inverse via functional calculus

Question

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!

Answer 1

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 $AH = I$.

Answer 2

This is not a proof of the fact, but you might find it more intuitive by considering what would happen with a simple number $h$ rather than a matrix $H$. In this case $$\int e^{-th} dt = -\frac{e^{-th}}{h} = -h^{-1}e^{-th}$$ Then $$\int_0^\infty e^{-th} dt = lim_{x \rightarrow \infty} \int_0^x e^{-th} dt = lim_{x \rightarrow \infty} (-h^{-1}e^{-xh} - (-h^{-1}e^{-0h})) = h^{-1}$$ So the situation with the matrix matches what would happen with a number.