The Hadamard Gamma function generalizes the factorial function to have no poles as a result of the reciprocal gamma function in its definition. It also is monotonic on $[-\infty,0]$ and $[2,\infty]$ for $x\in\Bbb R$. For a sixth in a series of summations of a single function, here are some related problems:
A closed form of $$\sum_\Bbb Z \text{sech}(x)
,\sum_\Bbb N \text{csch}(x)$$
Also here is the goal sum using Harmonic Numbers and the digamma function. The sum converges quickly with the $\pm 20$ th term being of magnitude $10^{-20}$.
$$\sum_{-\Bbb N^0}\text H(x)=\sum_0^\infty \text H(-x)=\frac12\sum_0^\infty\frac{H_\frac x2}{x!}-\frac12\sum_0^\infty\frac{H_\frac{x-1}{2}}{x!}= \sum_0^\infty\frac{ψ\left(\frac x2+1\right)- ψ\left(\frac {x+1}{2}\right)}{2 Γ(x+1)} $$
Here is the used integral representations of the Harmonic Numbers on the real axis where integration is done by each term to interchange operators:
$$\frac12\sum_0^\infty\frac{H_\frac x2}{x!}-\frac12\sum_0^\infty\frac{H_\frac{x-1}{2}}{x!} =\frac12 \int_0^1 \frac1{1-t}\sum_{x=0}^\infty \frac{1-t^\frac x2}{x!}dt-\frac12 \int_0^1 \frac1{1-t}\sum_{x=0}^\infty \frac{1-t^\frac {x-1}2}{x!}dt =\frac12 \int_0^1 \frac{1}{t-1}\sum_{x=0}^\infty\frac{t^\frac x2 -t^\frac{x-1}2}{x!}dt=\frac12\int_0^1 \frac{e^\sqrt t-\frac{e^\sqrt t}{\sqrt t}}{t-1}dt=\frac12\int_0^1\frac{e^\sqrt x}{x+\sqrt x}dx=\frac{\text{Ei}(2)-\text{Ei}(1)}{e}=\frac{Γ(0,-1,-2)}{e}=!(-1)-\frac{Γ(0,-2)}{e}=!(-1)+\frac{\text{Ei}(2)+i\pi}{e}=1.125386083083269719203240988…$$
It is interesting how $$\sum_0^\infty \text H(x)=\int_1^2\frac{e^{x-1}}{x}dx=\frac{\text{Ei}(2)-\text{Ei}(1)}e$$
In short, the main method used here was introducing an integral representation of the summand, switching by integrating term by term, taking the partial sum, and evaluating the resulting integral. How else can I prove that: $$\sum\limits_{-\infty}^0 \text H(x)= \frac{\text{Ei}(2)-\text{Ei}(1)}{e}? $$
Here is the Exponential Integral function,Subfactorial function, and Generalized Gamma function.
George Pólya once said:
“It is better to solve one problem five different ways, than to solve five problems one way.”
which leads me to saying that I originally posted this question to get an answer, but ended up solving it, so how can I solve it another way? Please correct me and give me feedback!
Just for fun, you can use the same integral technique to find the Gompertz Constant denoted $δ,G$ among other symbols. Note that $\Bbb N=1,2,3,…$ :
$$\sum_\Bbb N\frac{ψ(x)}{Γ(x)}=\int_0^1 \frac{e-e^t}{1-t}dt-e\gamma=-e\text{Ei}(-1)=\text G=.596347…$$
Best Answer
Define
$$f(t) = \sum_{n=0}^∞ H(-n)t^n.$$
By the functional equation $H(x + 1) = xH(x) + \frac{1}{Γ(1 - x)}$, we have
\begin{multline*} f(t) = \sum_{n=0}^∞ \left(-(n+1)H(-n-1) + \frac{1}{Γ(n+2)}\right)t^n \\ = -\sum_{n=1}^∞ nH(n)t^{n-1} + \sum_{n=0}^∞\frac{1}{(n+1)!}t^n = -f'(t) + \frac{e^t - 1}{t}. \end{multline*}
So
\begin{gather*} \frac{d}{dt} e^t f(t) = e^{t}(f'(t) + f(t)) = \frac{e^{2t} - e^t}{t}, \\ e^t f(t) = f(0) + \int_0^t \frac{e^{2t} - e^t}{t}\,dt = H(0) + (\mathrm{Ei}(2t) - \mathrm{Ei}(t) - \ln 2) = \mathrm{Ei}(2t) - \mathrm{Ei}(t), \\ \end{gather*}
giving the desired result at $t = 1$.