Closed Form of $\int x^n e^x~\mathrm{d}x$

Question

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.

Answer 1

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

$$p_k=(-1)^{n-k}\frac{n!}{k!}.$$

Answer 2

Using General Leibniz Rule for the $n$'th derivative of a product, we have

$$\begin{align} \int x^ne^x\,dx&=\left.\left(\frac{d^n}{db^n}\int e^{bx}\,dx\right)\right|_{b=1}\\\\ &=\left. \frac{d^n}{db^n}\left(\frac{e^{bx}}{b}\right)\right|_{b=1}+C\\\\ &=\left. \left(\sum_{k=0}^n \binom{n}{k}\left(\frac{d^{n-k}e^{bx}}{db^{n-k}}\right)\left(\frac{d^k b^{-1}}{db^k}\right)\right)\right|_{b=1}+C\\\\ &=\sum_{k=0}^n\binom{n}{k}(-1)^k k! x^{n-k}e^x+C \end{align}$$

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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.

Answer 3

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. $$ p_n(x)=\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}(n-k)!x^k=(-1)^n n!\sum_{k=0}^{n}(-1)^{k}\frac{x^k}{k!}. $$

Answer 4

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.