Analytic Continuation of the incomplete gamma function I have the following expression for $\alpha,z>0$:
\begin{equation}
\pi\mathrm i-\Gamma(\alpha)(-1)^\alpha\Gamma(1-\alpha,-z).
\end{equation}
In the context of the problem I am looking at this expression should be real-valued.  If we plot it as a function of $\alpha$, (I used $z=\pi$), we see that the imaginary part of the second term appears to be equal to $\pi\mathrm i$ (which it should be). If true, the imaginary terms cancel out and the value of the expression is real.

Is there any way I can use the analytic continuation of the incomplete gamma function to separate the real and imaginary parts of the second term?  I was able to do this for the case when $\alpha\in\Bbb N$ using this identity. Put another way,  I am looking for the expression for
\begin{equation}
\Re\{\Gamma(\alpha)(-1)^\alpha\Gamma(1-\alpha,-z)\}
\end{equation}
when $\alpha,z>0$.
 A: I am going to provide the solution for
\begin{equation}
\Re\left\{(-z)^{\alpha-1}\Gamma(1-\alpha,-z)\right\}.
\end{equation}
All that is needed to get the solution for the original question is a bit of algebra.

Now, we have to evaluate the real component of the second term for the cases of $\alpha\in\Bbb N$ and $\alpha\notin\Bbb N$ separately.
Starting with the noninteger $\alpha$ case, we us DLMF $8.7.3$ to rewrite the incomplete gamma term.  After a bunch of algebraic simplification we find
\begin{equation}
\Re\left\{(-z)^{\alpha-1}\Gamma(1-\alpha,-z)\right\}=%
-\sum_{k=0}^\infty \frac{z^k}{k!(1-\alpha+k)}%
-z^{\alpha-1}\Gamma(1-\alpha)\cos\pi\alpha.
\end{equation}
Then using $(1-\alpha+k)^{-1}=\frac{(1-\alpha)_k}{(1-\alpha)(2-\alpha)_k}$ we put the series term into the form of ${_1F_1}(a;b;z)$ to get 

\begin{equation}
\Re\left\{(-z)^{\alpha-1}\Gamma(1-\alpha,-z)\right\}=%
\frac{{_1F_1}(1-\alpha;2-\alpha;z)}{\alpha-1}%
-z^{\alpha-1}\Gamma(1-\alpha)\cos\pi\alpha,\quad\alpha\notin\Bbb N.
\end{equation}

For the integer $\alpha$ case we employ this identity and note that $z>0$, $\log(-z)=\log z+\pi\mathrm i$. Again, with a good amount of algebraic simplification we find

\begin{equation}
\Re\left\{(-z)^{\alpha-1}\Gamma(1-\alpha,-z)\right\}=%
\frac{z^{\alpha-1}}{\Gamma(\alpha)}%
\left(%
e^{z}\sum_{k=0}^{\alpha-2}\frac{k!}{z^{k+1}}%
-\operatorname{Ei}(z)%
\right),\quad \alpha\in\Bbb N.
\end{equation}

