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Suppose we are asked to prove this one using induction:

$$k! = \int_0^\infty e^{-x}x^{k} dx \,\,\, (*)$$

For $k=0$, it is clear after evaluating the appropriate improper integral that,

$$0! = \int_0^{\infty} e^{-x} dx = 1$$

For the induction case, is it possible just to do this?

Assuming $(*)$ holds now, we just multiply everything by $(k+1)$ to both sides so that we get,

$$k!(k+1) = k\int_0^\infty e^{-x}x^{k} dx + \int_0^\infty e^{-x}x^{k} dx = (k+1)!$$

Therefore by induction, $(*)$ is true for all natural number with zero included. Do we need to even do integration by part here? I thought it is only needed if we want to specifically prove the special recursive relation of Gamma function.

Further steps:

\begin{align}k!(k+1) &= (k+1)! = (k+1)\int_0^\infty e^{-x}x^{k} dx \\ &= -e^{-x}x^{k+1}\Big|_0^{\infty} + (k+1)\int_0^\infty e^{-x}x^{k} dx \\ &= \int_0^\infty e^{-x}x^{k+1} \end{align}

vTx
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  • Yes, do use integration by parts. – André Nicolas Oct 15 '14 at 05:49
  • If you would mind, my question is why do we need to do that if we already assume $()$ is true as part of the inductive step? Can't we just multiply $()$ with $(k+1)$ to imply that $k+1$ case holds. – vTx Oct 15 '14 at 05:52
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    It's not entirely clear without doing integration by parts that the term you have in the middle of your inductive step is the same as $\int _0 ^\infty e^{-x}t^{k+1}dx$, which is what you need to show – Alan Oct 15 '14 at 05:53
  • ok I just added further steps, is it ok now? – vTx Oct 15 '14 at 06:10
  • You should reorganize the i.b.p. argument; as written it looks like you prove that $k!(k + 1) = (k + 1)!$, which is essentially the definition of the factorial. – Travis Willse Oct 15 '14 at 06:26
  • So you used integration by parts. It would be much clearer (and more likely to be graded as correct) if you started with $\int_0^\infty x^{k+1}e^{-x},dx$, let $u=x^{k+1}$ and $dv=e^{-x},dx$, then went through the integration by parts concluding the result is $(k+1)\int_0^\infty x^ke^{-x},dx$, and then used the induction hypothesis. – André Nicolas Oct 15 '14 at 06:27
  • Ok with that procedure in mind, will it look like this? We start with $\int_0^\infty e^{-x}x^{k+1}$ and after doing the intermediate step as well as invoking induction hypothesis, we conclude that it is equal to $(k+1)!$ Is it also okay if I start with what I have added in my question, then ends in integral only? (So that I don't imply the definition of factorial) edited the first post – vTx Oct 15 '14 at 06:43

1 Answers1

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Looking at your amended answer, I see that it is correct, but I am a little bothered by your exposition. That is probably my problem.

When doing integration by parts, I find it helpful to explicitly state the parts. I also first do it as an indefinite integral and then assign the limits.

From first principles, $(uv)' =uv'+u'v $, so $uv =\int uv' + \int u'v $ or $\int uv' = uv - \int u'v $.

In this case, $uv' =e^{-x}x^{k}dx $. If we choose $u = e^{-x}$ and $v' = x^k dx$, then $u' = -e^{-x}dx$ and $v = \frac{x^{k+1}}{k+1}$, so the result is $\int e^{-x}x^{k}dx =e^{-x}\frac{x^{k+1}}{k+1}-\int (-e^{-x})\frac{x^{k+1}}{k+1}dx $ or $\int e^{-x}x^{k+1}dx =(k+1)\int e^{-x}x^{k}dx-e^{-x}x^{k+1} $.

At this point, when we assign the limits $0$ to $\infty$, we get your results in the form $\int_0^{\infty} e^{-x}x^{k+1}dx =(k+1)\int_0^{\infty} e^{-x}x^{k}dx-e^{-x}x^{k+1}\big|_0^{\infty} =(k+1)\int_0^{\infty} e^{-x}x^{k}dx $.

marty cohen
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  • Now I'm confused, it is ordinary integration by part right? Why are they suddenly not equal? $\int_0^\infty e^{-x}x^{k+1} = -e^{-x}x^{k+1}\Big|_0^{\infty} + (k+1)\int_0^\infty e^{-x}x^{k} dx $ Then the first term will vanish as the upper terminal approaches infinity. – vTx Oct 15 '14 at 07:03
  • You should go the other way: start with $\int e^{-x}x^{k+1}dx$, integrate this by parts, and get the expression involving $\int e^{-x}x^{k}dx$. – marty cohen Oct 15 '14 at 07:21
  • Ok but this is equal right? $\int_0^\infty e^{-x}x^{k+1}= -e^{-x}x^{k+1}\Big|_0^{\infty} + (k+1)\int_0^\infty e^{-x}x^{k} dx$ Is it because the calculation involves improper integral you cannot simply switch the expression in the equality? Can you elaborate what do you mean that it is not equal? – vTx Oct 15 '14 at 07:24
  • I completely rewrote my answer. I think I would have liked a few more steps in yours written out, but I now think it was correct. – marty cohen Oct 15 '14 at 16:34
  • Great! Thanks for your explanation. – vTx Oct 16 '14 at 03:32