Question: I need to evaluate:

$$ \Gamma(a,-x, \infty) = \int\_{-x}^{\infty} t^{a-1} e^{-t} dt$$

When I plug this into Wolfram, for instance $Gamma[\frac{1}{2}, -4]$ I get a complex number. How is this calculated?

Answer: 
Coincidentally there is a [paper](http://arxiv.org/abs/1608.04152) by A. Gil, D. Ruiz-Antolín, J. Segura, N. M. Temme from Aug. 14, which describes the computation of Tricomi's incomplete gamma function, this function $\gamma^{\*}(a, z)$ is real for real positive and negative values of $a$ and $z$ and can be defined as
$$\gamma^{\*}(a,z) = \frac{z^{-a}}{\Gamma(a)}\gamma(a,z)$$

If you have access to this function you can compute
$$\Gamma(a,z)=\Gamma(a)-\gamma(a,z)
=\Gamma(a)\Big(1-z^a\gamma^{\*}(a,z)\Big)$$

Here is the example with $a=1/2, z=-4$ from my own Pascal library:
$\gamma^{\*}(1/2,-4) = 9.2824012072877763$ and therefore 
$$\Gamma(1/2,-4) = \sqrt{\pi}(1-\sqrt{-4}\times 9.2824012072877763) \\=
1.7724538509055160-32.905255531014461 i$$

If you have access to Kummer's confluent hypergeometric $M$ function (sometimes called $\_1 F\_1$)
you can express the entire Tricomi function for all $a,z \in \mathbb{C}$
as (see [DLMF](http://dlmf.nist.gov/8.5))
$$
\gamma^{\*}(a,z) = e^{-z} \; \frac{M(1,a+1,z)}{\Gamma(a+1)}
= e^{-z} \sum\limits\_{n=0}^{\infty} \frac{z^n}{\Gamma(a+n+1)}.
$$
This confluent hypergeometric form is used by my routine for real $z<0,\,$ i.e.
$$\gamma^{\*}(1/2,-4)= \frac{e^4 M(1,3/2,-4)}{\Gamma(3/2)}= \\
54.5981500331442391 \times 0.150670194461895983 / 0.886226925452758014 = 9.2824012072877763$$
The upper incomplete gamma function is defined as such:
$$\Gamma(s,x) = \int\_x^\infty t^{s-1}e^{-t}dt$$  

In your case you are asking about
$$\Gamma(1/2,-4) = \int\_{-4}^\infty t^{-1/2}e^{-t}dt$$
Note that this is only real for $t>0$... for $t<0$ we have that $t^{1/2}$ is an imaginary number, therefore making the integral complex over this region. This should generalize to an arbitrary negative lower bound of integration.