Waste less time on Facebook — follow Brilliant.
×

Gamma Function

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

Note by Ethan Robinett
3 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

I 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 - 3 years, 4 months ago

Log in to reply

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!

Ethan Robinett - 3 years, 4 months ago

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 - 3 years ago

Log in to reply

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

Bogdan Simeonov - 3 years, 4 months ago

Log in to reply

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.

Ethan Robinett - 3 years, 4 months ago

Log in to reply

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

Bogdan Simeonov - 3 years, 4 months ago

Log in to reply

@Bogdan Simeonov Oh I see what you're saying. Yeah good catch I could've done that

Ethan Robinett - 3 years, 4 months ago

Log in to reply

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

Haroun Meghaichi - 3 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...