Intro:
I was looking for a fresh new problem to solve and was inspired by
The goal function is the Abramowitz function found in the NIST catalogue of library functions. Here is an interactive graph of the function. Maybe the Abramowitz function appears in Abramowitz and Stegun? The function is also standardized like all of the functions in the NIST DLMF:
$$\text f_m(x)\mathop=^\text{def}\int_0^\infty t^m e^{-t^2-\frac xt}dt,m\in\Bbb N$$
Special cases:
$$\text f_m(0)=\int_0^\infty t^m e^{-t^2}dt =\frac12 \Gamma\left(\frac{m+1}{2}\right),\text{Re}(m)>-1$$
The great result by Wolfram Alpha using the Meijer-G function:
$$\text f_0(x)=\int_0^\infty e^{-t^2-\frac xt}dt=\frac{x\,\text G^{3,0}_{0,3}\left(\frac{x^2}4\bigg|-\frac12,0,0\right)}{4\sqrt \pi}$$
Which can simplify into an expression with $\,_0\text F_2$ hypergeometric functions
Here is an attempt at an evaluation. There seems to be no Polynomial function generating function to make it simpler, so let’s use a regular series expansion integrated with the Gauss Hypergeometric function:
$$\text f_m(x)=\int_0^\infty t^m e^{-t^2-\frac xt}dt=\int_0^\infty t^m\sum_{n=0}^\infty \frac{(-1)^n (t^2+x t^{-1})^n}{n!}dt= \sum_{n=0}^\infty \frac{(-1)^nt^{m-n+1}}{n!(m-n+1)} \,_2\text F_1\left(\frac{m-n+1}3,-n;\frac{m-n+4}3;-\frac{t^3}x\right)\bigg|_0^\infty$$
Where the Gauss Hypergeometric function is just an Incomplete beta function, but the limits make the hypergeometric function diverge. Please correct me and give me feedback!
Note that an equivalent problem is solving it’s recurrence relation also found on NIST:
$$\text f_m(x)\mathop=^{m\ge 2}\frac{m-1}2\text f_{m-2}(x)+\frac x2 \text f_{m-3}(x)$$
also a paper called “ Evaluation of Abramowitz functions in the right half of the complex plane” states that a linear partial differential equation for the function is:
$$x\frac {d^3}{dx^3}\text f_m(x)+(1-m)\frac{d^2}{dx^2}\text f_m(x)+2\text f_m(x)=0 \implies \text f_m(x)=\frac12 c_2(m)x\,_0\text F_2\left(\frac32,1-\frac m2;-\frac{x^2}4\right)+c_1(m) \,_0\text F_2\left(\frac12,\frac{1-m}2;-\frac{x^2}4\right) +\left(\frac x2\right)^{m+1}c_3(m)\,_0\text F_2\left(\frac m2+1,\frac{m+3}2;-\frac{x^2}4\right)$$
and the very simple delay differential equation:
$$x\frac{d\,\text f_m(x)}{dx}=-\text f_{m-1}(x)$$
New Question
It has been figured out by @Gary and Claude Leibovici, partly from a CAS, that we have a new decomposition formula for $\,_0\text F_2$:
$$\boxed{\text f_m(x>0)=\frac12\Gamma\left(\frac{m+1}2\right)\,_0\text F_2\left(\frac12,\frac{1-m}2;-\frac{x^2}4\right)-\frac x2\Gamma\left(\frac m2\right) \,_0\text F_2\left(\frac32,1-\frac{m}2;-\frac{x^2}4\right)+x^{m+1}\Gamma(-m-1) \,_0\text F_2\left(\frac m2+1,\frac{m+3}2;-\frac{x^2}4\right)} $$
How does one get this solution from $xy’’’+(1-m)y’’+2y=0,y=y(x)$? One idea is a power series where the coefficients are a hypergeometric series. I know the question has changes, but this is a question, so a new one is here.