Show $(A-\lambda)^{-1}=-\int_0^\infty e^{-\lambda t}e^{tA}dt$ using spectral theorem

Question

Let $A$ be a bounded self-adjoint operator. Use the spectral theorem to show

$$(A-\lambda)^{-1}=-\int_0^\infty e^{-\lambda t}e^{tA}dt,\ \ \ \mathrm{for\ }\lambda>\|A\|$$

This seems like a beautiful and therefore maybe well know result. Do you where I could look it up?

Or can you give me any hints how to show this? I don't even know how to start. Thanks!

Spectral theorem for bounded self-adjoint operators:

Let $H$ be a Hilbert space and let $A \in L_b(H)$ ($L_b$ means linear and bounded) self-adjoint. Let $\mathcal A$ be the $\sigma$-Algebra of the Borel sets on $\sigma(A)$. Then there exists a unique spectral measure $E$ on $\sigma(A).$ This is the so called spectral measure of $A$ or the spectral decomposition of $A$, such that $$A=\int(t\mapsto t)dE=:\int t dE(t)$$ For a operator $T \in L_b(H)$ the following statements are equivalent:

$(i): TA=AT$

$(ii): TE(\Delta)=E(\Delta)T\ \ \ \forall \Delta \in \mathcal A$

We also used the notation $A=\int id_{\sigma(A)}dE$ instead of $A=\int(t\mapsto t)dE$

Answer 1

$$ (A-\lambda I)\int_0^{\infty}e^{-\lambda t}e^{tA}dt \\ = \int_0^{\infty}e^{-\lambda t}Ae^{-tA}+(-\lambda e^{-\lambda t})e^{tA} \\ = \int_0^{\infty}\frac{d}{dt}(e^{-\lambda t}e^{tA})dt = e^{-\lambda t}e^{tA}|_{t=0}^{\infty}= - I. $$