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Show that the gamma function

$$\Gamma(z)=\int_{0}^{\infty}e^{-t}t^{z-1}dt$$

Is holomorphic in $\{Re(z)>0\}$.

Ok, so I was told to try to solve this excercise by defining a sequence of functions

$$f_n=\int_{\frac{1}{n}}^{n}e^{-t}t^{z-1}dt$$

And try to bound the $n$th term by a converging integral (I guess) in any compact subset $K \subset \{Re(z)>0\}$. However, Im not sure what is that I have to bound, and how. Any hints?

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Joaquin Liniado
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  • first you want to show that for any $z$ in the open right half-plane the sequence $f_n(z)$ converges. then, to establish that this limit function is in fact a holomorphic function of $z$ you require that the convergence is uniform when $z$ is restricted to a compact subset of the right half-plane. technically this requires setting useful bounds for the nasty bits of the integral – David Holden Dec 09 '15 at 03:12
  • Yes I know! thats what I said precisely I don't know what is the nasty bit I should bound! – Joaquin Liniado Dec 09 '15 at 03:14
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    a useful fact is that $|t^{z-1}|=t^{\mathfrak{Re}(z)-1}$ – David Holden Dec 09 '15 at 03:36

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I suggest using Morera's Theorem instead, which states that a continuous function $f$ is holomorphic in an open region $\Omega$ if the integral of $f$ over every triangle $\Delta$ is $0$ whenever the triangle and its interior is contained in $\Omega$.

The integral function you have for $\Gamma(z)$ is continuous by the Lebesgue dominated convergence theorem. You can use Fubini's theorem to interchange orders of integration in order to prove that $$ \oint_{\Delta} \Gamma(z)dz = \int_{0}^{\infty}e^{-t}\left(\oint_{\Delta}e^{(-1+z)\ln t}dz\right)dt = 0. $$

Disintegrating By Parts
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