Inverse Laplace transform of Dirac delta function

Question

I am trying to understand how to identify, or at least derive some properties of the inverse Laplace transform of the Dirac delta function, i.e. a function $\eta$ s.t. $$ \int_0^\infty dt\, \eta(\omega,t) e^{-\omega’ t} = \delta(\omega-\omega’) \,. $$

Answer 1

The answer might not be definitive and depends on the context. Even if you intend to carry real computations ultimately, the Laplace transform is assumed to be defined with respect to a complex variable or, at the very least, it may be analytically continued to the complex plane for convergence purposes. In the present case, it allows us to "cheat" a little bit, because the Dirac delta function is simply represented by the inverse function in the complex plane thanks to Cauchy's formula. Concretely, one has $\delta(z) \equiv \frac{1}{2\pi iz}$, whose inverse transform is easily found as the Heaviside function. In consequence, one can conclude that $\eta(\omega,t) = \frac{e^{\omega t}}{2\pi i}u(t)$.

Alternatively and a bit more naively, you could also apply Mellin's inverse formula directly, i.e. $$ \eta(\omega,t) = \mathscr{L}^{-1}[\delta(\omega'-\omega)](t) = \int_{\omega-i\Bbb{R}}^{\omega+i\Bbb{R}} \frac{\mathrm{d}\omega'}{2\pi i}\, \delta(\omega'-\omega)e^{\omega't} = \frac{e^{\omega t}}{2\pi i}. $$