Let $-\Delta$ be the self-adjoint operator on $L^{2}(\mathbb{R}^n)$ with domain $\mathcal{D}(-\Delta)=H^2(\mathbb{R}^n)$, and with spectral measure $E_{\lambda}$. And let $\{U(t)\}_{t\geq 0}$ be the unitary group $e^{-it\Delta}$. How to give a integration of $\{U(t)\}_{t\geq 0}$?
In fact, $\forall A$ a unitary operator we have, $$U(t)=e^{iAt}=\int_{\mathbb{R}} e^{i\lambda t}dE_{\lambda} \ (*).$$
But how about the $(*)$ for $-\Delta$?
For the operator $-\Delta$, \begin{align} E_{\lambda}f &= \frac{1}{(2\pi)^{n/2}}\int_{|\xi|^2 \le \lambda}\hat{f}(\xi)e^{i\xi\cdot x}d\xi, \\ U(t)f &= \int_{0}^{\infty}e^{i\lambda t}dE_{\lambda}f \\ &= \frac{1}{(2\pi)^{n/2}}\int_{0}^{\infty}e^{i\lambda t}\int_{|\xi|^2=\lambda}\hat{f}(\xi)e^{i\xi\cdot x}d\xi d\lambda \\ &=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}\hat{f}(\xi)e^{i|\xi|^2t}e^{i\xi\cdot x}d\xi \end{align}
So $U(t)f = (\hat{f}(\xi)e^{i|\xi|^2t})^{\vee}$ is a Fourier multiplier that has only a radial dependence in the transform space. You can directly check that $\frac{d}{dt}U(t)f = -i\Delta f$.