Over the past few months, I've been chipping away trying to find a closed form expression for the following Newton series: $$f(x) = \sum_{n=0}^\infty \frac{1}{n!^2}x^\overline{n}$$ where $x^\overline{n} = \prod_{k=0}^{n-1} (x+k)$
The first thing I thought to try looking at it was to consider techniques used for the Bessel functions. That led to the following document (https://link.springer.com/content/pdf/bbm:978-3-0348-7229-4/1.pdf) where I tried to apply the technique used on page 145 to get $J_0(x)$, and while it managed to help find two similar results: $$\overline{J}_0(x) = \sum_{n=0}^\infty \frac{(-1/4)^n}{n!^2}x^\overline{2n} =\frac{1}{\pi}\int_0^\pi\left(\frac{(1+i\sin\theta)^{-x}+(1-i\sin\theta)^{-x}}{2}\right)d\theta $$ $$\overline{I}_0(x) = \sum_{n=0}^\infty \frac{(1/4)^n}{n!^2}x^\overline{2n} =\frac{1}{\pi}\int_0^\pi\left(\frac{(1+\sin\theta)^{-x}+(1-\sin\theta)^{-x}}{2}\right)d\theta $$ when applied to the one I wanted to initially solve, it failed to connect the function with anything I could give a closed form.
It was afterwards that I thought to see if there was any kind of finite ODE for the functions I was getting, and quickly saw a pattern. I found that $\overline{J}_0(x)$ solved $(x-1)y+\nabla{y}+(x-2)\nabla^2{y} = 0$, with the corresponding function inside its definite integral solved $\sin^2(\theta)y+\nabla^2y=0$. A similar pattern can be found with $\overline{I}_0(x)$, which solves $(1-x)y+\nabla{y}+(x-2)\nabla^2{y} = 0$ and was connected to $-\sin^2(\theta)y+\nabla^2y=0$. (where $\nabla{y}$ is representing the backwards difference of y)
After looking into it, I found that these both stem from the ablity to solve the following functional equation: $$\nabla^2y-a^2y=0;y=A\frac{(1+a)^{-x}+(1-a)^{-x}}{2} + B\frac{(1+a)^{-x}-(1-a)^{-x}}{2}$$
This didn't help for the original function however, because the functional equation for f(x) ($-f+2\nabla{f}+(x-2)\nabla^2f = 0$) connected with the solution to $-\sin(\theta)y + (1+\sin\theta)\nabla{y} + (2x-3)\nabla^2y = 0$, which only manages to introduce a variable into a case I don't know how to solve.
It is at this point that I'm at a loss as to what to do next. I've had ideas general ideas to go off of, find a different integral to apply with or look for solutions to the finite ODE, but those have been dead ends for the most part. The most I managed to prove is the following series solution:
$$qy+(p-q)\nabla{y}+(x-p-1)\nabla^2y=0;y(x)=\sum_{n=0}^\infty \frac{(-q)^n}{n!\cdot\Gamma(n+p)}x^\overline{n}$$
Which seems helpful, but I'm not sure what to do with it. I'm hoping that someone with more experience in the subject can clear this up, since most of what I've found has been through blind experimentation. Most of these formulas are also being double-checked with basic graphing calculators too, so if any of what I've presented here is wrong please also let me know.
Update: After working with contour integrals, with some trial and error I've managed to get the function to the following form:
$$f(x) = \frac{1}{2\pi}\int_0^{2\pi}\left(1-\frac{1}{re^{i\theta}}\right)^{x-1}\cdot e^{1-re^{i\theta}}d\theta $$
Which seems to work for any value $r>1$.
