Assume $A$ is a closed linear operator on a Banach space $X$ and is densely defined. Assume the spectral bound $s(A) = \sup\{Re\lambda: \lambda\in \sigma(A)\}$ is finite. For example, if $A$ is the generator of a semigroup $T(t)$ with growth bound $\omega$, i.e., $\|T(t)\|\le Ce^{\omega t}$, we have $s(A) \le \omega$. Let $\alpha > s(A)$ be a real numer. Let $R(\lambda, A) = (\lambda - A)^{-1}$ be the resolvent. Do we have the following spectral representation of the semigroup $e^{tA}$?
$$e^{tA} = \frac{1}{2\pi i}\int\_{\Gamma}e^{tz}R(z,A)dz,$$
where $\Gamma \subset \mathbb{C}$ is the vertical line $\{z\in\mathbb{C}, Re(z) = \alpha\}$. 



Spectral representation of closed operators with finite spectral bound
I've looked it up now. The formula in question does indeed hold in the following sense:


**Theorem.** Let $(e^{tA})\_{t \in [0,\infty)}$ be a $C\_0$-semigroup on a complex Banach space $X$. Let $\omega \in \mathbb{R}$ be a real number that is strictly larger than the growth bound of our semigroup and let, for each $k > 0$, $\Gamma\_k$ denote the complex line from $\omega - ik$ to $\omega +ik$. Then:


(i) The formula
$$
 (\*) \qquad e^{tA}x = \lim\_{k \to \infty} \frac{1}{2\pi i} \int\_{\Gamma\_k} e^{zt} R(z,A)x \, dz
$$
holds for each $x$ in the domain of $A$.


(ii) If $X$ is an UMD space, then $(\*)$ does even hold for each $x \in X$.


*Reference:* Proposition 3.12.1 and Theorem 3.12.2 in "Arendt, Batty, Hieber, Neubrander: Vector-Valued Laplace Transforms and Cauchy Problems (2011)".


The same reference contains a counterexample that shows that one cannot drop the assumption in (ii) that $X$ be a UMD space, in general (this is Example 3.12.3 there; hardly surprising, the counterexample is a shift semigroup...).


