# 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, 10 months ago

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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!

- 3 years, 10 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!

- 3 years, 10 months ago

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]

- 3 years, 7 months ago

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

- 3 years, 10 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.

- 3 years, 10 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 IBP

- 3 years, 10 months ago

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

- 3 years, 10 months ago

The first integral converges iff $$t>0$$.

- 3 years, 10 months ago