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I was motivated by this question on the various applications of integration by parts to prove the following integral:

$$\int^\infty_0 x^n e^{-x} \, dx = n!$$

Here's what I have done, I feel I am very close and am just having trouble with the simplification process.

$$\text{Let} \quad I_n = \lim_{t\to \infty} \int^t_0 x^n e^{-x} \, dx$$

then by letting $u = x^n$ and $v = e^{-x}$. We have

\begin{align} I_n &= \lim_{t\rightarrow \infty} \int^t_0 x^n \int e^{-x} \, dx - \int \left[\frac{du}{dx} \cdot \int e^{-x} dx\right] \, dx \\ I_n &= \lim_{t\rightarrow \infty} \int^t_0 -x^n \cdot e^{-x} + \int nx^{n-1} e^{-x} dx \\ I_n &= \lim_{t\rightarrow \infty} \int^t_0 -x^n \cdot e^{-x} + n \int x^{n-1}e^{-x} dx \\ \end{align}

If we continue this process ($n=3$) of integration by parts, we find something like the following:

\begin{align} I_n &= \lim_{t\rightarrow \infty} \int^t_0 -x^n \cdot e^{-x} + nx^{n-1}e^{-x} + n(n-1)x^{n-2}e^{-x}+n(n-1)(n-2)\int x^{n-3}e^{-x} dx \\ I_n &= \lim_{t\rightarrow \infty} \int^t_0 x^ne^{-x} \left[-1 +nx^{-1} + n(n-1)x^{-2}\right]+n(n-1)(n-2)I_{n-3} \end{align}

so, I can already see the factorial function forming and I'm assuming as you do this more and more, it will become more noticeable. The main problem, I'm having is simplifying the last line.

One question in two parts:

  1. Can someone tell me if I'm on the right track?
  2. and please, provide a small hint to finish the problem

Thanks a lot for your help!

Community
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Jeel Shah
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  • You made it nicely ! The only thing you did not notice is that $n(n-1)(n-2)$ is the beginning of $n!$ and that the left integral is zero at the limit. – Claude Leibovici Jul 05 '14 at 08:35
  • You're right! I kind of realized the factorial function forming and as a consequence realized that the left has to be zero but I didn't know what to do with right integral! I guess now with the answer provided I should be able to complete the last line. – Jeel Shah Jul 05 '14 at 08:40
  • Just start at $I_0$ and $I_1$. It will suffice. Cheers. – Claude Leibovici Jul 05 '14 at 08:51

1 Answers1

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You just need to do it once and from that, get a recursion relation: $$\begin{align*} I_n &= \int_{x=0}^\infty x^n e^{-x} \, dx \\ &= \lim_{t \to \infty} \Bigl[ -x^n e^{-x} \Bigr]_{x=0}^t + n \int_{x=0}^\infty x^{n-1} e^{-x} \, dx \\ &= \lim_{t \to \infty} - t^n e^{-t} + n I_{n-1} \\ &= n I_{n-1} . \end{align*}$$ Then it is trivial to calculate $I_0 = 1$, from which it is easy to see that $I_n = n!$ for nonnegative integers $n$. That's the least amount of information I can give away--you've essentially already solved the question, you just need to know when to evaluate and simplify and distill your calculations to make things clear.

heropup
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  • Does it suffice if I have the right limit evaluate to $0$ and then left over I have $I_n = n(n-1)(n-2)I_{n-3}$ then this implies that that we have $I_n = n!$. Does that make sense? – Jeel Shah Jul 05 '14 at 08:59
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    As my answer pointed out, it suffices to perform the integration by parts once. If you do it three times, you get a relationship between $I_n$ and $I_{n-3}$ that does not explicitly establish the intermediate values of $I_{n-1}$ and $I_{n-2}$, making your calculation unnecessarily more complicated. – heropup Jul 05 '14 at 16:30