Suppose $p_t(x)=\frac1{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}$ is the Gaussian density. $p_t(x)$ is also the Green kernel of the heat equation in 1D: $(\partial_t-\frac12\Delta)u=0$. The Fourier transform of $p_t(x)$ is $\hat p_t(\xi)=e^{-\frac{t}2|\xi|^2}$. Then the Laplace transform at 1 has the special bound $$\int_0^\infty e^{-t}|\hat p_t(\xi)|^2dt=\int_0^\infty e^{-t}e^{-t|\xi|^2}dt\lesssim \frac1{1+|\xi|^2}. $$ In this bound, we see that this integration decay as $|\xi|^{-2}$ when $|\xi|\to\infty$. Note also that the Laplace transform is closely related to resolvent of the operator $\Delta$.
Suppose now that we replace $\Delta$ by a "nice" (symmetric) operator $A$ with symbol $\psi(\xi)$. What are the characteristics of $A$ so that we have a similar bound $$\int_0^\infty e^{-\lambda t}|e^{-t\psi(\xi)}|^2dt\lesssim \frac1{1+|\xi|^a} $$ for some $\lambda>0$ and $a>0$. Is there a method to identify the decay rate $a$?
PS: I know this is probably well known in semigroup theory, but I couldn't find any exact reference.