Using the Dirac delta function $$f(t) = \delta(t-a)$$ to get the Mellin inverse transformation $$\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} e^{-sa}e^{st} ds$$
From there I made those changes. $s = iw$ and $ds = idw$
$$\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} ds = f(t) = \delta(t-a)$$
However, I don't see how to use the sifting property to solve this integral since the sifting property says $$\int_{-\infty}^{\infty} f(x) \delta(x) dx = f(0)$$
I'm wondering if I can verify the sifting property using $$\int_{-\infty}^{\infty} f(a) \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} ds = f(0)$$ where $f(a) = e^{-sa}$, thus $f(0) = 1$