In a demonstration of a theorem I have...$A:X\to X $ a bounded operator, $X $ a banach space and $$e^{tA}=\frac{1}{2\pi i}\int_{C_r} e^{\lambda t } R (\lambda ; A) \ d\lambda$$
With uniform norm, where $R (A, \lambda)=(\lambda I - A)^{-1} $, $\mu >r>\|A\| $ and $C_r $ is the complex circunference with center in the origin. The author so write... since outside $C_r $ the integrand is analytic and $$\|R (\lambda ; A)\|\leq C|\lambda |^{-1} $$ we can chift the patch of integration from $C_r$ from the line $Re \lambda =\mu $ by cauchy's theorem. I dont understed how did he use the cauchy theorem. Every hint I am thanks. I need show that
$$\frac{1}{2\pi i}\int_{C_r} e^{\lambda t } R (\lambda ; A) \ d\lambda=\frac{1}{2\pi i}\int_{\mu - i \infty }^{\mu + i\infty }e^{\lambda t } R (\lambda ; A) \ d\lambda$$