I am trying to obtain an analytical continuation for $\Gamma(z)$ into the region of the complex plane characterized by $\Re(z) \leq 0$ but am stuck.
Starting from the integral definition of $\Gamma(z)$ which is valid for $\Re(z) \geq 0$, re-write
\begin{equation}
\Gamma(z)=\int_{0}^{1} e^{-t} t^{z-1} dt +\int_{1}^{\infty} e^{-t} t^{z-1}dt \equiv f_1(z)+f_2(z).
\end{equation}
We first show that $f_2(z)$ is holomorphic in the entire complex plane by writing
$$
f_2(z)=\int_{1}^{\infty} e^{-t} t^{-1} e^{z\ln t}dt=\int_{1}^{\infty} \frac{e^{-t}}{t} e^{i\Im(z)\ln t} t^{\Re(z)}dt
$$
where I used $z=\Re(z)+i\Im(z)$ and $\exp{(z\ln t)}=\exp[{\ln t(\Re(z)+i\Im(z))}].$ Now we take the magnitude of $f_2(z)$ to obtain
$$
\big|f_2(z)\big|=\bigg| \int_{1}^{\infty} \frac{e^{-t}}{t} e^{i\Im(z)\ln t} t^{\Re(z)}dt\bigg|\leq \int_{1}^{\infty} e^{-t} t^{\Re(z)-1}dt
$$
and we can see that this converges uniformly $\forall \ z.$ Thus we conclude that $f_2(z)$ is a holomorphic $\forall \ z$ and it is an entire function. I am stuck here though, how to deal with the $f_1$ now? Thanks
Where the integral is defined, we can integrate by parts:
$$\begin{align} \int_0^1 e^{-t} t^{z-1}\,dt &= \frac{e^{-t}t^z}{z}\Biggr\rvert_0^1 + \frac{1}{z}\int_0^1 e^{-t} t^z\,dt\\ &= \frac{1}{ez} + \frac{1}{z}\int_0^1 e^{-t}t^z\,dt. \end{align}$$
The integral now is defined for $\Re z > -1$, and for $\Re z > 0$, we have the missing part of $\Gamma$, so
$$\Gamma(z) = \frac{1}{ez} + \frac{1}{z}\int_0^1 e^{-t} t^z\,dt + \int_1^\infty e^{-t} t^{z-1}\,dt$$
is valid for $\Re z > -1$. Integrate by parts more often, and you get a representation valid for $\Re z > -k$.