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

[happyparticle]

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.

 

[anuttarasammyak]

I see Wiki https://en.wikipedia.org/wiki/Mellin_transform says another formula of transformation.

 

[happyparticle]

I made a mistake, if someone can edit the post I would say the inverse Mellin transform.

 

[anuttarasammyak]

You use e but wiki uses x as
The Mellin transform of a function f is

Are they equivalent?

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

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