Closed Form of $\int x^n e^x~\mathrm{d}x$ My calculus teacher showed us how to solve $$\displaystyle\int x^n e^x~\mathrm{d}x$$ by iteratively doing integration by parts. I figured out that $$\displaystyle\int x^n e^x~\mathrm{d}x$$ is equal to $$x^n e^x - n\int x^{n-1} e^x~\mathrm{d}x.$$  You can then iteratively find out what the solution is for any $n$.  My question is whether or not there exists a closed form for this integral.  Any help would be much appreciated.
 A: Using General Leibniz Rule  for the $n$'th derivative of a product, we have
$$\begin{align}
\int x^ne^x\,dx&=\left.\left(\frac{d^n}{db^n}\int e^{bx}\,dx\right)\right|_{b=1}\\\\
&=\left. \frac{d^n}{db^n}\left(\frac{e^{bx}}{b}\right)\right|_{b=1}+C\\\\
&=\left. \left(\sum_{k=0}^n \binom{n}{k}\left(\frac{d^{n-k}e^{bx}}{db^{n-k}}\right)\left(\frac{d^k b^{-1}}{db^k}\right)\right)\right|_{b=1}+C\\\\
&=\sum_{k=0}^n\binom{n}{k}(-1)^k k! x^{n-k}e^x+C
\end{align}$$

Alternatively, using the recursive relationship, $I_n=x^ne^x-nI_{n-1}$, we have
$$\begin{align}
I_n&=x^ne^x-n(x^{n-1}e^x-(n-1)I_{n-2})\\\\
&=x^ne^x-nx^{n-1}e^x+n(n-1)I_{n-2}\\\\
&\vdots\\\\
&=(x^n-nx^{n-1}+n(n-1)x^{n-2}-n(n-1)(n-2)x^{n-3}\cdots+(-1)^nn!)e^x\\\\
&=\sum_{k=0}^n\binom{n}{k}(-1)^kk!x^{n-k}e^x
\end{align}$$
which is as expected modulo the integration constant.
A: It is clear that the antiderivative is a polynomial of degree $n$, let $P(x)$.
Then by derivation,
$$(P(x)'+P(x))e^x=x^ne^x$$ or
$$P'(x)+P(x)=x^n.$$
This yields the recurrence relation
$$p_{k-1}=-kp_k$$ with
$$p_n=1.$$
The solution is
$$p_k=(-1)^{n-k}\frac{n!}{k!}.$$
A: One may make a simple not of the two solutions provided. One states that
$$I_{n} = \int e^{x} \, x^{n} \,~\mathrm{d}x = (-1)^{n} \, n! \, e^{x} \, \sum_{k=0}^{n} \frac{(-x)^{k}}{k!}$$
is a solution and the other is
$$I_{n} = \int e^{x} \, x^{n} \,~\mathrm{d}x = e^{x} \, \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \, k! \, x^{n-k}.$$
By using the finite exponential (truncated exponential) function,
$$e_{n}(x) = \sum_{k=0}^{n} \frac{x^k}{k!},$$
then the first becomes
$$I_{n} = (-1)^{n} \, n! \, e^{x} \, e_{n}(-x).$$
The second can be placed in hypergeometric form as
$$I_{n} = x^{n} \, e^{x} \, {}_{2}F_{0}\left(-n, 1; --; \frac{1}{x}\right).$$
From this the formula
$$e_{n}(-x) = \frac{(-x)^{n}}{n!} \, {}_{2}F_{0}\left(-n, 1; --; \frac{1}{x}\right)$$
is obtained.
A: Let $$\int x^ne^x~\mathrm{d}x = p_n(x)e^x + C$$ for some polynomial $p_n(x)$, then from the integration by parts it follows that $p_n(x)=x^n-np_{n-1}(x)$, $n\ge 1$, and $p_0(x)=1$. Multiply both sides by $\dfrac{t^n}{n!}$ and sum over $n\ge 1$ to get
$$
\sum_{n\ge 1}{p_n(x)\frac{t^n}{n!}}=\sum_{n\ge 1}{x^n\frac{t^n}{n!}}-t\sum_{n\ge 1}{p_{n-1}(x)\frac{t^{n-1}}{(n-1)!}}.
$$
Let $P(x,t)=\displaystyle\sum_{n\ge 0}{p_n(x)\dfrac{t^n}{n!}}$, then $P(x,t)-1=(e^{xt}-1)-tP(x,t)$, i.e.
$$
P(x,t)=\frac{e^{xt}}{1+t}.
$$
It's easy to see that $\left[\dfrac{t^n}{n!}\right]e^{xt}=x^n$ and $\left[\dfrac{t^n}{n!}\right]\dfrac{1}{1+t}=(-1)^n n!$, so $p_n(x)$ is the exponential convolution of those sequences, i.e.
$$
p_n(x)=\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}(n-k)!x^k=(-1)^n n!\sum_{k=0}^{n}(-1)^{k}\frac{x^k}{k!}.
$$
