What is the $t$-anti-derivative $\int dt \; e^{i t}\mathrm{Ei}\big(- i [t+x]\big)\; t^{-n}$ for $n = 1,2,3$? Consider the functions for $n = 1, 2, 3$
$$
f_{n}(t,x) := \frac{e^{i t}\mathrm{Ei}\big(- i [t+x]\big)}{t^{n}} \ ,
$$
where $\mathrm{Ei}$ is the exponential integral function defined for complex $z \notin ( - \infty, 0]$ as
$$
\mathrm{Ei}(z) = -\int_{-z}^{\infty} dq \; \frac{e^{-q}}{q}
$$
Question: Is there a way to write down the $t$-anti-derivative for $f_{1,2,3}$? ie. what is the anti-derivative $\int dt \; e^{i t}\mathrm{Ei}\big(- i [t+x]\big)\; t^{-n}$ for $n=1,2,3$?
Mathematica doesn't know this one. Gradshteyn and Ryzhik do not have this anti-derivative either. My guess is that it involves a polylogarithm (since I am nesting integrals divided by $1/t$ in the above), but I cannot figure this out.
EDIT: Assume furthermore that $x>0$.
 A: You can expand as a series:
$$\int e^{it} t^{-n} \text{Ei}(-i(t+k))dt=\gamma\int e^{it}t^{-n}dt+\sum_{k=1}^\infty\sum_{m=0}^\infty\frac{(-i)^ki^m}{kk!m!}x^k\left(-x^{-1}\right)^{m-n+1}\text B_{-\frac tx}(m-n+1,k+1)$$
work in progress
using an expansion for $\text{Ei}(x)$ and the Incomplete Beta function. The Beta function will not have negative arguments if $n>0$, so there should be no convergence problems.
Attempt $2$ with the En function:
Integration by parts:
$$\int t^{-n} e^{it}\text{Ei}(-i(t+x))dt= e^{it}t^{-n} \frac{e^{-i(t+x)}}{t+x}+\int \frac{e^{i(t+x)}}{t+x} t^{1-n}\text E_n(-i t) dt= t^{-n} \frac{e^{-ix}}{t+x}-e^{ix}\int \frac{e^{it}}{t+x} t^{1-n}\text E_n(-i t) dt$$
Using a geometric series for $|t|<|x|>0$ assuming the interchanging of sum and integral:
$$t^{-n} \frac{e^{-ix}}{t+x}+e^{ix}\int \frac{e^{it}}{t+x} t^{1-n}\text E_n(-i t) dt = t^{-n} \frac{e^{-ix}}{t+x}+e^{ix}\int\left(\sum_{k=0}^\infty (-1)^k x^{-k-1} t^k\right)t^{1-n}\text E_n(-i t) \left(\sum_{m=0}^\infty \frac{(it)^m}{m!}\right)dt = t^{-n} \frac{e^{-ix}}{t+x}+e^{ix}\int \sum_{m,k\ge0}\frac{i^m (-1)^k x^{-k-1}}{m!}t^{1-n+k+m}\text E_n(-it)dt$$
Integration using this power En integral identity:
$$t^{-n} \frac{e^{-ix}}{t+x}+e^{ix}\int \sum_{m,k\ge0}\frac{i^m (-1)^k x^{-k-1}}{m!}t^{1-n+k+m}\text E_n(-it)dt= c+t^{-n} \frac{e^{-ix}}{t+x}+e^{ix}\sum_{m,k\ge0}\frac{i^m (-1)^k x^{-k-1}}{m!}\left(t^{k+m-n+2}\frac{\text E_n(-i t)-\text E_{n-k-m-1}(- i t)}{k+m+1}\right)$$
This solution should have no undefined values within the restrictions. Please correct me and give me feedback!
A: Few things before the answer-
$\textstyle\displaystyle\operatorname{Ei}(-i[t+x])=e^{-it}\int\frac{e^{-ix}}{t+x}dx$ [see this]
$\textstyle\displaystyle\int\frac{1}{t^n(t+x)}dt=\sum_{k=1}^{\infty}\frac{(-1)^kt^{k-n}}{(n-k)x^k}$ [see this]
$\textstyle\displaystyle\int\frac{e^{-ix}}{x^k}dx=-x^{1-k}\operatorname{E_k}(ix)$ [see this]
I am also going to assume that $\textstyle\displaystyle\int_X\frac{e^{-ix}}{t^{n+1}+xt^n}$ converges where $X$ is the whole of space, which basically means that the integral is interchangeable.

We have $f_n(t,x)=e^{it}t^{-n}\operatorname{Ei}(-i[t+x])$
We can write, $\textstyle\displaystyle{f_n(t,x)=e^{it}t^{-n}\bigg[e^{-it}\int\frac{e^{-ix}}{t+x}dx\bigg]}$
$\textstyle\displaystyle{\int{f_n(t,x)}dt=\int\frac{1}{t^n}\int\frac{e^{-ix}}{t+x}dxdt}$
$\textstyle\displaystyle{=\int e^{-ix}\int\frac{1}{t^n(t+x)}dtdx}$
$\textstyle\displaystyle{=\int e^{-ix}\sum_{k=1}^{\infty}\frac{(-1)^kt^{k-n}}{(n-k)x^k}dx}$
$\textstyle\displaystyle{=\sum_{k=1}^{\infty}\frac{(-1)^kt^{k-n}}{n-k}\int\frac{e^{-ix}}{x^k}dx}$
$\textstyle\displaystyle{=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}t^{k-n}x^{1-k}}{n-k}\operatorname{E_k}(ix)}$
Therefore, $\textstyle\displaystyle\int f_1(t,x)dt=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}t^{k-1}x^{1-k}}{1-k}\operatorname{E_k}(ix)$
$\textstyle\displaystyle\int f_2(t,x)dt=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}t^{k-2}x^{1-k}}{2-k}\operatorname{E_k}(ix)$
$\textstyle\displaystyle\int f_1(t,x)dt=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}t^{k-3}x^{1-k}}{3-k}\operatorname{E_k}(ix)$
