- #1

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- Homework Statement
- Sifting property of a Dirac delta inverse Mellin transformation

- Relevant Equations
- ##f(t) = \delta(t-a) ##

##\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} dw = f(t) = \delta(t-a)##

##\int_{-\infty}^{\infty} f(x) \delta(x) dx = f(0)##

Hi,

I have to verify the sifting property of ##\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} e^{-sa}e^{st} ds## which is the inverse Mellin transformation of the Dirac delta function ##f(t) = \delta(t-a) ##.

let ##s = iw## and ##ds = idw##

##\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} dw = f(t) = \delta(t-a)## (1)

The sifting property says that if n -- > ##\infty## then ##\int_{-\infty}^{\infty} f(x) \delta(x) dx --> f(0)##

From (1),

##2 \pi \delta(t-a) = \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} dw ##

Thus, my guess is that

##\int_{\infty}^{-\infty} f(t) \delta(t-a) dt = f(0)##

However, I don't see how to prove it.

I have to verify the sifting property of ##\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} e^{-sa}e^{st} ds## which is the inverse Mellin transformation of the Dirac delta function ##f(t) = \delta(t-a) ##.

let ##s = iw## and ##ds = idw##

##\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} dw = f(t) = \delta(t-a)## (1)

The sifting property says that if n -- > ##\infty## then ##\int_{-\infty}^{\infty} f(x) \delta(x) dx --> f(0)##

From (1),

##2 \pi \delta(t-a) = \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} dw ##

Thus, my guess is that

##\int_{\infty}^{-\infty} f(t) \delta(t-a) dt = f(0)##

However, I don't see how to prove it.