Details on the definition of the upper incomplete Gamma function Consider the Gamma function
$$
\Gamma(z)=\int_0^{\infty}e^{-t}t^{z-1}\,dt
$$
and the function is well defined for $\Re(z)>0$. If we integrate from another point we get the so called incomplete Gamma function
$$
\Gamma(z,x)=\int_x^{\infty}e^{-t}t^{z-1}\,dt
$$
I want to know what $x$ can be in this formula. Every reference I check is sloppy on the details. can $x$ take complex values? if so what path am I integrating in? 
 A: If $z$ is a whole number, then one can use a different formula for the incomplete Gamma function:
$$\Gamma(z,x)=(z-1)!e^{-x}\sum_{k=0}^{n-1}\frac{x^k}{k!}$$
Other such extensions follow easily through series expansions for any $x,z\in\mathbb C$:
$$\Gamma(z,x)=\Gamma(z)-z^{-1}x^ze^{-x}\ _1F_1(1;1+z;x)$$
Where $_1F_1$ is the Confluent Hypergeometric Function of the First Kind.
Taken from Wolfram.

As far as the actual integral itself, one usually takes any path between the two points that does not cross any singularities or branch cuts, and as $e^{-t}t^{z-1}$ has no singularities for $z>1$ (and no branch cuts if you choose your cuts right), then
$$\int_x^{+\infty}e^{-t}t^{z-1}\ dt=\int_Ce^{-t}t^{z-1}\ dt$$
for any path $C$ from $x$ to $+\infty$.
For example, take $C:e^{i\theta},\theta\in[0,\pi]$:
$$\begin{align}-2&=\int_{+1}^{-1}(x+1)\ dx\\&=\int_C(z+1)\ dz\\&=\int_0^\pi(e^{i\theta}+1)\ d(e^{i\theta})\\&=i\int_0^\pi(e^{2i\theta}+e^{i\theta})\ d\theta\\&=i\int_0^\pi(\cos(2\theta)+i\sin(2\theta)+\cos(\theta)+i\sin(\theta))\ d\theta\\&=i\left[2i\right]\\&=-2\end{align}$$
Try any other path starting at $1$ and going to $-1$ and you will get the same result.
Supposing that $\Re(x)>0$, one could use the following:
$$\Gamma(z,x)=\int_{\Re(x)}^{+\infty}e^{-t}t^{z-1}\ dt+i\int_0^{\Im(x)}e^{-(\Re(x)-it)}(\Re(x)-it)^{z-1}\ dt$$
which would be a path from $x$ to $\Re(x)$, then to $+\infty$.
A: Choosing the usual branch of $\log t$ analytic on $ \mathbb{C} \setminus (-\infty,0]$, so that $t^{s-1} = e^{(s-1)\log t}$ is analytic on $ \mathbb{C} \setminus (-\infty,0]$
If $z=a+ib \in \mathbb{C} \setminus (-\infty,0]$ and $s \in \mathbb{C}$ then 
$$\Gamma(s,z) = \int_z^{+\infty} t^{s-1} e^{-t}dt = \int_a^{+\infty} (t+ib)^{s-1} e^{-t-ib}dt \tag{1}$$
Note that $f(t) = t^{s-1} e^{-t}$ is analytic on $\mathbb{C} \setminus (-\infty,0]$ so it has a primitive $\lambda(t)$ analytic on $\mathbb{C} \setminus (-\infty,0]$. Check that $\displaystyle\lim_{\Re(t) \to \infty}\lambda(t) =\lambda(+\infty)$ has a well-defined unique value, so taking $C_z$ to be any curve going from $z$ to $+\infty$ we have $$\oint_{C_z} f(t)dt = \lambda(+\infty)-\lambda(z) = \Gamma(s,z)$$
and in particular we can choose $C_z$ to be the straight horizontal line $[a+ib,+\infty+ib)$ as in $(1)$
Choosing another branch of $\log t$ and $t^{s-1} = e^{(s-1)\log t}$ or equivalently continuing analytically $\Gamma(s,z)$ beyond the branch cut $(-\infty,0]$ you'll get another definition of the incomplete Gamma function, analytic on a different domain. This also proves that $\Gamma(s,z)$ has a branch point at $z=0$.
