In Pazy's Book page 26, the author gives a proof of Lemma 7.1, the lemma 7.1 says that: Let $B$ be a bounded linear operator. If $\gamma>\|B\|$ then $$e^{tB}=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{\gamma t}R(\lambda;B)d\lambda$$ The convergence in this formula is in the uniform operator topology and uniformly in t on bounded intervals.
In the proof of this lemma, the author get $$e^{tB}=\frac{1}{2\pi i}\int_{C_{r}}e^{\lambda t}R(\lambda;B)d\lambda$$ where $C_{r}$ be the circle of radius $r$ centered at the origin. Choose $r$ such that $\gamma>r>\|B\|$.
Then the author says that since outside $C_{r}$ the integral of the above formula is analytic and $\|R(\lambda;B)\|\leq C|\lambda|^{-1}$ we can shift the path of integral from $C_{r}$ to the line $Re z=\gamma$, using Cauchy's theorem.
I can just prove this is right when $t\in [\alpha,\beta]$ in which $0<\alpha<\infty,0<\beta<\infty$, that is to say, I can just check pazy's proof is right when $t$ in a compact set away from zero, but can not justify whether $t\in [0,T]$ is right. So, How can I check the "$t\in[0,T]$"? How to shift the path of integral from $C_{r}$ to the line $Re z=\gamma$ when $t\in [0,T],T<\infty$?
Any hints are welcome, Thank you very much!!