Closed solution of $\int\frac{x^n}{\sum_{k=0}^{n} \frac{x^{k}}{k!}}dx$ Is there a closed solution to
$$A(n,x) =\int\frac{x^n}{\sum\limits_{k=0}^{n} \frac{x^{k}}{k!}}dx?$$ 
I was thinking of looking at the roots of the denominator to try split it into partial fractions but the I couldn't find any general method. Just dividing the polynomials doesn't seem to do much and there doesn't seem to me to be any obvious substitution/by parts method. Perhaps I'm missing something right under my nose?
 A: The finite sum appearing in the denominator of the integral turns out to be what is known as the exponential sum function $\exp_n (x)$. It can be expressed in terms of the upper incomplete gamma function $\Gamma (a,x)$ as follows:
$$\exp_n (x) = \sum_{k = 0}^n \frac{x^k}{k!} =  \frac{e^x \Gamma (n + 1, x)}{\Gamma (n + 1)}.$$
Thus your indefinite integral reduces down to 
$$\int \frac{x^n}{\exp_n (x)} \, dx = n! \int \frac{e^{-x} x^n}{\Gamma (n + 1, x)} \, dx.$$
If we note that
$$\frac{d}{dx} \left (\Gamma (n + 1,x) \right ) = \frac{d}{dx} \int_x^\infty e^{-t} t^n \, dx = -e^{-x} x^n,$$
then we have
$$\int \frac{x^n}{\exp_n (x)} \, dx = -n! \ln[\Gamma(n + 1,x)] + C.$$ 

Update
We can write the final answer in a more manageable form involving the lower incomplete gamma function $\gamma(a,x)$ where
$$\gamma (a,x) = \int_0^x e^{-t} t^{a - 1} \, dt,$$
rather than in terms of $\Gamma (a, x)$ which is improper. 
Since $\gamma(a, x) + \Gamma (a, x) = \Gamma (a)$ where $\Gamma (a)$ is the normal gamma function, or in our case
$$\gamma (n + 1,x) + \Gamma (n + 1, x) = \Gamma (n + 1),$$
as the term $\Gamma (n + 1)$ is constant (independent of $x$) we have
$$\int \frac{x^n}{\exp_n (x)} \, dx = n! \cdot \ln[\gamma(n + 1,x)] + C.$$ 
A: Fortunately, the incomplete Gamma functions exist which can replace your summation term. They are defined as solutions to integrals, so whether this is "closed form" is debatable.
$$\sum_{k=0}^n \frac{x^k}{k!} = \frac{e^x}{n!} \cdot\Gamma(1+n,x)$$
From which your integral becomes
$$A(n,x)=n! \cdot\int \frac{\frac{x^n}{e^x}}{\Gamma(1+n,x)}\, \mathrm d x$$
It turns out that this integral is solvable. First, let's look at the derivative of $\Gamma(1+n,x)$:
$$\frac{\mathrm d}{\mathrm d x}\left(\Gamma(1+n,x)\right) = \frac{\mathrm d}{\mathrm d x} \left(\int_x^\infty t^{n} e^{-t}\, \mathrm d t\right) = -x^{n}e^{-x}$$
Well then, the factor on top of the $\Gamma$ denominator in our integral is equal to the $\Gamma$ function's derivative! So we have an integral of the form
$$\int - \frac{f'(x)} {f(x)}$$
whose solution is
$$-\log(f(x)) + C$$
and so our final answer is
$$A(n,x)=-n! \cdot \log \Big( \Gamma(1+n,x)\Big)$$
with the constant of integration, of course.
