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Consider these relations for the Fourier, Mellin, and Laplace transforms:

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

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

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{-xp} e^{yp} dp=\delta(x-y)$.

For me, the delta fct. results seem intuitive (also analogous to the orthogonality relationship for characters of character groups), and the eqns. encapsulate the properties of the transforms (try deriving the transform pairs and other relations from them, e.g. Plancheral, convolution, Poisson summation) and illustrate the transformations from one transform to another.

(Tried this as a comment initially, but had formatting problems.)

Consider these relations for the Fourier, Mellin, and Laplace transforms:

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

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

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{-xp} e^{yp} dp=\delta(x-y)$.

For me, the delta fct. results seem intuitive (also analogous to the orthogonality relationship for characters of character groups), and the eqns. encapsulate the properties of the transforms (try deriving the transform pairs and other relations from them, e.g., Plancherel, convolution, Poisson summation) and illustrate the transformations from one transform to another.

(Tried this as a comment initially, but had formatting problems.)

Consider these relations for the Fourier, Mellin, and Laplace transforms:

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

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

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{-xp} e^{yp} dp=\delta(x-y)$.

For me, the delta fct. results seem intuitive (also analogous to the orthogonality relationship for characters of character groups) and they encapsulate the properties of the transforms (try deriving the transform pairs and other relations from them, e.g. Plancheral, convolution, Poisson summation) and illustrate the transformations from one transform to another.

(Tried this as a comment initially, but had formatting problems.)

Consider these relations for the Fourier, Mellin, and Laplace transforms:

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

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

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{-xp} e^{yp} dp=\delta(x-y)$.

For me, the delta fct. results seem intuitive (also analogous to the orthogonality relationship for characters of character groups), and the eqns. encapsulate the properties of the transforms (try deriving the transform pairs and other relations from them, e.g. Plancheral, convolution, Poisson summation) and illustrate the transformations from one transform to another.

(Tried this as a comment initially, but had formatting problems.)

Consider these relations for the Fourier, Mellin, and Laplace transforms:

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

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

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{-xp} e^{yp} dp=\delta(x-y)$.

For me, the delta fct. results seem intuitive (also analogous to trace for character groups) and they encapsulate the properties of the transforms (try deriving the transform pairs and other relations from them, e.g. Plancheral, convolution, Poisson summation) and illustrate the transformations from one transform to another.

(Tried this as a comment initially, but had formatting problems.)

Consider these relations for the Fourier, Mellin, and Laplace transforms:

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

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

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{-xp} e^{yp} dp=\delta(x-y)$.

For me, the delta fct. results seem intuitive (also analogous to the orthogonality relationship for characters of character groups) and they encapsulate the properties of the transforms (try deriving the transform pairs and other relations from them, e.g. Plancheral, convolution, Poisson summation) and illustrate the transformations from one transform to another.

(Tried this as a comment initially, but had formatting problems.)