Mellin transform of Dirac delta function ##\delta(t-a)##

  • #1
happyparticle
360
19
Homework Statement:
Mellin transform of Dirac delta function ##\delta(t-a)##
Relevant Equations:
##F(s)= \int_0^{\infty} \delta (t-a)e^{-st} dt##

##f(t) = \frac{1}{2 \pi i} \int_{\gamma - i \omega}^{\gamma +i \omega} F(s) e^{st} ds##
Hi,
I found Laplace transform of this Dirac delta function which is ##F(s) = e^{-st}## since ##\int_{\infty}^{-\infty} f(t) \delta (t-a) dt = f(a)##
and that ##\delta(x) = 0## if ##x \neq 0##

Then the Mellin transform
##f(t) = \frac{1}{2 \pi i} \int_{\gamma - i \omega}^{\gamma +i \omega} e^{-sa} e^{st} ds##
Since there's no singularity I can't use the residue theorem, so I'm not sure what else can I use.
 

Answers and Replies

  • #3
happyparticle
360
19
I made a mistake, if someone can edit the post I would say the inverse Mellin transform.
 
  • #4
anuttarasammyak
Gold Member
1,851
954
You use e but wiki uses x as
The Mellin transform of a function f is

{\displaystyle \left\{{\mathcal {M}}f\right\}(s)=\varphi (s)=\int _{0}^{\infty }x^{s-1}f(x)\,dx.}
Are they equivalent?
1647860532674.png

Let ##f(x)=\delta(x-a)## a>0
[tex]\{Mf\}(s)= a^{s-1}[/tex]
[tex]\{M^{-1}\phi\}(x)= \frac{1}{2\pi i a}\int_{c-i\infty}^{c+i\infty} t^{-s} ds [/tex]
where t=x/a. It should be ##\delta(x-a)##.

I observe expression of delta(x) in Fourier transform[tex]\delta(x)=\frac{1}{2\pi i} \int_{-i\infty}^{i\infty}e^{px} dp[/tex]
is useful for the proof.
 
Last edited:

Suggested for: Mellin transform of Dirac delta function ##\delta(t-a)##

Replies
31
Views
1K
Replies
1
Views
586
  • Last Post
Replies
7
Views
1K
Replies
7
Views
479
Replies
2
Views
114
  • Last Post
Replies
3
Views
419
Replies
1
Views
276
Replies
7
Views
241
Replies
9
Views
630
Top