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How can I show that $\Gamma$ function is holomorphic ?

I have to show it by dominated convergence theorem or by Morera's Theorem

For $\Re(z)>0$ the $\Gamma$-function is defined as \begin{equation*} \displaystyle\Gamma(z)=\int_0^{\infty}x^{z-1}e^{-x}dx~\text{and let}~f:G\times I\to \mathbb C~\text{with}~f(z,x)=x^{z-1}e^{-x}dx \end{equation*}

then there is a compact set $K\subset G$ and an integrable function $\phi:I\to\mathbb R$ such that $|f(z,x)|\le\phi(x)$ for $z\in K, x\in I$, then $\Gamma$ is holomorphic. How can I choose $\phi$ ?

Or what is the condition here for a closed path to be ''suitable'' ?

Does somebody know why I cannot get the derivative ?

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  • $\begingroup$ So you mean the Gamma function only as defined by that integral, without its analytic continuation? $\endgroup$ – Timbuc May 6 '15 at 9:09
  • $\begingroup$ @Timbuc with the constraint that $\Re(z)>0$ $\endgroup$ – OBDA May 6 '15 at 9:10
  • $\begingroup$ @Ob Yesw, of course...otherwise the integral doesn't converge. $\endgroup$ – Timbuc May 6 '15 at 9:19
  • $\begingroup$ "there is a compact set $\;K\;$ ..."? For analicity as you want this must be true for all compact subsets in the given domain. $\endgroup$ – Timbuc May 6 '15 at 9:52
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Any compact set $K\subset\mathbb{C}$ is included in a bounded vertical strip :$\forall z\in K, \{0<a\leq\Re(z)\leq b<+\infty\}$.

$|x^{z-1}|\leq x^{\Re(z)-1}|\leq x^{\Re(b)-1}|$ $(x>0)$ implies $|\Gamma(z)|\leq \Gamma(\Re(z)) \leq \Gamma(b)$ $(z\in K)$

Convergence of the integral is then normal, uniform convergence follows and Weierstrass convergence theorem for holomorphic function applies.

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