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 February 14th, 2012, 04:38 PM #1 Senior Member   Joined: Sep 2008 Posts: 105 Thanks: 0 Gamma Function Show $\Gamma= \int^\infty_0 t^{z-1}e^{-t}\ dt$ is holomorphic in the right half plane by the following three steps: 1) Let $S_M\delta= {z \in \mathbb{C} : \delta < Re(z)=< M }=$ and show the function is holomorphic in each strip. 2) Let $\Gamma_\epsilon= \int^{1/\epsilon}_\epsilon t^{z-1}e^{-t}\ dt$ and show $\Gamma_\epsilon$ is holomorphic in $S_{M\delta}$ 3) Show that as $\epsilon -> 0$ we have $\Gamma_\epsilon -> \Gamma$uniformly on the compact subsets of the strip by obtaining an estimate $\Gamma_\epsilon - \Gamma= \int^\epsilon_0 t^{z-1}e^{-t}\ dt+\int^\infty_{1/\epsilon} t^{z-1}e^{-t}\ dt$
 February 15th, 2012, 04:13 PM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Gamma Function [color=#000000]Here is a simpler proof that might help you with your problem. Since the given integral is improper, we split the integral in the following way $\int_{0}^{\infty}e^{-t}t^{z-1}\;dt=\int_{0}^{1}e^{-t}t^{z-1}\;dt+\int_{1}^{\infty}e^{-t}t^{z-1}\;dt$. The integral $\int_{0}^{1}e^{-t}t^{z-1}\;dt$ converges absolutely for $\mathfrak{Re}(z)>0$, for every bounded and closed region $\overline{\mathbb{U}}$ in $\mathfrak{Re}(z)>0$, if $\min_{x\in\overline{\mathbb{U}}}x=\delta$ for $z=x+y\cdot i\in\overline{\mathbb{U}}$ we have $\hspace{350pt}\left|e^{-t}t^{z-1}\right|=e^{-t}t^{\mathfrak{Re}(z-1)}\leq t^{\delta-1}$ and since $\int_{0}^{1}t^{\delta-1}\;dt$ converges, by Weierstra?' criterion follows the uniform convergence of the integral on $\overline{\mathbb{U}}$. So we deduce that the integral $\int_{0}^{1}e^{-t}t^{z-1}\;dt$ converges uniformly and absolutely on every bounded region $\overline{\mathbb{U}}$. Now if $\overline{\mathbb{U}}$ is in the disc $|z|\leq M$, for $z\in\overline{\mathbb{U}}$, $\hspace{390pt}\left|e^{-t}t^{z-1}\right|\leq e^{-t}e^{M+1}$ using again the Weierstra? criterion we deduce (since $\int_{1}^{\infty}e^{-t}t^{M+1}\;dt$ converges), the integral $\int_{1}^{\infty}e^{-t}t^{z-1}\;dt$ converges uniformly and absolutely on every bounded region $\overline{\mathbb{U}}$. So ? is regular in $\mathfrak{Re}(z)>0$.[/color]

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