TITLE: Incomplete Gamma function of negative arguments
QUESTION [7 upvotes]: Playing with Mathematica, I found that $$\Re\left[\Gamma\left(-\frac{n}{2},x\right)\right]=\Gamma\left(-\frac{n}{2}\right)$$ for $x<0$ and $n$ any positive integer. 
Would it be possible to prove that?
On Wikipedia, I found a formula for the incomplete Gamma function when the first argument is a negative integer: $$\Gamma(-n,z)=\frac{1}{n!}\left(\frac{e^{-z}}{z^n}\sum_{k=0}^{n-1}(-1)^k(n-k-1)!z^k+(-1)^k+(-1)^n\Gamma(0,z)\right)$$ 
Maybe there is a similar formula when the first argument is a negative half-integer!

REPLY [5 votes]: By dlmf 8.4.6 
$$
\Gamma\left(\tfrac{1}{2},z\right)=2\int\limits_{\sqrt{z}}^{\infty}e^{-t^{2}}dt.
$$
For negative arguments we have
$$
\Gamma\left(\tfrac{1}{2},-|z|\right)=2\int\limits_{i\sqrt{|z|}}^{\infty}e^{-t^{2}}dt=2\int\limits_{i\sqrt{|z|}}^{i\sqrt{|z|}\,+\infty}e^{-t^{2}}dt=2\int_{0}^{+\infty}e^{-\left(t+i\sqrt{|z|}\right)^{2}}dt.
$$
$$
\text{Re}\left\{\Gamma\left(\tfrac{1}{2},-|z|\right)\right\}=2\int_{0}^{+\infty}e^{-\left(t^2-|z|\right)}\cos\left(2t\sqrt{|z|}\right)dt=2e^{|z|}\cdot\frac{\sqrt{\pi}}{2}e^{-|z|}=\Gamma\left(\frac12\right).
$$
By dlmf 8.8.2
$$
\Gamma\left(a+1,z\right)=a\Gamma\left(a,z\right)+z^{a}e^{-z}.
$$
From this one can easily deduce by induction that the claim is true for all $n\in\mathbb{Z}$:
$$\text{Re}\left\{\Gamma\left(n+\tfrac{1}{2},-|z|\right)\right\}=\Gamma\left(n+\tfrac{1}{2}\right).$$
Numerical checked confirmed this formula for non-negative $n$ too.