Show $ \lambda(P - \lambda^2)^{-1} = -i \int_0^\infty e^{-it\lambda} \cos(t \sqrt{P}) dt$ for positive, self-adjoint $P$, and $\text{Im} \lambda < 0$

Question

Let $P$ be a positive, self-adjoint operator on an Hilbert space $(H, \langle \cdot, \cdot \rangle)$. Then for each $\lambda \in \mathbb{C}$ with $\text{Im} \lambda < 0$, $\lambda^2$ belongs to the resolvent set of $P$.

I would like to show $$\lambda (P - \lambda^2)^{-1} = -i \int_0^\infty e^{-it\lambda} \cos(t \sqrt{P}) dt, \qquad \text{Im} \lambda < 0. $$

This identity is "well known" and is quoted without proof or reference in an article I'm reading. Hints or solutions are greatly appreciated.

Answer 1

It suffices to show that, for $u$ in the domain of the operator $P$, $$\frac{-i}{\lambda} \int_0^\infty e^{-it\lambda} \cos(t \sqrt{P}) dt (P- \lambda^2)u = u$$.

One has \begin{equation*} \begin{split} \frac{-i}{\lambda} \int_0^\infty e^{-it\lambda} \cos(t \sqrt{P}) dt (P- \lambda^2)u &= \lim_{R \to \infty} \int_0^R e^{-it\lambda} \cos(t \sqrt{P}) dt (P- \lambda^2)u \\ &= \lim_{R \to \infty} \int_0^R e^{-it\lambda} \cos(t \sqrt{P})(P- \lambda^2)u dt \\ &= \lim_{R \to \infty} \int_0^R e^{-it\lambda} (-\partial_t^2- \lambda^2) \cos(t \sqrt{P})u dt \end{split} \end{equation*}

From here, one needs to integrate by parts twice. The resulting integrals cancel each other and the boundary terms at $t = R$ vanish in the limit that $R \to \infty$. One can check we are left only with

\begin{equation*} \begin{split} \frac{-i}{\lambda} \int_0^\infty e^{-it\lambda} \cos(t \sqrt{P}) dt (P- \lambda^2)u &= \frac{i}{\lambda} e^{-it\lambda}\sin(t \sqrt{P})\sqrt{P}u \rvert_{t= 0} \\ &+ e^{-it\lambda}\cos(t \sqrt{P})u \rvert_{t= 0}\\ &=u. \end{split} \end{equation*}