inverse via functional calculus Let $H$ be a positive definit self-adjoint operator. Can someone explain to me why $$\int_{0}^{\infty}{e^{-tH}dt}=H^{-1}?$$ Thanks in advance!

Let $A = \int_0^\infty e^{-tH}dt$. It suffices to show that $AH = HA = I$ (if $H$ is an operator over a finite-dimensional space, it suffices to show that $HA = I$) where $I$ is the identity operator.
We have
$$
HA = H \int_0^\infty e^{-tH}dt = \int_0^\infty He^{-tH}dt = \int_0^\infty \frac{d}{dt}[-e^{-tH}]\,dt\\
= e^{-(0)H} - \lim_{t \to \infty}e^{-tH} = I - O = I.
$$
We can similarly show that $AH = I$.
