Inverse of (i*I+A) for self adjoint operator

Question

If we have that $A: H \rightarrow H$ is a bounded self-adjoint operator on a Hilbert space, then the spectrum of $A$ is entirely real, i.e. $\sigma(A) \subseteq \mathbb{R}$. Hence we know that $i$ is in the resolvent set of $A$, and in particular $(iI+A)^{-1}$ exists where $I$ is the identity. Is there any way to find $(iI+A)^{-1}$ explicitly? I tried to write it in terms of a geometric series (something like $-i\sum_{n=0}^\infty (iA)^n$) like we did for showing the resolvent set is open, but that obviously won't converge unless $\|A\| < 1$. Is there any other way that the inverse can be solved for explicitly?

Answer 1

You might try something like this: $$ (iI+A)^{-1} = -i(I-iA)^{-1} = -i\int_0^{\infty}e^{-t(I-iA)}dt = -i\int_0^{\infty}e^{-t}e^{itA}dt $$ You can expand $e^{itA}=\sum_{n=0}^{\infty}\frac{1}{n!}i^nt^nA^n$ integrate using the Gamma function: $$ \Gamma(\alpha)=\int_0^{\infty}e^{-t}t^{-1+\alpha}dt,\;\;\Re\alpha > 0. $$