For each $t\in [0,1]$, let $T_t\colon L^2(\mathbb{R})\to L^2(\mathbb{R})$ be the operator that shifts everything to the right by $t$, i.e. $$T_t f(x)=f(x-t)$$ for all $f\in L^2(\mathbb{R})$.
Then $t\mapsto T_t$ defines a map $$\gamma\colon[0,1]\to\mathcal{B}(L^2(\mathbb{R})).$$ Note that $\gamma$ is not continuous.
Question: Is $\gamma$ Bochner-integrable, i.e. does the integral $$\int_{[0,1]}T_t\,dt$$ converge in $\mathcal{B}(L^2(\mathbb{R}))$?
Thoughts: Since each operator $T_t$ has norm, the question is whether $\gamma$ is strongly measurable. In particular, there is a question of whether there exists a subset of $J\subseteq[0,1]$ of measure $1$ such that $\gamma(J)$ is separable. Certainly $\gamma([0,1])$ is not separable, and I suspect that no such subset $J$ exists, but I'm not sure how to show this.
NOT AN ANSWER - EXTENDED COMMENT.
This question suggested me the following formal manipulation. The semigroup $T_t$ can be written, formally, as $$ T_t=e^{t\frac{\partial}{\partial x}}=\sum_{n=0}^\infty \frac{t^n}{n!}\frac{d^n}{dx^n}.$$ Integrating in $t\in[0, 1]$, again formally, we obtain $$ \int_0^1 T_t\, dt = \sum_{n=0}^\infty \frac{1}{n!(n+1)}\frac{d^n}{dx^n}.$$ It seems to me that the comments are suggesting that $T_t$ is NOT integrable. Therefore this last expression should make no sense. But I cannot see why.