23 Reformatted, added important example of transform pair

Two equations that encapsulate the properties of the Fourier and Mellin transforms:

$\int^{\infty}_{-\infty}{exp(2$\int^{\infty}_{-\infty}{\exp(2 \pi ifx)exp(-2 ifx)\exp(-2 \pi ify)df} = \delta(x-y)$delta(x-y)$$\frac{1}{2\pi \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(ln(x)-ln(y))= delta(\ln(x)-\ln(y))= y \delta(x-y). delta(x-y).$$ The transformations from one equation to the other are obvious. The delta function results are intuitive and an extrapolation of the discrete case for the orthogonality relationships of the characters of character groups. The transform pairs, Plancherel and convolution theorems, and other relations are easy to derive from these two. (Note that whereas$e^{sz}$is an eigenfunction for$d/dz$and so the Laplace/Fourier transforms are appropriate for devising an operator calculus for$f(d/dz)$,$z^s$is an eigenfunction of$zd/dz$and so the Mellin transform is more appropriate for$f(zd/dz)$.) Ramanujan's Master Formula/Theorem (see Wikipedia, particularly the Hardy refer.) gives a somewhat intuitive perspective on the Mellin transform as providing an "interpolation" of the coefficients of the Taylor series of certain classes of functions, as discussed in the intro of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. E.g.,$\int^{\infty}_{0}{f(x) x^{s-1}/(s-1)! $\int^{\infty}_{0}f(x)\frac{x^{s-1}}{(s-1)!} dx } = g(-s)$ g(-s)$$and \frac{1}{2\pi \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} (\pi/sin(\pi \frac{\pi}{\sin(\pi s)) } g(-s) x^{-s}/(-s)! \frac{x^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(n) (-x)^{n}/n! \frac{(-x)^{n}}{n!} = f(x) f(x)$$

for the transform pairs

$f(x)=exp(-x)$ f(x)=\exp(-x)$and$g(-s)= 1(\sigma>0)$and$f(x)=1/(1+x)$f(x)=\frac{1}{1+x}$ and $g(-s)= (-s)!$ $(0<\sigma<1$ and $abs(x)<1)$.<1)f(x)=\exp(-x^2)$and$g(-s)= \cos(\pi\frac{ s}{2})\frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} (\sigma>0)$. (Note the appearance again of the Dirac delta fct. and its derivatives as$x^{-n-1}/(-n-1)!$.)\frac{x^{-n-1}}{(-n-1)!}$.)

A simple way to derive the formulas in your question is by looking at the inverse Mellin transform rep of the Dirac delta function. See my short note on the Inverse Mellin Transform and the Dirac Delta Function. See also some applications in Dirac's Delta Function and Riemann's Jump Function J(x) for the Primes and The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.

Edwards in Riemann's Zeta Function in Ch. 10 Fourier Analysis Sec. 10.1 Invariant Operators on R+ and Their Transforms gives a nice, more group-theoretic intro to the Mellin transform in line with other comments in this stream.

22 Fixed index

Two equations that encapsulate the properties of the Fourier and Mellin transforms:

$\int^{\infty}_{-\infty}{exp(2 \pi ifx)exp(-2 \pi ify)df} = \delta(x-y)$

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(ln(x)-ln(y))= y \delta(x-y)$.

The transformations from one equation to the other are obvious. The delta function results are intuitive and an extrapolation of the discrete case for the orthogonality relationships of the characters of character groups. The transform pairs, Plancherel and convolution theorems, and other relations are easy to derive from these two.

(Note that whereas $e^{sz}$ is an eigenfunction for $d/dz$ and so the Laplace/Fourier transforms are appropriate for devising an operator calculus for $f(d/dz)$, $z^s$ is an eigenfunction of $zd/dz$ and so the Mellin transform is more appropriate for $f(zd/dz)$.)

Ramanujan's Master Formula/Theorem (see Wikipedia, particularly the Hardy refer.) gives a somewhat intuitive perspective on the Mellin transform as providing an "interpolation" of the coefficients of the Taylor series of certain classes of functions, as discussed in the intro of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. E.g.,

$\int^{\infty}_{0}{f(x)x^{s-1}/(s-1)! \int^{\infty}_{0}{f(x) x^{s-1}/(s-1)! dx} = g(-s)$ and

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} (\pi/sin(\pi s)) g(-s)x^{-s}/(-s)g(-s) x^{-s}/(-s)! ds$

$ds = \sum_{k=0}^{\infty} sum_{n=0}^{\infty} g(n) (-x)^{n}/n! = f(x)$

for the transform pairs

$f(x)=exp(-x)$ and $g(-s)= 1$ $(\sigma>0)$ and

$f(x)=1/(1+x)$ and $g(-s)= (-s)!$ $(0<\sigma<1$ and $abs(x)<1)$.

(Note the appearance again of the Dirac delta fct. and its derivatives as $x^{-n-1}/(-n-1)!$.)

A simple way to derive the formulas in your question is by looking at the inverse Mellin transform rep of the Dirac delta function. See my short note on the Inverse Mellin Transform and the Dirac Delta Function. See also some applications in Dirac's Delta Function and Riemann's Jump Function J(x) for the Primes and The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.

Edwards in Riemann's Zeta Function in Ch. 10 Fourier Analysis Sec. 10.1 Invariant Operators on R+ and Their Transforms gives a nice, more group-theoretic intro to the Mellin transform in line with other comments in this stream.

Two equations that encapsulate the properties of the Fourier and Mellin transforms:

$\int^{\infty}_{-\infty}{exp(2 \pi ifx)exp(-2 \pi ify)df} = \delta(x-y)$

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(ln(x)-ln(y))= y \delta(x-y)$.

The transformations from one equation to the other are obvious. The delta function results are intuitive and an extrapolation of the discrete case for the orthogonality relationships of the characters of character groups. The transform pairs, Plancherel and convolution theorems, and other relations are easy to derive from these two.

(Note that whereas $e^{sz}$ is an eigenfunction for $d/dz$ and so the Laplace/Fourier transforms are appropriate for devising an operator calculus for $f(d/dz)$, $z^s$ is an eigenfunction of $zd/dz$ and so the Mellin transform is more appropriate for $f(zd/dz)$.)

Ramanujan's Master Formula/Theorem (see Wikipedia, particularly the Hardy refer.) gives a somewhat intuitive perspective on the Mellin transform as providing an "interpolation" of the coefficients of the Taylor series of certain classes of functions, as discussed in the intro of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. E.g.,

$\int^{\infty}_{0}{f(x)x^{s-1}/(s-1)! dx} = g(-s)$ and

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} (\pi/sin(\pi s)) g(-s)x^{-s}/(-s)! ds$

$= \sum_{k=0}^{\infty} g(n) (-x)^{n}/n! = f(x)$

for the transform pairs

$f(x)=exp(-x)$ and $g(-s)= 1$ $(\sigma>0)$ and

$f(x)=1/(1+x)$ and $g(-s)= (-s)!$ $(0<\sigma<1$ and $abs(x)<1)$.

(Note the appearance again of the Dirac delta fct. and its derivatives as $x^{-n-1}/(-n-1)!$.)

A simple way to derive the formulas in your question is by looking at the inverse Mellin transform rep of the Dirac delta function. See my short note on the Inverse Mellin Transform and the Dirac Delta Function. See also some applications in Dirac's Delta Function and Riemann's Jump Function J(x) for the Primes and The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.

Edwards in Riemann's Zeta Function in Ch. 10 Fourier Analysis Sec. 10.1 Invariant Operators on R+ and Their Transforms gives a nice, more group-theoretic intro to the Mellin transform in line with other comments in this stream.

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