I came across the following series involving the Digamma function $\Psi$:
\begin{equation}
\sum^{\infty}_{k=0} \Psi(k+1) \frac{z^k}{k!},
\end{equation}
where z < 0. Plugging it into Wolfram Alpha gave me
\begin{equation}
\sum^{\infty}_{k=0} \Psi(k+1) \frac{z^k}{k!} = e^{z} ( \ln(z) + \Gamma(0, z)),
\end{equation}
with the upper incomplete Gamma function $\Gamma(a, z)$. I found out that $\Gamma(0, z)$ is related to the exponential integral via
\begin{equation} \Gamma(0, z) = \begin{cases} - Ei(-z) - i \pi &\text{, for } z < 0 \\ -Ei(-z) &\text{, for } z > 0 \end{cases} \end{equation} The closest relation I could find to relating the exponential integral to the series over the Digamma functions is (5.1.10) and (5.1.11) in Abramowitz Stegun ( http://people.math.sfu.ca/~cbm/aands/page_229.htm).
Question: How do I obtain the result given by Wolfram Alpha?