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?
Best Answer
Here is a derivation based on $$\psi(k+1)=-\gamma+\sum_{n=1}^{k}\frac{1}{n}=-\gamma+\int_0^1\frac{1-t^k}{1-t}\,dt$$ (let me use the conventional notation) and the formula $$\gamma=\int_0^1\frac{1-e^{-x}}{x}\,dx-\int_1^\infty\frac{e^{-x}}{x}\,dx.$$ We get \begin{align}\sum_{k=0}^{\infty}\psi(k+1)\frac{z^k}{k!}&=-\gamma e^z+\int_0^1\frac{e^z-e^{zt}}{1-t}\,dt\\&=e^z\left(-\gamma+\int_0^1\frac{1-e^{-z(1-t)}}{1-t}\,dt\right)\\&=e^z\left(-\gamma+\int_0^z\frac{1-e^{-x}}{x}\,dx\right),\\\int_0^z\frac{1-e^{-x}}{x}\,dx&=\int_0^1\frac{1-e^{-x}}{x}\,dx+\int_1^z\frac{dx}{x}-\int_1^z\frac{e^{-x}}{x}\,dx\\&=\int_0^1\frac{1-e^{-x}}{x}\,dx+\int_1^z\frac{dx}{x}-\int_1^\infty\frac{e^{-x}}{x}\,dx+\int_z^\infty\frac{e^{-x}}{x}\,dx\\&=\gamma+\ln z+\Gamma(0,z),\end{align} with implied agreement of chosen branches of $\ln z$ and $\Gamma(0,z)$ ($=$ paths from $1$ to $z$ and from $z$ to $\infty$).