monomocoso
New member
- Joined
- Jan 25, 2012
- Messages
- 31
Show \(\displaystyle \Gamma = \int^\infty_0 t^{z-1}e^{-t}\ dt \) is holomorphic in the right half plane by the following three steps:
1) Let \(\displaystyle S_M\delta = {z \in \mathbb{C} : \delta < Re(z) < M }\) and show the function is holomorphic in each strip.
2) Let \(\displaystyle \Gamma_\epsilon = \int^{1/\epsilon}_\epsilon t^{z-1}e^{-t}\ dt\) and show \(\displaystyle \Gamma_\epsilon\) is holomorphic in \(\displaystyle S_{M\delta}\)
3) Show that as \(\displaystyle \epsilon -> 0\) we have \(\displaystyle \Gamma_\epsilon -> \Gamma\) uniformly on the compact subsets of the strip by obtaining an estimate \(\displaystyle \Gamma_\epsilon - \Gamma = \int^\epsilon_0 t^{z-1}e^{-t}\ dt+\int^\infty_{1/\epsilon} t^{z-1}e^{-t}\ dt\)
1) Let \(\displaystyle S_M\delta = {z \in \mathbb{C} : \delta < Re(z) < M }\) and show the function is holomorphic in each strip.
2) Let \(\displaystyle \Gamma_\epsilon = \int^{1/\epsilon}_\epsilon t^{z-1}e^{-t}\ dt\) and show \(\displaystyle \Gamma_\epsilon\) is holomorphic in \(\displaystyle S_{M\delta}\)
3) Show that as \(\displaystyle \epsilon -> 0\) we have \(\displaystyle \Gamma_\epsilon -> \Gamma\) uniformly on the compact subsets of the strip by obtaining an estimate \(\displaystyle \Gamma_\epsilon - \Gamma = \int^\epsilon_0 t^{z-1}e^{-t}\ dt+\int^\infty_{1/\epsilon} t^{z-1}e^{-t}\ dt\)