# Laplace transform for damped waves

In Zeidler's book on QFT, page 94 there is a definition for a Laplace transform that reads

$$$$(\mathcal{L} f)(\mathcal{E}) = \int_0^\infty e^{i\mathcal{E}t/\hbar}f(t) dt,$$$$

where $$\mathcal{E}_0=E_0-i\Gamma_0=E_0-i\hbar \gamma_0$$ and $$\mathcal{E}=E-i\Gamma$$. Consider a model for a "truncated" damped wave

$$$$y(t)=A\theta(t)e^{-\gamma_0t}e^{-iw_0t},$$$$

where $$\theta(x)$$ is the Heaviside function and $$\gamma,w\in\mathbb{R}$$. According to the definition of the Laplace transform, the $$y(t)$$ becomes

$$$$A\int_0^\infty e^{i\mathcal{E}t/\hbar}e^{-i\mathcal{E}_0t}dt = A\int_0^\infty e^{-i(E_0-E)t/\hbar}e^{\gamma t-\gamma_0 t}dt.$$$$

I'm confused with this last expression. If I understand correctly, we are considering "truncated" waves such that $$\gamma_0 \leq 0$$ (this is written in page 93). Then, in order for the integral to converge we need $$\gamma>-\gamma_0$$. If this is true, the exponent $$\gamma t-\gamma_0 t$$ does not go to the zero as $$t\rightarrow \infty$$. Did I screwed up in a sign, or the definition of the Lapace transform has a wrong sign?

• Can you fix your initial definition? $\mathcal{E}_0$ appears nowhere in it. – probably_someone Sep 3 at 21:00
• Done, the missing definition was for $\mathcal{E}$. – user2820579 Sep 3 at 21:41