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edited May 15 2012 at 8:46
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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.
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edited Mar 22 2012 at 4:06
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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.
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edited Mar 22 2012 at 3:49
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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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edited Mar 22 2012 at 3:30
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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 reference thereinrefer.) 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
$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} (\pi/sin(\pi s)) g(-s)x^{-s}/(-s)!) g(-s)x^{-s}/(-s)! ds$
$= \sum_{k=0}^{\infty} g(n) x^{n}/n! (-x)^{n}/n! = f(x)$
for the transform pairs
$f(x)=exp(-x) f(x)=exp(-x)$ and $g(-s)= 1 (\sigma>0)$ 1$ $(\sigma>0)$ and
$f(x)=1/(1+x) f(x)=1/(1+x)$ and $g(-s)= (-s)! (0<\sigma<1 -s)!$ $(0<\sigma<1$ and abs(x)$abs(x)<1)$.
(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 the other comments in this stream.
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edited Mar 22 2012 at 3:14
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The Two equations that encapsulate the properties of the Fourier and Mellin Transform is used 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 obtain conditions the other are obvious. The delta function results are intuitive and an extrapolation of the discrete case for the validity 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 (RMF). I think reflecting backwards from RMF to Wikipedia, particularly the Mellin Transform Hardy reference therein) gives a somewhat intuitive handle perspective on the functionality Mellin transform as providing an "interpolation" of the transformcoefficients of the Taylor series of certain classes of functions, as discussed in the introduction intro of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in 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 Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life transform pairs $f(x)=exp(-x) and work. See also references in Mathworld on RMTg(-s)= 1 (\sigma>0)$ 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. Update: 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 representation 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 the other comments in this stream. 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 $f(zd/dz)$ . Two equations which encapsulate the properties of the Fourier and Mellin transforms: $\int^{\infty}_{-\infty}{exp(2 \pi ifx)exp(-2 \pi ify)df} = \delta(x-y)$ $\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(ln(x)-ln(y))= y \delta(x-y)$.
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edited Feb 14 2012 at 3:13
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
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 $f(xd/dx)$ f(zd/dz)$ .
Two equations which encapsulate the properties of the Fourier and Mellin transforms:
$\int^{\infty}_{-\infty}{exp(2 \pi ifx)exp(-2 \pi ify)df} = \delta(x-y)$
$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(ln(x)-ln(y))= y \delta(x-y)$ delta(x-y)$.
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edited Feb 14 2012 at 3:07
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
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 $f(xd/dx)$ .
Two equations which encapsulate the properties of the Fourier and Mellin transforms:
$\int^{\infty}_{-\infty}{exp(2 \pi ifx)exp(-2 \pi ify)df} = \delta(x-y)$
$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(ln(x)-ln(y))= y \delta(x-y)$
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edited Feb 14 2012 at 3:02
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
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 $f(xd/dx)$ .
Two equations which encapsulate the properties of the Fourier and Mellin transforms:
$\int^{\infty}_{-\infty}{exp(2 \delta (x-y) $
pi ifx)exp(-2 \pi ify)df} = \delta(x-y)$
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edited Feb 14 2012 at 2:56
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
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 $f(xd/dx)$ .
Two equations which encapsulate the properties of the Fourier and Mellin transforms:
$ \delta (x-y) $
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edited Feb 13 2012 at 11:24
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
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 $f(xd/dx)$ .
For me, the two equations which encapsulate the properties of the Fourier and Mellin transforms are
$\int^{\infty}_{-\infty}{exp(\2 \pi ifx)exp(-\2 \pi ify)df} = \delta(x-y)$ and
$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(ln(x)-ln(y))= y \delta(x-y)$ .
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edited Feb 13 2012 at 11:14
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
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 $f(xd/dx)$ .
For me, the two equations which encapsulate the properties of the Fourier and Mellin transforms are
$\int^{\infty}{-\infty}{exp(\2 \int^{\infty}_{-\infty}{exp(\2 \pi ifx)exp(-\2 \pi ify)df}= ify)df} = \delta(x-y)$ and
$= \frac{1}{2\pi \frac{1}{2\pi i} \int^{c+i\infty}{c-i\infty} int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds = ds= \delta(ln(x)-ln(y))= y\delta(x-y)$y \delta(x-y)$ .
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edited Feb 13 2012 at 11:08
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
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 $f(xd/dx)$ .
For me, the two equations which encapsulate the properties of the Fourier and Mellin transforms are
$\int^{\infty}{-\infty}{exp(\2 \pi ifx)exp(-\2 \pi ify)df}= \delta(x-y)$ and
$= \frac{1}{2\pi i} \int^{c+i\infty}{c-i\infty} x^{-s} y^{s} ds = \delta(ln(x)-ln(y))= y\delta(x-y)$.
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edited Nov 30 2011 at 22:43
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
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 $f(d/dz) = g(z)$, f(d/dz)$, $z^s$ is an eigenfunction of $zd/dz$ and so the Mellin transform is more appropriate for $f(xd/dx) = g(x)$ f(xd/dx)$ .
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edited Nov 28 2011 at 2:41
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
Note that whereas $e^(sz)$ e^{sz}$ is an eigenfunction for $d/dz$ and so the Laplace/Fourier transforms are appropriate for devising an operator calculus for $f(d/dz) = g(z)$, $z^(-s)$ z^s$ is an eigenfunction of $z/dz$ zd/dz$ and so the Mellin transform is more appropriate for $f(xd/dx) = g(x)$ .
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edited Nov 28 2011 at 2:35
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
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 $f(d/dz)= f(d/dz) = g(z)$ , $z^(-s)$ is an eigenfunction of $z/dz$ and so the Mellin transform is more appropriate for $f(xd/dx)= f(xd/dx) = g(x)$ .
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edited Nov 28 2011 at 2:29
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation 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 the other comments in this stream.
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 $f(d/dz)= g(z)$, $z^(-s)$ is an eigenfunction of $z/dz$ and so the Mellin transform is more appropriate for $f(xd/dx)= g(x)$.
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edited Nov 22 2011 at 6:27
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edited Nov 22 2011 at 5:19
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation of the Dirac delta function. See my note Dirac’s Delta Function Note on the Inverse Mellin Transform and Riemann’s Jump Function J(x) for the PrimesDirac Delta Function.
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 the other comments in this stream.
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edited Nov 22 2011 at 5:11
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edited Nov 14 2011 at 1:22
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation of the Dirac delta function. See the first draft of my note on The Inverse Mellin Transform and the Dirac Delta Function.
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 the other comments in this stream.
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edited Nov 13 2011 at 19:12
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The Mellin Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
Update: A simple way to derive the formulas in your question is by looking at the inverse Mellin transform representation of the Dirac delta function. See the first draft of my note on The Mellin Transform.
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 the other comments in this stream.
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edited Nov 3 2011 at 13:44
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The Mellin transform Transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures's lectures on subjects suggested by his life and work. See also references in Mathworld on RMT.
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answered Nov 3 2011 at 13:38
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The Mellin transform is used to obtain conditions for the validity of Ramanujan's Master Formula/Theorem (RMF). I think reflecting backwards from RMF to the Mellin Transform gives a somewhat intuitive handle on the functionality of the transform, as discussed in the introduction of "Ramanujan's Master Theorem ..." by Olafsson and Pasquale. I became interested in the Mellin Transform some years ago after reading Hardy's Ramanujan: twelve lectures's on subjects suggested by his life and work. See also references in Mathworld on RMT.
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