Mellin transform Information & Mellin transform Links at HealthHaven.com

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 line 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 Hjalmar Mellin.

[edit] Importance of the fundamental strip

Domains of f(x)=1/(1+x) and its image function.

Fundamental strip of the Mellin transform of exp(-x) - 1 + x, i.e. the Gamma function.

A Mellin transform should never be computed without its fundamental strip, which tells us where the image function converges. This strip is key to the Mellin inversion process, which arises in number theoretic applications of the transform and in the study of harmonic sums, frequently encountered in computer science. The basic idea is to compute the Mellin transform of a sum and invert it thereafter, thus obtaining an asymptotic expansion. However the Mellin inversion integral is computed over a line parallel to the imaginary axis that lies in the fundamental strip. Without knowing where the strip lies, the integral cannot be computed, more precisely, one does not know which residues contribute to its value.

The fundamental strip arises from the analysis of the convergence properties of the Mellin integral:

$\int_0^\infty x^s f(x)\frac{dx}{x}.$

We split the integral into two parts, as follows:

$\left( \int_0^1 + \int_1^\infty \right) x^s f(x)\frac{dx}{x}.$

Assuming f(x) is locally integrable along the positive real line, the first integral must remain bounded at zero, and the second, at infinity (understood as "in the limit" if e.g. Riemann integrability is used). Letting s = σ + it, we find that

$\left| \int_0^1 x^s f(x)\frac{dx}{x} \right| \le \int_0^1 x^\sigma |f(x)| \frac{dx}{x}$

and

$\left| \int_1^\infty x^s f(x)\frac{dx}{x} \right| \le \int_1^\infty x^\sigma |f(x)| \frac{dx}{x}.$

Now suppose f(x) = O(x^u) at x=0. The first bounding integral converges if

$\sigma + u - 1 > -1 \quad \mbox{or} \quad \sigma > -u.$

Furthermore suppose that f(x) = O(x^v) at infinity. The second bounding integral converges if

$\sigma + v - 1 < -1 \quad \mbox{or} \quad \sigma < -v.$

These two constraints on s define two half planes, the first a left half plane and the second one a right half plane. The intersection of the two half planes is the fundamental strip, denoted 〈−u,−v〉. It frequently happens that the image function can be analytically continued to the whole plane, which makes it possible to compute the inversion integral by shifting the line of integration to the left or to the right. The original Mellin integral, however, remains restricted to the fundamental strip.

Summary: if f(x) is locally integrable along the positive real line, and

$f(x)_{x\rightarrow 0+} = O(x^u) \quad \mbox{and} \quad f(x)_{x\rightarrow +\infty} = O(x^v)$

then its Mellin transform $\varphi(s)$ converges in the fundamental strip 〈−u,−v〉 and the corresponding Mellin inversion integral is taken along a line parallel to the imaginary axis in this strip.

[edit] Computing the fundamental strip

As an example, consider the transform pair

$f(x) = \frac{1}{1+x} \quad \mbox{and} \quad \varphi(s) = \frac{\pi}{\sin \pi s}.$

By inspection, we have

$f(x)_{x\rightarrow 0+} = O(1) = O(x^0) \quad \mbox{and} \quad f(x)_{x\rightarrow +\infty} \sim \frac{1}{x} = O(x^{-1})$

and the fundamental strip is 〈0,1〉. This is illustrated in the first diagram at the beginning of this section.

As a second example, consider the transform pair

$f(x) = \exp(-x) -1 + x \quad \mbox{and} \quad \varphi(s) = \Gamma(s).$

We have the following series expansion around x=0:

$f(x) = \sum_{n\ge 2} \frac{(-1)^n \, x^n}{n!},$

which implies that

$f(x)_{x\rightarrow 0+} = O(x^2).$

At infinity, we have

$f(x)_{x\rightarrow +\infty} \sim x = O(x^1)$

so that the fundamental strip is 〈-2,-1〉. This is shown in the second diagram.

[edit] Example calculation: the Mellin transform of f(x)=1/(1+x)

Keyhole contour used to calculate the Mellin transform of f(x)=1/(1+x).

This section contains an example of how to calculate a particular Mellin transform, that of $f(x) = 1/(1+x)\,$ , given by the integral

$\varphi(s) = \int_0^\infty \frac{x^{s-1}}{1+x} \, dx$

in the fundamental strip $\langle 0, 1\rangle,$ and where $s = \sigma + it\,$ .

We use the Cauchy residue theorem with

$g(z) = \frac{z^{s-1}}{1+z}$

and the keyhole contour shown at right. The simple pole at $z = - 1$ is shown in blue. The contour consists of four segments, a small circle of radius r, a large one of radius R and two line segments. We choose the branch of the logarithm that has a branch cut along the positive real line, with a branch point at zero. The cut is shown in red in the diagram. The range of the argument of $\log\, z$ is from zero (inclusive) to $2π$ .

So now let's consider the contribution to the contour integral made by each segment.

1. The first segment, denoted by $\Gamma_1\,$ , runs parallel to the positive real axis and above the cut. It starts at r and goes up to R. We will let R go to infinity and r go to zero, and let the two line segments approach the cut from above and from below. The integral along $\Gamma_1\,$ is $\varphi(s)$ (in the limit).

2. The second line segment, denoted $\Gamma_3\,$ , although it runs arbitrarily close to $\Gamma_1\,$ (albeit in the opposite direction) turns out to contribute a different amount to the integral than $\Gamma_1\,$ - in fact a multiple of $\varphi(s)$ in the limit. This can be shown by starting with

$g(z) = \frac{z^{s-1}}{1+z} = \frac{\exp(\log(z) (s-1))}{1+z}$

Because $z$ is complex we use the equation for complex logarithms

$\log(z) = \log \left|z \right| + i\arg(z)$

where $arg(z)$ is the angle subtended by z when expressed in polar coordinates and since $\Gamma_3\,$ runs under the real line

$\,\arg(z) = 2\pi$ ,

So now $g (z)$ along $\Gamma_3\,$ becomes

$\exp(2\pi i (s-1)) \, \frac{\exp(\log (|z|) (s-1))}{1+z} \, = \, \exp(2\pi i s) \, \frac{\exp(\log (|z|) (s-1))}{1+z},$

so that the integral is $\,- \exp(2\pi i s) \varphi(s).$

3. The integral along the big circle, denoted $\Gamma_2\,$ , is evaluated with the ML inequality, which states that

$\left| \int_C g(z) dz \right| \le ML$

where $M$ is the maximum modulus of $g (z)$ on the curve $C$ and $L$ is the length of $C .$

On $\Gamma_2\,$ we have

$\left| z^{s-1} \right| = \left| \exp( \log(z) (s-1) ) \right| = \left| \exp( (\log(R) + i \theta) (s-1) ) \right|,$

where $0 \le \theta < 2\pi,$ according to the branch of the logarithm we are using. Hence

$\left| z^{s-1} \right| = R^{\sigma-1} \exp(-t\theta) \le R^{\sigma-1} \exp(2\pi|t|).$

This gives the bound

$\left| \int_{\Gamma_2} g(z) dz \right| \le 2\pi R \frac{R^{\sigma-1}}{R-1} \exp(2\pi|t|) = 2\pi \frac{R^\sigma}{R-1} \exp(2\pi|t|).$

We know that $\sigma <1\,$ , because s lies in the fundamental strip. Hence the integral vanishes as R goes to infinity.

4. The integral along the small circle, denoted $\Gamma_4\,$ , is also evaluated with the ML inequality, giving the upper bound

$\left| \int_{\Gamma_4} g(z) dz \right| \le 2\pi r \frac{r^{\sigma-1}}{1-r} \exp(2\pi|t|) = 2\pi \frac{r^\sigma}{1-r} \exp(2\pi|t|).$

We have $\sigma >0\,$ because s lies in the fundamental strip, and hence this integral vanishes also, as r goes to zero.

The residue of $g (z)$ at the simple pole at $z = - 1$ is

$\lim_{z \rightarrow -1} (z+1) \frac{z^{s-1}}{1+z} = (-1)^{s-1} = \exp(\pi i (s-1)) = - \exp (\pi i s).$

Putting all of the segment integrals together, the Cauchy residue theorem yields

$\left(\int_{\Gamma_1} + \int_{\Gamma_2} + \int_{\Gamma_3} + \int_{\Gamma_4} \right) g(z) dz = - 2\pi i \exp (\pi i s),$

so that in the limit,

$\varphi(s) \, (1 - \exp(2\pi i s)) = - 2\pi i \exp (\pi i s)$

or

$\varphi(s) = \pi \frac{2 i \exp (\pi i s)}{\exp(2\pi i s) - 1} = \pi \frac{2 i}{\exp(\pi i s) - \exp(-\pi i s)} = \frac{\pi}{\sin \pi s}, \quad \mbox{QED}.$

This integral was discussed on the newsgroup es.ciencia.matematicas, where an image of the contour used above under the exponential map was used and the article is here.

[edit] 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 x^s 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 $d x$ , which is translation invariant, so that $d (x + a) = d x$ .

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

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

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.

[edit] Cahen-Mellin integral

For $c > 0$ , $\Re(y)>0$ and $y - s$ on the principal branch, one has

$e^{-y}= \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(s) y^{-s}\;ds$

where $Γ(s)$ is the gamma function. This integral is known as the Cahen-Mellin integral^[1].

[edit] Unitarity

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

$\tilde\mathcal{M}\colon L^2(0,\infty)\to L^2(-\infty,\infty), \{\tilde\mathcal{M}f\}(s) := \frac{1}{\sqrt{2\pi}}\int_0^{\infty} x^{-\frac{1}{2}+is} f(x)\,dx.$

This operator is usually denoted by just plain $\mathcal{M}$ and called the "Mellin transform", but $\tilde\mathcal{M}$ is used here to distinguish from the definition used elsewhere in this article. In other words we have set

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

The Mellin inversion theorem then shows that $\tilde\mathcal{M}$ is invertible with inverse

$\tilde\mathcal{M}^{-1}\colon L^2(-\infty,\infty) \to L^2(0,\infty), \{\tilde\mathcal{M}^{-1}\varphi\}(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} x^{-\frac{1}{2}-is} \varphi(s)\,ds.$

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

[edit] Applications

The Mellin Transform is widely used in computer science 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. 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 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.

[edit] Examples

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.

[edit] Notes

^ G.H. Hardy and J.E. Littlewood, "Contributions to the Theory of the Riemann Zeta-Function and the Theory of the Distribution of Primes", Acta Mathematica, 41(1916) pp.119-196. (See notes therein for further references to Cahen's and Mellin's work, including Cahen's thesis.)

[edit] References

Paris, R. B., and Kaminsky, D., Asymptotics and Mellin-Barnes Integrals, Cambridge University Press, 2001.
A. D. Polyanin and A. V. Manzhirov, Handbook of Integral Equations, CRC Press, Boca Raton, 1998. ISBN 0-8493-2876-4
P. Flajolet, X. Gourdon, P. Dumas, Mellin transforms and asymptotics: Harmonic sums, Theoretical Computer Science, 144(1-2):3-58, June 1995
Tables of Integral Transforms at EqWorld: The World of Mathematical Equations.
Weisstein, Eric W., "Mellin Transform" from MathWorld.

[edit] External links

Marko Riedel, Applications of the Mellin-Perron Formula in Number Theory.
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).

[0] G.H. Hardy and J.E. Littlewood, "Contributions to the Theory of the Riemann Zeta-Function and the Theory of the Distribution of Primes", Acta Mathematica, 41(1916) pp.119-196. (See notes therein for further references to Cahen's and Mellin's work, including Cahen's thesis.)

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