The Gamma function has no zeros How can I prove the Gamma function has no zeros in its holomorphy domain $\Bbb C\setminus\Bbb Z_{\le0}$ using only its integral definition $\Gamma(z)=\int_0^{+\infty}t^{z-1}e^{-t}\,dt$ valid when $\Re z>0$ and the functional equation $\Gamma(z+1)=z\Gamma(z)$?
From the integral definition we can find easily the holomorphic extension; thus it would be enough to prove that $\Gamma\neq0$ in $\{\Re z>0\}$, using thus the integral form. But I can't prove neither this.
Can someone help me?
EDIT: This question is not a duplicate because all the solution given use more "advanced" tools. Here I'm asking to prove that Gamma has no zeros using ONLY its integral representation
 A: Assume that $\Gamma(\alpha) = 0$ for some $\Re(\alpha) > 0$. Then for any $s \geq 0$, the substitution $ t = (1+s)x$ gives
$$ 0 = \frac{\Gamma(\alpha)}{(1+s)^{\alpha}} = \int_{0}^{\infty} x^{\alpha-1} e^{-x} e^{-sx} \, \mathrm{d}x. \tag{1}$$


*

*This is already enough to give a contradiction since the right-hand side is the Laplace transform of $x \mapsto x^{\alpha-1} e^{-x}$ and hence cannot be identically zero.

*If we avoid the use of Laplace transform, still we can derive a contradiction. Let $0 < \sigma < \Re(\alpha)$. For this $\sigma$, we know that $\Gamma(\sigma) > 0$. Then multiply both sides of $\text{(1)}$ by $s^{\sigma-1}/\Gamma(\sigma)$ and integrate w.r.t. $s$ on $[0, \infty)$. By the Fubini's theorem, this yields
\begin{align*}
0
& = \int_{0}^{\infty} x^{\alpha-1} e^{-x} \left( \frac{1}{\Gamma(\sigma)} \int_{0}^{\infty} s^{\sigma-1} e^{-sx} \, \mathrm{d}y \right) \, \mathrm{d}x \\
&= \int_{0}^{\infty} x^{\alpha-\sigma-1} e^{-x} \, \mathrm{d}x \\
&= \Gamma(\alpha-\sigma).
\end{align*}
This shows that $\Gamma(z) = 0$ along the line segment joining $i\Im(\alpha)$ and $\alpha$. Then the identity theorem tells us that $\Gamma(z)$ is identically zero for $\Re(z) > 0$, which is impossible.
A: $$
\eqalign{
  & \Gamma (z) = \int_{t\, = \,0}^{\,\infty } {e^{\, - z} t^{\,z - 1} dt}  = \int_{t\, = \,0}^{\;1} {e^{\, - z} t^{\,z - 1} dt}  + \int_{t\, = \,1}^{\;\infty } {e^{\, - z} t^{\,z - 1} dt}  =   \cr 
  &  = \sum\nolimits_{t\, = \,0}^{\;\infty } {{{\left( { - 1} \right)^{\, - k} } \over {k!(z + k)}}}  + \int_{t\, = \,1}^{\;\infty } {e^{\, - z} t^{\,z - 1} dt}  \cr} 
$$
where $\int_{t\, = \,1}^{\;\infty } {e^{\, - z} t^{\,z - 1} dt} $ is an entire function.
This a well known development, you can find further details in many publications on Gamma function (e.g.: in "The theory of analytic functions, a brief course - A. I. Markushevich ")
