Here we will prove: \( (t-1)! = \displaystyle \int_{0}^{\infty} x^{t-1}e^{-x} \,dx\)

with \(t \in \mathbb{R}\) , \(t \notin \mathbb{Z}^{-}\)

Which is the definition of the Gamma Function: \(\Gamma(t) = (t-1)! = \displaystyle \int_{0}^{\infty} x^{t-1}e^{-x} \,dx\)

Proof:

For brevity, let \(I=\displaystyle \int_{0}^{\infty} x^{t-1}e^{-x} \,dx\)

We proceed to integrate by parts, letting: \(u=x^{t-1}\) and \(dv=e^{-x}dx\)

\(\Rightarrow\) \(du=(t-1)x^{t-2}dx\) , \(v=-e^{-x}\)

\(\Rightarrow\) \(I= -x^{t-1}e^{-x} \displaystyle \mid_{0}^{\infty} + (t-1)\displaystyle \int_{0}^{\infty} x^{t-2}e^{-x}\,dx\)

Integrating by parts again: \(u=x^{t-2}\) and \(dv=e^{-x}dx\)

\(\Rightarrow\) \(du=(t-2)x^{t-3}\) , \(v=-e^{-x}\)

\(\Rightarrow\) \(I=- x^{t-1}e^{-x}-(t-1)x^{t-2}e^{-x} \displaystyle \mid_{0}^{\infty} + (t-1)(t-2)\displaystyle \int_{0}^{\infty} x^{t-3}e^{-x}\,dx\)

We now recognize a pattern that will continue (this pattern can easily be seen using the table method of integration by parts). It is clear that:

\(I=-e^{-x}[x^{t-1}+(t-1)x^{t-2}+(t-1)(t-2)x^{t-3}+ .... + (t-1)!]_{0}^{\infty}\)

Then: \(I= \displaystyle \lim_{x\to\infty} -e^{-x}[x^{t-1}+(t-1)x^{t-2}+(t-1)(t-2)x^{t-3}+ .... + (t-1)!]\)

\(+ (t-1)!\)

Now consider: \(\displaystyle \lim_{x\to\infty} -e^{-x}[x^{t-1}+(t-1)x^{t-2}+(t-1)(t-2)x^{t-3}+ .... + (t-1)!]\)

It is tempting to transform this limit to the form \(\frac{0}{0}\) and use L'Hopital's Rule. We will use L'Hopital's Rule, but we will transform this to the form \(\frac{\infty}{\infty}\).

We have: \( \displaystyle \lim_{x\to\infty} -e^{-x}[x^{t-1}+(t-1)x^{t-2}+(t-1)(t-2)x^{t-3}+ .... + (t-1)!]\)

\(= \displaystyle \lim_{x\to\infty} \frac{[x^{t-1}+(t-1)x^{t-2}+(t-1)(t-2)x^{t-3}+ .... + (t-1)!]}{\frac{1}{-e^{-x}}}\)

\(=\frac{\infty}{\infty}\)

Now, if we differentiate the numerator and denominator \(t+1\) times (this is the definition of L'Hopital's Rule), we have:

\(\frac{d^{t+1}}{dx^{t+1}} (x^{t-1}+(t-1)x^{t-2}+(t-1)(t-2)x^{t-3}+ .... + (t-1)!) = 0\)

\(\frac{d^{t+1}}{dx^{t+1}} \frac{1}{-e^{-x}} = \frac{1}{-e^{-x}}\)

Then we have: \(\displaystyle \lim_{x\to\infty} \frac{0}{\frac{1}{-e^{-x}}} = \frac{0}{\infty} = 0\)

Note that \(\frac{0}{\infty} \) is not an indeterminate form.

Recall that \(I= \displaystyle \lim_{x\to\infty} -e^{-x}[x^{t-1}+(t-1)x^{t-2}+(t-1)(t-2)x^{t-3}+ .... + (t-1)!]\)

\(+ (t-1)!\)

We just evaluated the limit at infinity and found it to be equal to zero. Then we have:

\(I=(t-1)!\)

\(\Rightarrow\) \( \displaystyle \int_{0}^{\infty} x^{t-1}e^{-x} \,dx = (t-1)!\)

Which was to be proved.

QED

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## Comments

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TopNewestI like the idea of creating a recursion using integration by parts. But, instead, you should use induction for complete rigor, rather than saying "from here we can recognize the pattern." Good article!

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I always thought it was a little "incomplete" to say something like that in a proof. All the teachers I've had in the past have done it that way, so that's how I got used to doing it, but it never really sat right with me. I never considered using induction, but thanks for the suggestion. That actually makes a lot more sense to me. I'm still relatively "new" in the proof-writing department so I really appreciate any guidance anyone is willing to give. Thanks!

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Is there any way(there has to be) to do it in reverse, I mean is there any way to prove that \((t-1)! = \int _{ 0 }^{ \infty }{ { x }^{ t-1 }{ e }^{ -x } } dx\) and not \(\int _{ 0 }^{ \infty }{ { x }^{ t-1 }{ e }^{ -x } } dx = (t-1)!\) {I hope you got it]

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You could have just used the relation gamma(t)=(t-1)gamma(t-1).

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I wanted to show that the integral actually converged to that. The relation you're talking about wouldn't exist if someone didn't establish that the integral was equal to the factorial.

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No, I meant just using the integration by parts only one time and the integral definition of the gamma function.Look at your note after the first IBP

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The first integral converges iff \(t>0\).

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