The following definition and Theorem are given in the book "A short course on operator semigroup" by the author "K-J Engel and R Nagel".
Sectoral operator: A closed linear operator $(A,D(A))$ in Banach space $X$ is called sectoral (of angle $\delta$) if there exists $0<\delta\leq\frac{\pi}{2}$ such that sector $\sum_{\frac{\pi}{2}+\delta}:=\{\lambda\in \mathbb{C}:| arg\lambda |<\frac{\pi}{2}+\delta\} \backslash\{0\}$ is contained in the resolvent set $\rho(A),$ and if for each $\epsilon\in (0,\delta)$, there exists $M_{\epsilon}\geq1$ such that$$\|R(\lambda,A)\|\leq\frac{M_{\epsilon}}{|\lambda|},$$ for $0\neq\lambda\in \overline{\sum_{\frac{\pi}{2}+\delta-\epsilon}}. $ Again, we know the following theorem
Theorem An operator $(A,D(A))$ on a Banach space $X$ generates a bounded analytic semigroup $\Big\{T(z):z\in\sum_{\delta}\cup\{0\}\Big\}$ on $X$ if and only if A is densly defined and sectoral operator.
Where $T(0)=I$ and for $z\in\sum_{\delta}$, $$T(z)=\frac{1}{2\pi i}\int_{\gamma}e^{\mu z}R(\mu,A)d\mu$$ where $\gamma$ is any piecewise smooth curve in $\sum_{\frac{\pi}{2}+\delta}$ going from $\infty e^{-i(\frac{\pi}{2}+\delta')}$ to $\infty e^{i(\frac{\pi}{2}+\delta')}$, for some $\delta'\in(|argz|,\delta)$.
In the chapter 5 of the book "R. deLaubenfels, Existence Families, Functional calculi and Evollution Equations, springer-verlag, Berlin,1994" at page no 69, it is given that $T(z)$ are one to one and it's range is dense in $X$.
My question: For $t>0$, What is the inverse operator of $T(t)$ from it's range set in term of contour integration? Please help me.
I was expecting $$T(-t)=\frac{1}{2\pi i}\int_{\gamma}e^{-\mu t}R(\mu,A)d\mu$$ would be the inverse of T(t) but what i found , it can not be. please help me.