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

## 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! – Cody Johnson · 2 years, 9 months ago

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– Ethan Robinett · 2 years, 9 months ago

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!Log in to reply

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] – Kartik Sharma · 2 years, 6 months ago

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You could have just used the relation gamma(t)=(t-1)gamma(t-1). – Bogdan Simeonov · 2 years, 9 months ago

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– Ethan Robinett · 2 years, 9 months ago

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.Log in to reply

– Bogdan Simeonov · 2 years, 9 months ago

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 IBPLog in to reply

– Ethan Robinett · 2 years, 9 months ago

Oh I see what you're saying. Yeah good catch I could've done thatLog in to reply

The first integral converges iff \(t>0\). – Haroun Meghaichi · 2 years, 9 months ago

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