# Complex $\Gamma$-function is holomorphic?

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 ?

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

Any compact set $$K\subset\mathbb{C}$$ is included in a bounded vertical strip :$$\forall z\in K, \{0.
$$|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)$$