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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.,

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

$$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} g(-s) \frac{x^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(n) \frac{(-x)^{n}}{n!} = f(x)$$

for the transform pairs

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

$f(x)=\frac{1}{1+x}$ and $g(-s)= (-s)!$ $(0<\sigma<1$ and $abs(x)<1)$

$f(x)=\exp(-x^2)$ and $g(-s)= \cos(\pi\frac{ s}{2})\frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} $ $(\sigma>0)$.

From a similar perspective, the iconic Euler (Mellin) integral for the gamma function for $Real(s) > 0$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-t\;p} \; dt = p^{-s}$$

provides the scaffolding for understanding and utilizing the interplay among the Mellin transform, its inverse, operator calculus, and interpolation.

A natural interpolation of the derivative as the fractional integroderivative of fractional calculus is obtained by using the Mellin transform to interpolate the op coefficients of the op e.g.f. $\displaystyle e^{tD_x} \;,$ i.e., the shift op, for the integer powers of the derivative:

$$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; dt \; H(x) g(x) = D_x^{-s} H(x) g(x) = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; H(x) g(x)\; dt$$

$$\displaystyle = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; g(x-t) dt \; . $$

Then specifically acting on the power function for $\displaystyle \alpha > -1$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; (x-t)^\alpha dt = \int_0^x \frac{t^{s-1}}{(s-1)!} \; (x-t)^\alpha \; dt $$

$$\displaystyle = \int_0^x \frac{t^{s-1}}{(s-1)!} \; \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \frac{\alpha!}{(\alpha-k)} \; \frac{t^k}{k!} \; dt = \frac{1}{(s-1)!} \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \binom{\alpha}{k} \; \frac{t^{s+k}}{s+k} \; |_{t=0}^{x}$$

$$\displaystyle = x^{\alpha + s} \; (-s)! \; \sum_{k \ge 0} \; \binom{\alpha}{k} \; \frac{sin(\pi (s+k))}{\pi (s+k)} = x^{\alpha +s} \frac{\alpha!}{(\alpha+s)!} \; = D_x^{-s} x^\alpha \; .$$

The last summation converges with no restriction on $s$. So, we see that the Mellin transform does indeed interpolate the coefficients of the e.g.f. generated by the binomial theorem expansion $\displaystyle x^{\alpha-k} \frac{\alpha!}{(\alpha-k)}$ to $\displaystyle x^{\alpha+s} \frac{\alpha!}{(\alpha+s)}$ to give an interpolation of the coefficients of the shift op $ D_x^k$ to $ D_x^{-s}$ consistent with fractional calculus.

The same method can be used to interpolate

$$\displaystyle (x \; D_x \;x)^n = x^n D_x^n x^n = x^n \; n!\; L_n(-:xD_x:) , $$

where $ L$ denotes the Laguerre polynomials and $(:xD_x:)^k = x^kD_x^k$ by definition, leading to

$$ \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-txD_xx} \; dt \; H(x) x^\alpha = (xD_xx)^{-s}\; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; \frac{x^\alpha}{(1+xt)^{\alpha+1}} \; dt = x^{\alpha-s} \frac{(\alpha-s)!}{\alpha!} = x^{-s} D_x^{-s} x^{-s} \; x^\alpha $$

for $ 0 < Real(s) < \alpha +1 \; .$

Or, give the analytic continuation for a Mellin transform related to a class of differential operators encompassing the Witt Lie algebra:

$$ (x^{1+y}D_x)^{-s} \; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H[\frac{x}{(1+y\;t\;x^y)^{1/y}}] \frac{x^\alpha}{(1+y\;t\;x^y)^{\alpha/y}} \; dt $$

$$= H(y) \; x^{\alpha-sy} y^{-s} \frac{(-s+\alpha/y-1)!}{(\alpha/y-1)!} \;+ \; H(-y) \; x^{\alpha+s|y|} |y|^{-s} \frac{(\alpha/|y|)!}{(\alpha/|y|+s)!} \;.$$

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.,

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

$$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} g(-s) \frac{x^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(n) \frac{(-x)^{n}}{n!} = f(x)$$

for the transform pairs

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

$f(x)=\frac{1}{1+x}$ and $g(-s)= (-s)!$ $(0<\sigma<1$ and $abs(x)<1)$

$f(x)=\exp(-x^2)$ and $g(-s)= \cos(\pi\frac{ s}{2})\frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} $ $(\sigma>0)$.

From a similar perspective, the iconic Euler (Mellin) integral for the gamma function for $Real(s) > 0$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-t\;p} \; dt = p^{-s}$$

provides the scaffolding for understanding and utilizing the interplay among the Mellin transform, its inverse, operator calculus, and interpolation.

A natural interpolation of the derivative as the fractional integroderivative of fractional calculus is obtained by using the Mellin transform to interpolate the op coefficients of the op e.g.f. $\displaystyle e^{tD_x} \;,$ i.e., the shift op, for the integer powers of the derivative:

$$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; dt \; H(x) g(x) = D_x^{-s} H(x) g(x) = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; H(x) g(x)\; dt$$

$$\displaystyle = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; g(x-t) dt \; . $$

Then specifically acting on the power function for $\displaystyle \alpha > -1$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; (x-t)^\alpha dt = \int_0^x \frac{t^{s-1}}{(s-1)!} \; (x-t)^\alpha \; dt $$

$$\displaystyle = \int_0^x \frac{t^{s-1}}{(s-1)!} \; \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \frac{\alpha!}{(\alpha-k)} \; \frac{t^k}{k!} \; dt = \frac{1}{(s-1)!} \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \binom{\alpha}{k} \; \frac{t^{s+k}}{s+k} \; |_{t=0}^{x}$$

$$\displaystyle = x^{\alpha + s} \; (-s)! \; \sum_{k \ge 0} \; \binom{\alpha}{k} \; \frac{sin(\pi (s+k))}{\pi (s+k)} = x^{\alpha +s} \frac{\alpha!}{(\alpha+s)!} \; = D_x^{-s} x^\alpha \; .$$

The last summation converges with no restriction on $s$. So, we see that the Mellin transform does indeed interpolate the coefficients of the e.g.f. generated by the binomial theorem expansion $\displaystyle x^{\alpha-k} \frac{\alpha!}{(\alpha-k)}$ to $\displaystyle x^{\alpha+s} \frac{\alpha!}{(\alpha+s)}$ to give an interpolation of the coefficients of the shift op $ D_x^k$ to $ D_x^{-s}$ consistent with fractional calculus.

The same method can be used to interpolate

$$\displaystyle (x \; D_x \;x)^n = x^n D_x^n x^n = x^n \; n!\; L_n(-:xD_x:) , $$

where $ L$ denotes the Laguerre polynomials and $(:xD_x:)^k = x^kD_x^k$ by definition, leading to

$$ \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-txD_xx} \; dt \; H(x) x^\alpha = (xD_xx)^{-s}\; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; \frac{x^\alpha}{(1+xt)^{\alpha+1}} \; dt = x^{\alpha-s} \frac{(\alpha-s)!}{\alpha!} = x^{-s} D_x^{-s} x^{-s} \; x^\alpha $$

for $ 0 < Real(s) < \alpha +1 \; .$

Or, give the analytic continuation for a Mellin transform related to a class of differential operators encompassing the Witt Lie algebra:

$$ (x^{1+y}D_x)^{-s} \; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H[\frac{x}{(1+y\;t\;x^y)^{1/y}}] \frac{x^\alpha}{(1+y\;t\;x^y)^{\alpha/y}} \; dt $$

$$= H(y) \; x^{\alpha-sy} y^{-s} \frac{(-s+\alpha/y-1)!}{(\alpha/y-1)!} \;+ \; H(-y) \; x^{\alpha+s|y|} |y|^{-s} \frac{(\alpha/|y|)!}{(\alpha/|y|+s)!} \;.$$

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.,

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

$$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} g(-s) \frac{x^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(n) \frac{(-x)^{n}}{n!} = f(x)$$

for the transform pairs

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

$f(x)=\frac{1}{1+x}$ and $g(-s)= (-s)!$ $(0<\sigma<1$ and $abs(x)<1)$

$f(x)=\exp(-x^2)$ and $g(-s)= \cos(\pi\frac{ s}{2})\frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} $ $(\sigma>0)$.

From a similar perspective, the iconic Euler (Mellin) integral for the gamma function for $Real(s) > 0$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-t\;p} \; dt = p^{-s}$$

provides the scaffolding for understanding and utilizing the interplay among the Mellin transform, its inverse, operator calculus, and interpolation.

A natural interpolation of the derivative as the fractional integroderivative of fractional calculus is obtained by using the Mellin transform to interpolate the op coefficients of the op e.g.f. $\displaystyle e^{tD_x} \;,$ i.e., the shift op, for the integer powers of the derivative:

$$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; dt \; H(x) g(x) = D_x^{-s} H(x) g(x) = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; H(x) g(x)\; dt$$

$$\displaystyle = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; g(x-t) dt \; . $$

Then specifically acting on the power function for $\displaystyle \alpha > -1$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; (x-t)^\alpha dt = \int_0^x \frac{t^{s-1}}{(s-1)!} \; (x-t)^\alpha \; dt $$

$$\displaystyle = \int_0^x \frac{t^{s-1}}{(s-1)!} \; \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \frac{\alpha!}{(\alpha-k)} \; \frac{t^k}{k!} \; dt = \frac{1}{(s-1)!} \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \binom{\alpha}{k} \; \frac{t^{s+k}}{s+k} \; |_{t=0}^{x}$$

$$\displaystyle = x^{\alpha + s} \; (-s)! \; \sum_{k \ge 0} \; \binom{\alpha}{k} \; \frac{sin(\pi (s+k))}{\pi (s+k)} = x^{\alpha +s} \frac{\alpha!}{(\alpha+s)!} \; = D_x^{-s} x^\alpha \; .$$

The last summation converges with no restriction on $s$. So, we see that the Mellin transform does indeed interpolate the coefficients of the e.g.f. generated by the binomial theorem expansion $\displaystyle x^{\alpha-k} \frac{\alpha!}{(\alpha-k)}$ to $\displaystyle x^{\alpha+s} \frac{\alpha!}{(\alpha+s)}$ to give an interpolation of the coefficients of the shift op $ D_x^k$ to $ D_x^{-s}$ consistent with fractional calculus.

The same method can be used to interpolate

$$\displaystyle (x \; D_x \;x)^n = x^n D_x^n x^n = x^n \; n!\; L_n(-:xD_x:) , $$

where $ L$ denotes the Laguerre polynomials and $(:xD_x:)^k = x^kD_x^k$ by definition, leading to

$$ \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-txD_xx} \; dt \; H(x) x^\alpha = (xD_xx)^{-s}\; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; \frac{x^\alpha}{(1+xt)^{\alpha+1}} \; dt = x^{\alpha-s} \frac{(\alpha-s)!}{\alpha!} = x^{-s} D_x^{-s} x^{-s} \; x^\alpha $$

for $ 0 < Real(s) < \alpha +1 \; .$

Or, give the analytic continuation for a Mellin transform related to a class of differential operators encompassing the Witt Lie algebra:

$$ (x^{1+y}D_x)^{-s} \; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H[\frac{x}{(1+y\;t\;x^y)^{1/y}}] \frac{x^\alpha}{(1+y\;t\;x^y)^{\alpha/y}} \; dt $$

$$= H(y) \; x^{\alpha-sy} y^{-s} \frac{(-s+\alpha/y-1)!}{(\alpha/y-1)!} \;+ \; H(-y) \; x^{\alpha+s|y|} |y|^{-s} \frac{(\alpha/|y|)!}{(\alpha/|y|+s)!} \;.$$

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.,

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

$$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} g(-s) \frac{x^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(n) \frac{(-x)^{n}}{n!} = f(x)$$

for the transform pairs

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

$f(x)=\frac{1}{1+x}$ and $g(-s)= (-s)!$ $(0<\sigma<1$ and $abs(x)<1)$

$f(x)=\exp(-x^2)$ and $g(-s)= \cos(\pi\frac{ s}{2})\frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} $ $(\sigma>0)$.

From a similar perspective, the iconic Euler (Mellin) integral for the gamma function for $Real(s) > 0$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-t\;p} \; dt = p^{-s}$$

provides the scaffolding for understanding and utilizing the interplay among the Mellin transform, its inverse, operator calculus, and interpolation.

A natural interpolation of the derivative as the fractional integroderivative of fractional calculus is obtained by using the Mellin transform to interpolate the op coefficients of the op e.g.f. $\displaystyle e^{tD_x} \;,$ i.e., the shift op, for the integer powers of the derivative:

$$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; dt \; H(x) g(x) = D_x^{-s} H(x) g(x) = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; H(x) g(x)\; dt$$

$$\displaystyle = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; g(x-t) dt \; . $$

Then specifically acting on the power function for $\displaystyle \alpha > -1$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; (x-t)^\alpha dt = \int_0^x \frac{t^{s-1}}{(s-1)!} \; (x-t)^\alpha \; dt $$

$$\displaystyle = \int_0^x \frac{t^{s-1}}{(s-1)!} \; \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \frac{\alpha!}{(\alpha-k)} \; \frac{t^k}{k!} \; dt = \frac{1}{(s-1)!} \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \binom{\alpha}{k} \; \frac{t^{s+k}}{s+k} \; |_{t=0}^{x}$$

$$\displaystyle = x^{\alpha + s} \; (-s)! \; \sum_{k \ge 0} \; \binom{\alpha}{k} \; \frac{sin(\pi (s+k))}{\pi (s+k)} = x^{\alpha +s} \frac{\alpha!}{(\alpha+s)!} \; = D_x^{-s} x^\alpha \; .$$

The last summation converges with no restriction on $s$. So, we see that the Mellin transform does indeed interpolate the coefficients of the e.g.f. generated by the binomial theorem expansion $\displaystyle x^{\alpha-k} \frac{\alpha!}{(\alpha-k)}$ to $\displaystyle x^{\alpha+s} \frac{\alpha!}{(\alpha+s)}$ to give an interpolation of the coefficients of the shift op $ D_x^k$ to $ D_x^{-s}$ consistent with fractional calculus.

The same method can be used to interpolate

$$\displaystyle (x \; D_x \;x)^n = x^n D_x^n x^n = x^n \; n!\; L_n(-:xD_x:) , $$

where $ L$ denotes the Laguerre polynomials and $(:xD_x:)^k = x^kD_x^k$ by definition, leading to

$$ \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-txD_xx} \; dt \; H(x) x^\alpha = (xD_xx)^{-s}\; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; \frac{x^\alpha}{(1+xt)^{\alpha+1}} \; dt = x^{\alpha-s} \frac{(\alpha-s)!}{\alpha!} = x^{-s} D_x^{-s} x^{-s} \; x^\alpha $$

for $ 0 < Real(s) < \alpha +1 \; .$

Or, give the analytic continuation for a Mellin transform related to a class of differential operators encompassing the Witt Lie algebra:

$$ (x^{1+y}D_x)^{-s} \; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H[\frac{x}{(1+y\;t\;x^y)^{1/y}}] \frac{x^\alpha}{(1+y\;t\;x^y)^{\alpha/y}} \; dt $$

$$= H(y) \; x^{\alpha-sy} y^{-s} \frac{(-s+\alpha/y-1)!}{(\alpha/y-1)!} \;+ \; H(-y) \; x^{\alpha+s|y|} |y|^{-s} \frac{(\alpha/|y|)!}{(\alpha/|y|+s)!} \;.$$

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.,

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

$$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} g(-s) \frac{x^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(n) \frac{(-x)^{n}}{n!} = f(x)$$

for the transform pairs

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

$f(x)=\frac{1}{1+x}$ and $g(-s)= (-s)!$ $(0<\sigma<1$ and $abs(x)<1)$

$f(x)=\exp(-x^2)$ and $g(-s)= \cos(\pi\frac{ s}{2})\frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} $ $(\sigma>0)$.

From a similar perspective, the iconic Euler (Mellin) integral for the gamma function for $Real(s) > 0$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-t\;p} \; dt = p^{-s}$$

provides the scaffolding for understanding and utilizing the interplay among the Mellin transform, its inverse, operator calculus, and interpolation.

A natural interpolation of the derivative as the fractional integroderivative of fractional calculus is obtained by using the Mellin transform to interpolate the op coefficients of the op e.g.f. $\displaystyle e^{tD_x} \;,$ i.e., the shift op, for the integer powers of the derivative:

$$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; dt \; H(x) g(x) = D_x^{-s} H(x) g(x) = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; H(x) g(x)\; dt$$

$$\displaystyle = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; g(x-t) dt \; . $$

Then specifically acting on the power function for $\displaystyle \alpha > -1$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; (x-t)^\alpha dt = \int_0^x \frac{t^{s-1}}{(s-1)!} \; (x-t)^\alpha \; dt $$

$$\displaystyle = \int_0^x \frac{t^{s-1}}{(s-1)!} \; \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \frac{\alpha!}{(\alpha-k)} \; \frac{t^k}{k!} \; dt = \frac{1}{(s-1)!} \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \binom{\alpha}{k} \; \frac{t^{s+k}}{s+k} \; |_{t=0}^{x}$$

$$\displaystyle = x^{\alpha + s} \; (-s)! \; \sum_{k \ge 0} \; \binom{\alpha}{k} \; \frac{sin(\pi (s+k))}{\pi (s+k)} = x^{\alpha +s} \frac{\alpha!}{(\alpha+s)!} \; = D_x^{-s} x^\alpha \; .$$

The last summation converges with no restriction on $s$. So, we see that the Mellin transform does indeed interpolate the coefficients of the e.g.f. generated by the binomial theorem expansion $\displaystyle x^{\alpha-k} \frac{\alpha!}{(\alpha-k)}$ to $\displaystyle x^{\alpha+s} \frac{\alpha!}{(\alpha+s)}$ to give an interpolation of the coefficients of the shift op $ D_x^k$ to $ D_x^{-s}$ consistent with fractional calculus.

The same method can be used to interpolate

$$\displaystyle (x \; D_x \;x)^n = x^n D_x^n x^n = x^n \; n!\; L_n(-:xD_x:) , $$

where $ L$ denotes the Laguerre polynomials and $(:xD_x:)^k = x^kD_x^k$ by definition, leading to

$$ \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-txD_xx} \; dt \; H(x) x^\alpha = (xD_xx)^{-s}\; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; \frac{x^\alpha}{(1+xt)^{\alpha+1}} \; dt = x^{\alpha-s} \frac{(\alpha-s)!}{\alpha!} = x^{-s} D_x^{-s} x^{-s} \; x^\alpha $$

for $ 0 < Real(s) < \alpha +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.,

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

$$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} g(-s) \frac{x^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(n) \frac{(-x)^{n}}{n!} = f(x)$$

for the transform pairs

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

$f(x)=\frac{1}{1+x}$ and $g(-s)= (-s)!$ $(0<\sigma<1$ and $abs(x)<1)$

$f(x)=\exp(-x^2)$ and $g(-s)= \cos(\pi\frac{ s}{2})\frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} $ $(\sigma>0)$.

From a similar perspective, the iconic Euler (Mellin) integral for the gamma function for $Real(s) > 0$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-t\;p} \; dt = p^{-s}$$

provides the scaffolding for understanding and utilizing the interplay among the Mellin transform, its inverse, operator calculus, and interpolation.

A natural interpolation of the derivative as the fractional integroderivative of fractional calculus is obtained by using the Mellin transform to interpolate the op coefficients of the op e.g.f. $\displaystyle e^{tD_x} \;,$ i.e., the shift op, for the integer powers of the derivative:

$$\displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; dt \; H(x) g(x) = D_x^{-s} H(x) g(x) = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; H(x) g(x)\; dt$$

$$\displaystyle = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; g(x-t) dt \; . $$

Then specifically acting on the power function for $\displaystyle \alpha > -1$

$$ \displaystyle \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; (x-t)^\alpha dt = \int_0^x \frac{t^{s-1}}{(s-1)!} \; (x-t)^\alpha \; dt $$

$$\displaystyle = \int_0^x \frac{t^{s-1}}{(s-1)!} \; \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \frac{\alpha!}{(\alpha-k)} \; \frac{t^k}{k!} \; dt = \frac{1}{(s-1)!} \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \binom{\alpha}{k} \; \frac{t^{s+k}}{s+k} \; |_{t=0}^{x}$$

$$\displaystyle = x^{\alpha + s} \; (-s)! \; \sum_{k \ge 0} \; \binom{\alpha}{k} \; \frac{sin(\pi (s+k))}{\pi (s+k)} = x^{\alpha +s} \frac{\alpha!}{(\alpha+s)!} \; = D_x^{-s} x^\alpha \; .$$

The last summation converges with no restriction on $s$. So, we see that the Mellin transform does indeed interpolate the coefficients of the e.g.f. generated by the binomial theorem expansion $\displaystyle x^{\alpha-k} \frac{\alpha!}{(\alpha-k)}$ to $\displaystyle x^{\alpha+s} \frac{\alpha!}{(\alpha+s)}$ to give an interpolation of the coefficients of the shift op $ D_x^k$ to $ D_x^{-s}$ consistent with fractional calculus.

The same method can be used to interpolate

$$\displaystyle (x \; D_x \;x)^n = x^n D_x^n x^n = x^n \; n!\; L_n(-:xD_x:) , $$

where $ L$ denotes the Laguerre polynomials and $(:xD_x:)^k = x^kD_x^k$ by definition, leading to

$$ \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-txD_xx} \; dt \; H(x) x^\alpha = (xD_xx)^{-s}\; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; \frac{x^\alpha}{(1+xt)^{\alpha+1}} \; dt = x^{\alpha-s} \frac{(\alpha-s)!}{\alpha!} = x^{-s} D_x^{-s} x^{-s} \; x^\alpha $$

for $ 0 < Real(s) < \alpha +1 \; .$

Or, give the analytic continuation for a Mellin transform related to a class of differential operators encompassing the Witt Lie algebra:

$$ (x^{1+y}D_x)^{-s} \; x^\alpha = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H[\frac{x}{(1+y\;t\;x^y)^{1/y}}] \frac{x^\alpha}{(1+y\;t\;x^y)^{\alpha/y}} \; dt $$

$$= H(y) \; x^{\alpha-sy} y^{-s} \frac{(-s+\alpha/y-1)!}{(\alpha/y-1)!} \;+ \; H(-y) \; x^{\alpha+s|y|} |y|^{-s} \frac{(\alpha/|y|)!}{(\alpha/|y|+s)!} \;.$$

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.