# Series involving Digamma relates to Exponential Integral

I came across the following series involving the Digamma function $$\Psi$$:

$$$$\sum^{\infty}_{k=0} \Psi(k+1) \frac{z^k}{k!},$$$$ where z < 0. Plugging it into Wolfram Alpha gave me
$$$$\sum^{\infty}_{k=0} \Psi(k+1) \frac{z^k}{k!} = e^{z} ( \ln(z) + \Gamma(0, z)),$$$$ with the upper incomplete Gamma function $$\Gamma(a, z)$$. I found out that $$\Gamma(0, z)$$ is related to the exponential integral via

$$$$\Gamma(0, z) = \begin{cases} - Ei(-z) - i \pi &\text{, for } z < 0 \\ -Ei(-z) &\text{, for } z > 0 \end{cases}$$$$ 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?

• Updated the question. Sep 23, 2019 at 11:19
• Hint: observe that $\psi(k+1) = H_k -\gamma$, and that the harmonic number can be written as $H_n=\int_0^1 \frac{1-x^n}{1-x}$. Can you continue from here? Sep 23, 2019 at 12:06

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 $$\infty$$).
$$\psi(k+1)=\int_0^\infty \left(\frac{1}{te^t} - \frac{1}{(e^t-1)e^{kt}}\right)\,dt$$
$$\sum_{k\geq 0}\psi(k+1)\frac{z^k}{k!}=\int_0^\infty\sum_{k\geq 0} \left(\frac{1}{te^t} - \frac{1}{(e^t-1)e^{kt}}\right)\frac{z^k}{k!}\,dt =\int_{0}^{+\infty}\left(\frac{e^z}{t e^t}-\frac{e^{z e^{-t}}}{(e^t-1)}\right)\,dt$$ and the RHS equals, via $$e^t\mapsto u$$, $$\int_{1}^{+\infty}\left(\frac{e^z}{u^2\log u}-\frac{e^{z/u}}{u(u-1)}\right)\,du$$ or, via $$u\mapsto\frac{1}{v}$$, $$\int_{0}^{1}\left(\frac{e^{vz}}{1-v}-\frac{e^z}{\log v}\right)\,dv=e^z\int_{0}^{1}\left(\frac{e^{-wz}}{w}-\frac{1}{\log(1-w)}\right)\,dw.$$ Now the incomplete $$\Gamma$$ function can be recognized in $$\int_{0}^{1}\frac{1-e^{-wz}}{w}\,dw=\gamma+\Gamma(0,z)+\log(z)$$.