Mellin Transform

What is the Mellin Transform?

The Mellin transform is a type of integral transform. It is defined as follows:

$$\hat{f}(s)={\int }_0^{\mathrm{\infty }}f(x)x^{s-1}dx$$

where $s\in \mathbb{C}$.

In the context of transforming functions, we also use the notation:

$$\mathcal{M}:f\mapsto \hat{f}$$

Integral Transforms

A transformation of functions using the following expression is called an integral transform:

$$\hat{f}(y)={\int }_a^{b}f(x)K(x,y)dx$$

Here $K(x,y)$ is called the kernel.

There are various types of integral transforms, with notable examples including the Fourier transform and the Laplace transform.

Mellin inversion formula

The inverse transform of the Mellin transform is given by:

$$f(x)=\frac{1}{2\pi i}{\int }_{\sigma -i\mathrm{\infty }}^{\sigma +i\mathrm{\infty }}\hat{f}(s)x^{-s}ds$$

where $\sigma \in \mathbb{R}$.

The contour of integration on the right-hand side is the same as the Bromwich integral that appears in the Laplace inverse transform.

This formula can be derived from the known Fourier transform as follows.

By substituting $x=e^u$ into the definition integral:

$$\hat{f}(s)={\int }_{u=-\mathrm{\infty }}^{\mathrm{\infty }}f(e^u)e^{u(s-1)}{e}^{u}du={\int }_{u=-\mathrm{\infty }}^{\mathrm{\infty }}f(e^u)e^{us}du$$

Further,

by letting $s=\sigma -it$, we get:

$$\hat{f}(\sigma -it)={\int }_{u=-\mathrm{\infty }}^{\mathrm{\infty }}f(e^u)e^{u(\sigma -it)}du={\int }_{u=-\mathrm{\infty }}^{\mathrm{\infty }}f(e^u)e^{\sigma u}{e}^{-itu}du$$

In this expression, let:

$$g(u)=f(e^u)e^{\sigma u}$$

$$G(t)=\hat{f}(\sigma -it)$$

Then, $G(t)$ is the Fourier transform of $g(u)$.

By applying the inverse Fourier transform, we have:

$$g(u)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}G(t)e^{itu}dt$$

Thus,

$$f(e^u)e^{\sigma u}=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\hat{f}(\sigma -it)e^{itu}dt$$

Hence,

$$f(e^u)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\hat{f}(\sigma -it)e^{itu}{e}^{-\sigma u}dt=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\hat{f}(\sigma -it)e^{-u(\sigma -it)}dt=\frac{1}{2\pi }{\int }_{s=\sigma +i\mathrm{\infty }}^{\sigma -i\mathrm{\infty }}\hat{f}(s)e^{-us}(-\frac{1}{i}ds)=\frac{1}{2\pi i}{\int }_{\sigma -i\mathrm{\infty }}^{\sigma +i\mathrm{\infty }}\hat{f}(s)e^{-us}ds$$

Thus,

$$f(x)=\frac{1}{2\pi i}{\int }_{\sigma -i\mathrm{\infty }}^{\sigma +i\mathrm{\infty }}\hat{f}(s)x^{-s}ds$$

which provides the Mellin inversion formula.