Let $z\in \mathbb C$ with $\text{Re}(z)>0$ (that is, $z=x+iy, x>0$), it is known that $$\Gamma(z) = \int_0^{\infty} t^{z-1}e^{-t}dt.$$
Let $n\in\mathbb Z \setminus \{0\},$ and define $I_n(z)=\int_{n}^{\infty} t^{z-1}e^{-t} dt \ \ \ (\text{Re}(z)>0).$
Question: What can we say about $I_n(z)$? Can expect to evolute $I_n$? What can we say about $I_n$ as $n\to \pm \infty$?
As Travis mentions, this is the (upper) incomplete Gamma function:
$$\Gamma(z,s)=\int_s^{+\infty}t^{z-1}e^{-t}\ dt$$
It is known that $\lim_{s\to+\infty}\Gamma(z,s)=0$ and $\lim_{s\to-\infty}\Gamma(z,s)$ does not exist for any $z$. The limit to infinity may easily be verified with a squeeze on upper bounds.
By integration by parts, it can be seen that
$$\Gamma(z+1,s)=z\Gamma(z,s)+s^ze^{-s}$$
If $z\in\mathbb N$, closed form may be found by letting $t-s=u$,
$$\Gamma(z,s)=e^{-s}\int_0^{+\infty}(u+s)^{z-1}e^{-u}\ du$$
and then binomial expanding into the Gamma function. Values at half integers will reduce to the error function, and negative integers reduce to the exponential integral.
A nice asymptotic expansion also reveals that
$${\displaystyle {\frac {\Gamma (s,x)}{x^{s-1}e^{-x}}}\rightarrow 1}\text{ as }{\displaystyle x\rightarrow +\infty }$$
$${\displaystyle {\frac {\Gamma (s,x)}{x^{s}}}\rightarrow -{\frac {1}{s}}}\text{ as }{\displaystyle x\rightarrow 0}$$
$${\displaystyle \Gamma (s,z)\sim z^{s-1}e^{-z}\sum _{k=0}{\frac {\Gamma (s)}{\Gamma (s-k)}}z^{-k}}$$