How to integrate $ \int x^n e^x dx$?
How can I solve this indefinite integral for an arbitrary integer $n>0$?
$$ \int{x^n e^x dx}$$
I could partially integrate it for small $n$, but that's not really a solution.
Edit: (TB) This question is closely related to: Is there a closed form solution for $\int x^n e^{cx}$?, but it is more elementary, because $n$ is an integer here.
Solution 1:
Hint: Use integration by parts.
EDIT: Try several values of $n$. $$ \int {x e^x dx} = (x - 1)e^x + C $$ $$ \int {x^2 e^x dx} = (x^2 - 2x + 2)e^x + C. $$ $$ \int {x^3 e^x dx} = (x^3 - 3x^2 + 6x - 6)e^x + C. $$ $$ \int {x^4 e^x dx} = (x^4 - 4x^3 + 12x^2 - 24x + 24)e^x + C. $$ $$ \int {x^5 e^x dx} = (x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120)e^x + C. $$ Conclude that $$ \int {x^n e^x dx} = \bigg[\sum\limits_{k = 0}^n {( - 1)^{n - k} \frac{{n!}}{{k!}}x^k } \bigg]e^x + C. $$
Solution 2:
You could use the generating function approach.
$$ \eqalign{\int_0^X e^{tx} e^x \ dx &= \frac{e^{(1+t)X} - 1}{1+t}\cr
&= \sum_{k=0}^\infty (-1)^k t^k \left(e^X -1 + \sum_{j=1}^\infty e^X \,\frac{X^j}{j!} t^j\right)\cr
&= \sum_{n=1}^\infty \left((-1)^n (e^X - 1) + \sum_{j=1}^n (-1)^{n-j} \frac{X^j}{j!} e^X \right) t^n\cr}$$
But also
$$ \int_0^X e^{tx} e^x \, dx = \sum_{n=0}^\infty \frac{t^n}{n!} \int_0^X x^n e^x \, dx$$
Equating coefficients of $t^n$ from both sides,
$$ \int_0^X x^n e^x\, dx = (-1)^n n! (e^X - 1) +
\sum_{j=1}^n (-1)^{n-j} \frac{n!}{j!} X^j e^X $$
Solution 3:
I find it a little difficult for me to guess the solution by trying several $n$. I would like to do it as following:
$$\begin{align}\int x^ne^xdx&=x^ne^x+(-1)n\int x^{n-1}e^xdx,\qquad n\geq 1\\ \int x^0e^xdx&=e^x\end{align}$$
Then you get the recurrence relation:
$$\begin{align}a_n(x)&=x^ne^x+(-1)na_{n-1}(x),\qquad n\geq 1\\ a_0(x)&=e^x\end{align}$$
With the recursive formula, it may be easier to find the pattern of the result.