Consider a separable Hilbert space $\mathcal H$ and the bounded linear operators $B(\mathcal H)$.
Consider a function $T: [0, \infty) \to B(\mathcal H)$, under what assumptions on $T(t)$ is it true that
$$\big(\int\_0^c T(t) \, dt \big) (x) = \int\_0^c T(t)x \, dt \ , \ \ \ \forall x \in \mathcal H , c \in (0, \infty)$$  

for the Bochner integral? I am aware of a similar question for semigroups of BLO on Banach spaces,[The Bochner integral about a semigroup of bounded linear operators on a Banach space](https://mathoverflow.net/q/271377/161092), but I am interested in general operator-valued functions.



Action of Bochner integral of operator-valued functions on vectors
I understand you assume that $T:[0,c]\to\mathcal{B(H)}$ is Bochner integrable in order to write $\int\_0^c T(t)dt$ as Bochner integral. Then for any $x\in\mathcal H$ the map $ [0,c]\ni t\mapsto \mathcal H$ is also Bochner integrable and the identity you wrote holds. More generally: for a measure situation $(X,\mathcal S,\mu)$, a couple of B-spaces $\mathbb E$ and $\mathbb F$, a bounded linear operator $L:\mathbb E\to \mathbb F$, and a Bochner integrable map $f:X\to \mathbb E$, the composition $L\circ f:X\to \mathbb F$ is Bochner integrable and $\int\_X L f(u) d\mu(u)=L\int\_X f(u))d\mu(u)$ (in your case $L$ is the evaluation map $\mathcal{B(H)}\ni A\mapsto Ax\in\mathcal H$).


(The proof is immediate if $f:=v\chi\_S$ with $v\in\mathbb E$ and $S\in\mathcal S$; by linearity it generalizes immediately to integrable simple functions $f:X\to \mathbb E$; it further generalizes immediately to $f\in \mathcal L^1(\mu,\mathbb E)$, by the very definition of Bochner integrable function and integral).


