# Integral operator identity involving square root of an operator

I would be most thankful if you could help me prove the following operator identity. Let $A$ and $B$ be two completely continuous Hermitian operators on a Hilbert space $H$, such that $A$ and $B$ do not commute. If $A$ is a positive operator, then show that

 $\left\langle x,\left(\sqrt{A}B-B\sqrt{A}\right)x\right\rangle=\frac{1}{\pi}% \int_{{0}}^{{\infty}}\left\langle x,\left(B\frac{1}{A+\omega}-\frac{1}{A+% \omega}B\right)x\right\rangle\sqrt{\omega}d\omega$

It is known that for $z$ and $z_{0}$ real and positive numbers one has

 $\sqrt{z}-\sqrt{z_{0}}=\frac{1}{\pi}\int_{{0}}^{{\infty}}\left(\frac{1}{z_{0}+% \omega}-\frac{1}{z+\omega}\right)\sqrt{\omega}d\omega$

Tahnk you!