Find $\int x^ne^xdx$

Question

Find $\displaystyle \int x^ne^xdx$.

I haven't been able to figure this out. Do I just keep repeatedly applying integration by parts?

This is a duplicate of that question of mine, in fact it's a less general case (take $b=-1$). — YoTengoUnLCD, Dec 22 '15 at 03:56
If $n$ is assumed to be a positive integer, as suggested by the question but not made explicit, then the situation is significantly better than the one outlined in the proposed duplicate. — hardmath, Dec 22 '15 at 04:46
Regarding repeated integration by parts, see also this: http://math.stackexchange.com/questions/70974/lesser-known-integration-tricks/71195#71195 — Hans Lundmark, Dec 22 '15 at 09:54

score 4 · Answer 1 · answered Dec 22 '15 at 03:48

4

May I recommend integration by undetermined coefficients, where given some $n$, we know the solution will take the form $P(x)e^x + C$, where $P(x)$ is some polynomial of degree $n$. (It really can't be anything else). So write $P(x)$ as $a_0 + a_1x + ... + a_nx^n$, take the derivative, and back out the coefficients.

Or if you want to get practice, use integration by parts. But that's really nasty.

answered Dec 22 '15 at 03:48

Aurey

1,332
8
17

I read for a brief moment "underdefined coefficients" :P – Zelos Malum Dec 22 '15 at 03:51

score 3 · Answer 2 · answered Dec 22 '15 at 03:56

Finding an explicit formula for the antiderivative will involve the incomplete Gamma function.

What you could do (and what is usually standard) is get a recurrence relation.

Let $$I_n = \int x^n e^x dx, \ n\in \mathbb{N}\cup{0}$$ Integration by parts yields

$$I_n = x^n e^x -n \int x^{n-1} e^x dx = x^n e^x - n I_{n-1}.$$

The first term in this recurrence relation is $$I_0 = e^x.$$

score 2 · Answer 3 · answered Dec 22 '15 at 04:43

2

Here is another approach for definite integrals:

Let $I(t) = \int_a^b e^{tx} dx = {1 \over t} (e^{tb}-e^{ta})$.

Then $I^{(n)}(1) = \int_a^b x^n e^{x} dx$.

answered Dec 22 '15 at 04:43

copper.hat

172,524

This is a very efficient approach and the one that I would use. +1 Happy holidays! - Mark – Mark Viola Dec 22 '15 at 05:13
@Dr.MV: Happy holidays to you too! – copper.hat Dec 22 '15 at 06:16

score 1 · Answer 4 · answered Dec 22 '15 at 05:59

Notice, integrating by parts as follows $$\int x^ne^x\ dx$$ $$=x^n\int e^x\ dx-\int nx^{n-1}e^x\ dx$$ $$=x^ne^x-nx^{n-1}e^x+n\int (n-1)x^{n-2}e^x\ dx$$ $$=x^ne^x-nx^{n-1}e^x+n(n-1)x^{n-2}e^x-n(n-1)\int (n-2)x^{n-3}e^x\ dx$$ $$................$$ $=\left(x^n-nx^{n-1}+n(n-1)x^{n-2}-n(n-1)(n-2)x^{n-3}+\ldots +(-1)^{n+1}n!\right)e^x+C$

$\forall \ \ n\in N$

Find $\int x^ne^xdx$

4 Answers4