \left(\frac{y}{x} \right)\frac{dx}{x}$$

$$\mathcal{M}(f \times g)=(\mathcal{M}f)(\mathcal{M}g)$$