The Integral $\int\limits_0^1$ $\frac{e^x-\sum^{n-1}_{i=0} \frac{x^i}{i!}}{x^n}dx$ $\def \ein{\operatorname{Ein}}$
$\def \ei{\operatorname{Ei}}$

Evaluate $$\int\limits_0^1 \frac{e^x-\sum^{n-1}_{i=0} \frac{x^i}{i!}}{x^n}dx,$$
for $n\in\mathbb N$.

Specific case:
Case ($n=1$):
$$\int_0^1 \frac{e^x-1}{x}\,dx$$
Which is the exponential integral $\operatorname{ein}$
$$\ein(z)=-\int_0^z \frac{e^{-x}-1}{x}\,dx$$
Substituting $u=-x$
$$\ein(z)=-\int_0^{-z} \frac{e^{y}-1}{y}\,dy$$
$$-\ein(-1)=\ei(1)-\gamma-\ln(1)=\ei(1)-\gamma$$
But this doesn't seem to be generalizable as the $\ein$ and $\ei$ series seem to be uniquely similar enabling this.
I've also tried using wolfram alpha, which gives:
$$$$\begin{array}{|c|c|} \hline
n & \text{The Intergral} \\ \hline
1 & \ei(1)-\gamma \\ \hline
2 & \ei(1)-\gamma +2-e \\ \hline
3 & \frac{1}{2}(\ei(1)-\gamma+\frac{9}{2}-2e) \\ \hline
4 & \frac{1}{6}(\ei(1)-\gamma+\frac{59}{6}-4e) \\ \hline
5 & \frac{1}{24}(\ei(1)-\gamma+\frac{626}{24}-10e) \\ \hline
6 & \frac{1}{120}(\ei(1)-\gamma+\frac{10954}{120}-34e) \\ \hline
\end{array}$$$$
the oies was helpful in identifiying the $e$ coefficient to be the left factorial  (which is consistent with pax's answer)
And was unable to identify anything else.
 A: 
One can use the Taylor series expansion of exp(x) and solve as shown in picture. The summation notation at last I believe can be solved with beta function(https://en.m.wikipedia.org/wiki/Beta_function#:~:text=In%20mathematics%2C%20the%20beta%20function,0%2C%20Re%20y%20%3E%200.) It converges for n≥1.
A: For $n>1$ by integration by parts
$$\int_0^1 \frac{e^x-\sum_{i=0}^{n-1}\frac{x^i}{i!}}{x^n}dx=\int_0^1 \frac{e^x-\sum_{i=0}^{n-2}\frac{x^i}{i!}}{(n-1)x^{n-1}}dx+[-\frac{e^x-\sum_{i=0}^{n-1}\frac{x^i}{i!}}{(n-1)x^{n-1}}]_0^1=$$
$$\frac{1}{(n-1)}\int_0^1 \frac{e^x-\sum_{i=0}^{n-2}\frac{x^i}{i!}}{x^{n-1}}dx-\frac{e-\sum_{i=0}^{n-1}\frac{1}{i!}}{(n-1)}.$$
So in particular, if we denote your integral as
$$I_n=\int_0^1 \frac{e^x-\sum_{i=0}^{n-1}\frac{x^i}{i!}}{x^n}dx,$$
$$I_{n+1}=\frac{1}{n}I_{n}-\frac{e-\sum_{i=0}^{n}\frac{1}{i!}}{n}$$
One can solve this recurence as standard.
A: Assume that $n\geq 2$. Motivated by the OP's table, substituting the ansatz
$$
I_n = \frac{1}{{(n - 1)!}}\left( {\operatorname{Ei}(1) - \gamma  + q_n  - !(n - 1)e} \right)
$$
into the recurrence by @pax, we obtain
$$
q_{n + 1}  = q_n  + (n - 1)!\sum\limits_{i = 0}^n {\frac{1}{{i!}}}  = q_n  + \frac{{\left\lfloor {n!e} \right\rfloor }}{n}.
$$
This leads to
$$
q_n  = \sum\limits_{k = 1}^{n - 1} {\frac{{\left\lfloor {k!e} \right\rfloor }}{k}} .
$$
Thus
$$
\int_0^1 {\frac{1}{{x^n }}\left( {e^x  - \sum\limits_{i = 0}^{n - 1} {\frac{{x^i }}{{i!}}} } \right)dx}  = \frac{1}{{(n - 1)!}}\left( {\operatorname{Ei}(1) - \gamma  + \sum\limits_{k = 1}^{n - 1} {\frac{{\left\lfloor {k!e} \right\rfloor }}{k}}  - !(n - 1)e} \right)$$
for $n\geq 2$. Note that when empty sums are interpreted as $0$, this formula holds also when $n=1$.
A: Note that you can use this main Regularized Incomplete Gamma function series definition Q(a,z) to solve. Also use the Lower Regularized Incomplete Gamma function P(a,z) noting that $P(a,z)+Q(a,z)=1$:
$$\int_0^1\frac{e^x-\sum\limits_{k=0}^{n-1}\frac{x^k}{k!}}{x^n}dx=\int_0^1\frac{e^x-e^x Q(n,x)}{x^n}dx=\int_0^1\frac{e^x(1-Q(n,x))}{x^n}dx=\int_0^1 \frac{e^x P(n,x)}{x^n} dx$$
Using the derived gamma formulas and Generalized Exponential Integral function properties:
$$\int_0^1 \frac{e^x P(n,x)}{x^n} dx=\int_0^1 \frac {e^x}{x^n}dx-\int_0^1\frac{e^x Q(n,x)}{x^n}dx=(-1)^n x^n x^{-n}Γ(1-n,-x)\mathop|\limits_0^1-\int_0^1\frac{e^x Q(n,x)}{x^n}dx\mathop=^\text{for convergent n}(-1)^n[e\ !(-n) -(-n)!]-\int_0^1\frac{e^x Q(n,x)}{x^n}dx$$
The !y represents the subfactorial/derangement function. There seems to be no known special function closed form for either Integral. Note that the integral is with respect to $x$, so we are not integrating a gamma function directly, just its “incomplete part”. Please correct me and give me feedback!
