Mellin transform

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

${\displaystyle \left\{{\mathcal {B}}f\right\}(s)=\left\{{\mathcal {M}}f(-\ln x)\right\}(s)}$

and conversely we can get the Mellin transform from the two-sided Laplace transform by

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

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

We also may define the Fourier transform in terms of the Mellin transform and vice versa; in terms of the Mellin transform and of the two-sided Laplace transform defined above

${\displaystyle \left\{{\mathcal {F}}f\right\}(-s)=\left\{{\mathcal {B}}f\right\}(-is)=\left\{{\mathcal {M}}f(-\ln x)\right\}(-is)\ .}$

We may also reverse the process and obtain

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

The Mellin transform also connects the Newton series or binomial transform together with the Poisson generating function, by means of the Poisson–Mellin–Newton cycle.

The Mellin transform may also be viewed as the Gelfand transform for the convolution algebra of the locally compact abelian group of positive real numbers with multiplication.

For ${\displaystyle \alpha ,\beta \in \mathbb {R} }$, let the open strip ${\displaystyle \langle \alpha ,\beta \rangle }$ be defined to be all ${\displaystyle s\in \mathbb {C} }$ such that ${\displaystyle s=\sigma +it}$ with ${\displaystyle \alpha <\sigma <\beta .}$ The fundamental strip of ${\displaystyle {\mathcal {M}}f(s)}$ is defined to be the largest open strip on which it is defined. For example, for ${\displaystyle a>b}$ the fundamental strip of

${\displaystyle f(x)={\begin{cases}x^{a}&x<1,\\x^{b}&x>1,\end{cases}}}$

is ${\displaystyle \langle -a,-b\rangle .}$ As seen by this example, the asymptotics of the function as ${\displaystyle x\to 0^{+}}$ define the left endpoint of its fundamental strip, and the asymptotics of the function as ${\displaystyle x\to +\infty }$ define its right endpoint. To summarize using Big O notation, if ${\displaystyle f}$ is ${\displaystyle O(x^{a})}$ as ${\displaystyle x\to 0^{+}}$ and ${\displaystyle O(x^{b})}$ as ${\displaystyle x\to +\infty ,}$ then ${\displaystyle {\mathcal {M}}f(s)}$ is defined in the strip ${\displaystyle \langle -a,-b\rangle .}$[5]

An application of this can be seen in the gamma function, ${\displaystyle \Gamma (s).}$ Since ${\displaystyle f(x)=e^{-x}}$ is ${\displaystyle O(x^{0})}$ as ${\displaystyle x\to 0^{+}}$ and ${\displaystyle O(x^{k})}$ for all ${\displaystyle k,}$ then ${\displaystyle \Gamma (s)={\mathcal {M}}f(s)}$ should be defined in the strip ${\displaystyle \langle 0,+\infty \rangle ,}$ which confirms that ${\displaystyle \Gamma (s)}$ is analytic for ${\displaystyle \Re (s)>0.}$

The properties in this table may be found in Bracewell (2000) and Erdélyi (1954).

Properties of the Mellin transform

Function	Mellin transform	Fundamental strip	Comments
${\displaystyle f(x)}$	${\displaystyle {\tilde {f}}(s)=\{{\mathcal {M}}f\}(s)=\int _{0}^{\infty }f(x)x^{s}{\frac {dx}{x}}}$	${\displaystyle \alpha <\Re s<\beta }$	Definition
${\displaystyle x^{\nu }\,f(x)}$	${\displaystyle {\tilde {f}}(s+\nu )}$	${\displaystyle \alpha -\Re \nu <\Re s<\beta -\Re \nu }$
${\displaystyle f(x^{\nu })}$	${\displaystyle {\frac {1}{|\nu |}}\,{\tilde {f}}\left({\frac {s}{\nu }}\right)}$	${\displaystyle \alpha <\nu ^{-1}\,\Re s<\beta }$	${\displaystyle \nu \in \mathbb {R} ,\;\nu \neq 0}$
${\displaystyle f(x^{-1})}$	${\displaystyle {\tilde {f}}(-s)}$	${\displaystyle -\beta <\Re s<-\alpha }$
${\displaystyle x^{-1}\,f(x^{-1})}$	${\displaystyle {\tilde {f}}(1-s)}$	${\displaystyle 1-\beta <\Re s<1-\alpha }$	Involution
${\displaystyle {\overline {f(x)}}}$	${\displaystyle {\overline {{\tilde {f}}({\overline {s}})}}}$	${\displaystyle \alpha <\Re s<\beta }$	Here ${\displaystyle {\overline {z}}}$ denotes the complex conjugate of ${\displaystyle z}$.
${\displaystyle f(\nu x)}$	${\displaystyle \nu ^{-s}{\tilde {f}}(s)}$	${\displaystyle \alpha <\Re s<\beta }$	${\displaystyle \nu >0}$, Scaling
${\displaystyle f(x)\,\ln x}$	${\displaystyle {\tilde {f}}'(s)}$	${\displaystyle \alpha <\Re s<\beta }$
${\displaystyle f'(x)}$	${\displaystyle -(s-1)\,{\tilde {f}}(s-1)}$	${\displaystyle \alpha +1<\Re s<\beta +1}$
${\displaystyle \left({\frac {d}{dx}}\right)^{n}\,f(x)}$	${\displaystyle (-1)^{n}\,{\frac {\Gamma (s)}{\Gamma (s-n)}}{\tilde {f}}(s-n)}$	${\displaystyle \alpha +n<\Re s<\beta +n}$
${\displaystyle x\,f'(x)}$	${\displaystyle -s\,{\tilde {f}}(s)}$	${\displaystyle \alpha <\Re s<\beta }$
${\displaystyle \left(x\,{\frac {d}{dx}}\right)^{n}\,f(x)}$	${\displaystyle (-s)^{n}{\tilde {f}}(s)}$	${\displaystyle \alpha <\Re s<\beta }$
${\displaystyle \left({\frac {d}{dx}}\,x\right)^{n}\,f(x)}$	${\displaystyle (1-s)^{n}{\tilde {f}}(s)}$	${\displaystyle \alpha <\Re s<\beta }$
${\displaystyle \int _{0}^{x}f(y)\,dy}$	${\displaystyle -s^{-1}\,{\tilde {f}}(s+1)}$	${\displaystyle \alpha -1<\Re s<\min(\beta -1,0)}$	Valid only if the integral exists.
${\displaystyle \int _{x}^{\infty }f(y)\,dy}$	${\displaystyle s^{-1}\,{\tilde {f}}(s+1)}$	${\displaystyle \max(\alpha -1,0)<\Re s<\beta -1}$	Valid only if the integral exists.
${\displaystyle \int _{0}^{\infty }f_{1}\left({\frac {x}{y}}\right)\,f_{2}(y)\,{\frac {dy}{y}}}$	${\displaystyle {\tilde {f}}_{1}(s)\,{\tilde {f}}_{2}(s)}$	${\displaystyle \max(\alpha _{1},\alpha _{2})<\Re s<\min(\beta _{1},\beta _{2})}$	Multiplicative convolution
${\displaystyle x^{\mu }\int _{0}^{\infty }y^{\nu }\,f_{1}\left({\frac {x}{y}}\right)\,f_{2}(y)\,dy}$	${\displaystyle {\tilde {f}}_{1}(s+\mu )\,{\tilde {f}}_{2}(s+\mu +\nu +1)}$		Multiplicative convolution (generalized)
${\displaystyle x^{\mu }\int _{0}^{\infty }y^{\nu }\,f_{1}(x\,y)\,f_{2}(y)\,dy}$	${\displaystyle {\tilde {f}}_{1}(s+\mu )\,{\tilde {f}}_{2}(1-s-\mu +\nu )}$		Multiplicative convolution (generalized)
${\displaystyle f_{1}(x)\,f_{2}(x)}$	${\displaystyle {\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }{\tilde {f}}_{1}(r)\,{\tilde {f}}_{2}(s-r)\,dr}$	${\displaystyle {\begin{aligned}\alpha _{2}+c&<\Re s<\beta _{2}+c\\\alpha _{1}&<c<\beta _{1}\end{aligned}}}$	Multiplication. Only valid if integral exists. See Parseval's theorem below for conditions which ensure the existence of the integral.

In the study of Hilbert spaces, the Mellin transform is often posed in a slightly different way. For functions in ${\displaystyle L^{2}(0,\infty )}$ (see Lp space) the fundamental strip always includes ${\displaystyle {\tfrac {1}{2}}+i\mathbb {R} }$, so we may define a linear operator ${\displaystyle {\tilde {\mathcal {M}}}}$ as

${\displaystyle {\tilde {\mathcal {M}}}\colon L^{2}(0,\infty )\to L^{2}(-\infty ,\infty ),}$

${\displaystyle \{{\tilde {\mathcal {M}}}f\}(s):={\frac {1}{\sqrt {2\pi }}}\int _{0}^{\infty }x^{-{\frac {1}{2}}+is}f(x)\,dx.}$

In other words, we have set

${\displaystyle \{{\tilde {\mathcal {M}}}f\}(s):={\tfrac {1}{\sqrt {2\pi }}}\{{\mathcal {M}}f\}({\tfrac {1}{2}}+is).}$

This operator is usually denoted by just plain ${\displaystyle {\mathcal {M}}}$ and called the "Mellin transform", but ${\displaystyle {\tilde {\mathcal {M}}}}$ is used here to distinguish from the definition used elsewhere in this article. The Mellin inversion theorem then shows that ${\displaystyle {\tilde {\mathcal {M}}}}$ is invertible with inverse

${\displaystyle {\tilde {\mathcal {M}}}^{-1}\colon L^{2}(-\infty ,\infty )\to L^{2}(0,\infty ),}$

${\displaystyle \{{\tilde {\mathcal {M}}}^{-1}\varphi \}(x)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }x^{-{\frac {1}{2}}-is}\varphi (s)\,ds.}$

Furthermore, this operator is an isometry, that is to say ${\displaystyle \|{\tilde {\mathcal {M}}}f\|_{L^{2}(-\infty ,\infty )}=\|f\|_{L^{2}(0,\infty )}}$ for all ${\displaystyle f\in L^{2}(0,\infty )}$ (this explains why the factor of ${\displaystyle 1/{\sqrt {2\pi }}}$ was used).

In probability theory, the Mellin transform is an essential tool in studying the distributions of products of random variables.[8] If X is a random variable, and X⁺ = max{X,0} denotes its positive part, while X⁻ = max{−X,0} is its negative part, then the Mellin transform of X is defined as[9]

${\displaystyle {\mathcal {M}}_{X}(s)=\int _{0}^{\infty }x^{s}dF_{X^{+}}(x)+\gamma \int _{0}^{\infty }x^{s}dF_{X^{-}}(x),}$

where γ is a formal indeterminate with γ² = 1. This transform exists for all s in some complex strip D = {s : a ≤ Re(s) ≤ b} , where a ≤ 0 ≤ b.[9]

The Mellin transform ${\displaystyle {\mathcal {M}}_{X}(it)}$ of a random variable X uniquely determines its distribution function F_X.[9] The importance of the Mellin transform in probability theory lies in the fact that if X and Y are two independent random variables, then the Mellin transform of their product is equal to the product of the Mellin transforms of X and Y:[10]

${\displaystyle {\mathcal {M}}_{XY}(s)={\mathcal {M}}_{X}(s){\mathcal {M}}_{Y}(s)}$

In the Laplacian in cylindrical coordinates in a generic dimension (orthogonal coordinates with one angle and one radius, and the remaining lengths) there is always a term:

${\displaystyle {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)=f_{rr}+{\frac {f_{r}}{r}}}$

For example, in 2-D polar coordinates the Laplacian is:

${\displaystyle \nabla ^{2}f={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}}}$

and in 3-D cylindrical coordinates the Laplacian is,

${\displaystyle \nabla ^{2}f={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \varphi ^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}.}$

This term can be treated with the Mellin transform,[11] since:

${\displaystyle {\mathcal {M}}\left(r^{2}f_{rr}+rf_{r},r\to s\right)=s^{2}{\mathcal {M}}\left(f,r\to s\right)=s^{2}F}$

For example, the 2-D Laplace equation in polar coordinates is the PDE in two variables:

${\displaystyle r^{2}f_{rr}+rf_{r}+f_{\theta \theta }=0}$

and by multiplication:

${\displaystyle {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}}=0}$

with a Mellin transform on radius becomes the simple harmonic oscillator:

${\displaystyle F_{\theta \theta }+s^{2}F=0}$

with general solution:

${\displaystyle F(s,\theta )=C_{1}(s)\cos(s\theta )+C_{2}(s)\sin(s\theta )}$

Now let's impose for example some simple wedge boundary conditions to the original Laplace equation:

${\displaystyle f(r,-\theta _{0})=a(r),\quad f(r,\theta _{0})=b(r)}$

these are particularly simple for Mellin transform, becoming:

${\displaystyle F(s,-\theta _{0})=A(s),\quad F(s,\theta _{0})=B(s)}$

These conditions imposed to the solution particularize it to:

${\displaystyle F(s,\theta )=A(s){\frac {\sin(s(\theta _{0}-\theta ))}{\sin(2\theta _{0}s)}}+B(s){\frac {\sin(s(\theta _{0}+\theta ))}{\sin(2\theta _{0}s)}}}$

Now by the convolution theorem for Mellin transform, the solution in the Mellin domain can be inverted:

${\displaystyle f(r,\theta )={\frac {r^{m}\cos(m\theta )}{2\theta _{0}}}\int _{0}^{\infty }\left({\frac {a(x)}{x^{2m}+2r^{m}x^{m}\sin(m\theta )+r^{2m}}}+{\frac {b(x)}{x^{2m}-2r^{m}x^{m}\sin(m\theta )+r^{2m}}}\right)x^{m-1}\,dx}$

where the following inverse transform relation was employed:

${\displaystyle {\mathcal {M}}^{-1}\left({\frac {\sin(s\varphi )}{\sin(2\theta _{0}s)}};s\to r\right)={\frac {1}{2\theta _{0}}}{\frac {r^{m}\sin(m\varphi )}{1+2r^{m}\cos(m\varphi )+r^{2m}}}}$

where ${\displaystyle m={\frac {\pi }{2\theta _{0}}}}$.

The Mellin Transform is widely used in computer science for the analysis of algorithms[12] because of its scale invariance property. The magnitude of the Mellin Transform of a scaled function is identical to the magnitude of the original function for purely imaginary inputs. This scale invariance property is analogous to the Fourier Transform's shift invariance property. The magnitude of a Fourier transform of a time-shifted function is identical to the magnitude of the Fourier transform of the original function.

This property is useful in image recognition. An image of an object is easily scaled when the object is moved towards or away from the camera.

In quantum mechanics and especially quantum field theory, Fourier space is enormously useful and used extensively because momentum and position are Fourier transforms of each other (for instance, Feynman diagrams are much more easily computed in momentum space). In 2011, A. Liam Fitzpatrick, Jared Kaplan, João Penedones, Suvrat Raju, and Balt C. van Rees showed that Mellin space serves an analogous role in the context of the AdS/CFT correspondence.[13][14][15]

• Perron's formula describes the inverse Mellin transform applied to a Dirichlet series.
• The Mellin transform is used in analysis of the prime-counting function and occurs in discussions of the Riemann zeta function.
• Inverse Mellin transforms commonly occur in Riesz means.
• The Mellin transform can be used in audio timescale-pitch modification .

Following list of interesting examples for the Mellin transform can be found in Bracewell (2000) and Erdélyi (1954):

Selected Mellin transforms

Function ${\displaystyle f(x)}$	Mellin transform ${\displaystyle {\tilde {f}}(s)={\mathcal {M}}\{f\}(s)}$	Region of convergence	Comment
${\displaystyle e^{-x}}$	${\displaystyle \Gamma (s)}$	${\displaystyle 0<\Re s<\infty }$
${\displaystyle e^{-x}-1}$	${\displaystyle \Gamma (s)}$	${\displaystyle -1<\Re s<0}$
${\displaystyle e^{-x}-1+x}$	${\displaystyle \Gamma (s)}$	${\displaystyle -2<\Re s<-1}$	And generally ${\displaystyle \Gamma (s)}$ is the Mellin transform of[16] ${\displaystyle e^{-x}-\sum _{n=0}^{N-1}{\frac {(-1)^{n}}{n!}}x^{n},{\text{ for }}-N<\Re s<-N+1}$
${\displaystyle e^{-x^{2}}}$	${\displaystyle {\tfrac {1}{2}}\Gamma ({\tfrac {1}{2}}s)}$	${\displaystyle 0<\Re s<\infty }$
${\displaystyle \mathrm {erfc} (x)}$	${\displaystyle {\frac {\Gamma ({\tfrac {1}{2}}(1+s))}{{\sqrt {\pi }}\;s}}}$	${\displaystyle 0<\Re s<\infty }$
${\displaystyle e^{-(\ln x)^{2}}}$	${\displaystyle {\sqrt {\pi }}\,e^{{\tfrac {1}{4}}s^{2}}}$	${\displaystyle -\infty <\Re s<\infty }$
${\displaystyle \delta (x-a)}$	${\displaystyle a^{s-1}}$	${\displaystyle -\infty <\Re s<\infty }$	${\displaystyle a>0,\;\delta (x)}$ is the Dirac delta function.
${\displaystyle u(1-x)=\left\{{\begin{aligned}&1&&\;{\text{if}}\;0<x<1&\\&0&&\;{\text{if}}\;1<x<\infty &\end{aligned}}\right.}$	${\displaystyle {\frac {1}{s}}}$	${\displaystyle 0<\Re s<\infty }$	${\displaystyle u(x)}$ is the Heaviside step function
${\displaystyle -u(x-1)=\left\{{\begin{aligned}&0&&\;{\text{if}}\;0<x<1&\\&-1&&\;{\text{if}}\;1<x<\infty &\end{aligned}}\right.}$	${\displaystyle {\frac {1}{s}}}$	${\displaystyle -\infty <\Re s<0}$
${\displaystyle u(1-x)\,x^{a}=\left\{{\begin{aligned}&x^{a}&&\;{\text{if}}\;0<x<1&\\&0&&\;{\text{if}}\;1<x<\infty &\end{aligned}}\right.}$	${\displaystyle {\frac {1}{s+a}}}$	${\displaystyle -\Re a<\Re s<\infty }$
${\displaystyle -u(x-1)\,x^{a}=\left\{{\begin{aligned}&0&&\;{\text{if}}\;0<x<1&\\&-x^{a}&&\;{\text{if}}\;1<x<\infty &\end{aligned}}\right.}$	${\displaystyle {\frac {1}{s+a}}}$	${\displaystyle -\infty <\Re s<-\Re a}$
${\displaystyle u(1-x)\,x^{a}\ln x=\left\{{\begin{aligned}&x^{a}\ln x&&\;{\text{if}}\;0<x<1&\\&0&&\;{\text{if}}\;1<x<\infty &\end{aligned}}\right.}$	${\displaystyle {\frac {1}{(s+a)^{2}}}}$	${\displaystyle -\Re a<\Re s<\infty }$
${\displaystyle -u(x-1)\,x^{a}\ln x=\left\{{\begin{aligned}&0&&\;{\text{if}}\;0<x<1&\\&-x^{a}\ln x&&\;{\text{if}}\;1<x<\infty &\end{aligned}}\right.}$	${\displaystyle {\frac {1}{(s+a)^{2}}}}$	${\displaystyle -\infty <\Re s<-\Re a}$
${\displaystyle {\frac {1}{1+x}}}$	${\displaystyle {\frac {\pi }{\sin(\pi s)}}}$	${\displaystyle 0<\Re s<1}$
${\displaystyle {\frac {1}{1-x}}}$	${\displaystyle {\frac {\pi }{\tan(\pi s)}}}$	${\displaystyle 0<\Re s<1}$
${\displaystyle {\frac {1}{1+x^{2}}}}$	${\displaystyle {\frac {\pi }{2\sin({\tfrac {1}{2}}\pi s)}}}$	${\displaystyle 0<\Re s<2}$
${\displaystyle \ln(1+x)}$	${\displaystyle {\frac {\pi }{s\,\sin(\pi s)}}}$	${\displaystyle -1<\Re s<0}$
${\displaystyle \sin(x)}$	${\displaystyle \sin({\tfrac {1}{2}}\pi s)\,\Gamma (s)}$	${\displaystyle -1<\Re s<1}$
${\displaystyle \cos(x)}$	${\displaystyle \cos({\tfrac {1}{2}}\pi s)\,\Gamma (s)}$	${\displaystyle 0<\Re s<1}$
${\displaystyle e^{ix}}$	${\displaystyle e^{i\pi s/2}\,\Gamma (s)}$	${\displaystyle 0<\Re s<1}$
${\displaystyle J_{0}(x)}$	${\displaystyle {\frac {2^{s-1}}{\pi }}\,\sin(\pi s/2)\,\left[\Gamma (s/2)\right]^{2}}$	${\displaystyle 0<\Re s<{\tfrac {3}{2}}}$	${\displaystyle J_{0}(x)}$ is the Bessel function of the first kind.
${\displaystyle Y_{0}(x)}$	${\displaystyle -{\frac {2^{s-1}}{\pi }}\,\cos(\pi s/2)\,\left[\Gamma (s/2)\right]^{2}}$	${\displaystyle 0<\Re s<{\tfrac {3}{2}}}$	${\displaystyle Y_{0}(x)}$ is the Bessel function of the second kind
${\displaystyle K_{0}(x)}$	${\displaystyle 2^{s-2}\,\left[\Gamma (s/2)\right]^{2}}$	${\displaystyle 0<\Re s<\infty }$	${\displaystyle K_{0}(x)}$ is the modified Bessel function of the second kind

• Mellin inversion theorem
• Perron's formula
• Ramanujan's master theorem

• Philippe Flajolet, Xavier Gourdon, Philippe Dumas, Mellin Transforms and Asymptotics: Harmonic sums.
• Antonio Gonzáles, Marko Riedel Celebrando un clásico, newsgroup es.ciencia.matematicas
• Juan Sacerdoti, Funciones Eulerianas (in Spanish).
• Mellin Transform Methods, Digital Library of Mathematical Functions, 2011-08-29, National Institute of Standards and Technology
• Antonio De Sena and Davide Rocchesso, A FAST MELLIN TRANSFORM WITH APPLICATIONS IN DAFX