I need to find the closed-form expression of $f_n(x)$ that satisfies the following integral recurrence:
$$f_n(x)=\int_0^\infty f_{n-1}\left(x-a\right) \, \dfrac{1}{2} \mathrm{e}^{-\frac{a}{2}} da -2\operatorname{Ei}\left(\dfrac{x}{2}\right)\,\mathrm{e}^{-\frac{x}{2}}$$
with $f_0(x)=0$, where $\operatorname{Ei}$ is the exponential integral, defined as
$$\operatorname{Ei}(x) = \int_{-\infty}^x \frac{e^t}t$$
with $\operatorname{Ei}'(x) = \frac{e^x}x$.
What I know is that the function $f_n(x)$ has the following form:
$$f_n(x)= \left [a(n)_0+a(n)_2x+\dots a(n)_{n-2}x^{n-2}\right] -\left [b(n)_0+b(n)_2x+\dots b(n)_{n-1}x^{n-1}\right]\mathrm{e}^{-\frac{x}{2}}\operatorname{Ei}\left(\dfrac{x}{2}\right).$$
Also, from the above integral recurrence I derived the following recurrence:
$$\fbox{$2f'_n(x)+f_n(x)=f_{n-1}(x)-2/x$},$$
which can be used to find recurrences among the polynomial coefficients.
Could you find the closed-form expressions of the vectors: $$a(n)=\left(a(n)_0,\dots, a(n)_{n-2} \right)\in \mathbb R^{n-1} \\ b(n)=\left (b(n)_0,\dots, b(n)_{n-1} \right ) \in \mathbb R^{n}$$ appearing in the above two polynomials of degrees $n-2$ and $n-1$ based on the above recurrence?
The vectors for $n=1, 2, 3, 4$ are as follows:
$a(1)= - \,, \quad b(1)=(2) \\ a(2)=(2), \quad b(2)=(2, 1) \\ a(3)=(3,\frac{1}{2}), \quad b(3)=(2,1, \frac{1}{4}) \\ a(4)=(\frac{11}{3},\frac{2}{3},\frac{1}{12}), \quad b(4)=(2,1, \frac{1}{4},\frac{1}{24}).$