Gamma function

From Conservapedia

$\frac{d}{dx} \sin x=?\,$

This article/section deals with mathematical concepts appropriate for a student in late high school or early university.

The Gamma funciton is defined as

$\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt$

Relations and values

Factorial

Using integration by parts,

$\Gamma(z) = \int_{0}^{\infty}t^{z-1}d(e^{-t}dt)$

$= \left[t^{z-1}(-e^{-t}) \right]_0^{\infty}- \int_{0}^{\infty}(z-1)t^{(z-2)}(-e^{-t})dt.$

At $t = 0$ , $t z - 1 ( - e - t)$ goes to 0. Using L'Hopital's rule, it's easy to show that $\lim_{t\to \infty} \frac{t^{z-1}}{e^{t}} = 0$ .

So,

$=(z-1) \int_{0}^{\infty}t^{(z-1)-1}e^{-t}dt$

= (z - 1)Γ(z - 1).

Thus,

Γ(z) = (z - 1)Γ(z - 1).

Using this, and the the fact that $Γ(1) = 1$ , then we can get the factorial function. For $n$ a positive integer,

Γ(n) = (n - 1)!.

z=1/2

For z=1/2,

$\Gamma(1/2) = \int_{0}^{\infty}t^{-1/2}e^{-t}dt$

Substituting $x = t 1 / 2$ ,

$=\int_{0}^{\infty} (\frac{1}{x}) e^{-x^2} (2xdx )$

$=2 \int_{0}^{\infty} e^{-x^2}dx$

$= \int_{-\infty}^{\infty} e^{-x^2}dx$

where in the last line we used the fact that $e^{-x^2}$ is an even function. The integral is called the Gaussian integral and has a well known value of $\sqrt\pi$ .

Thus,

$\Gamma(1/2) = \sqrt\pi.$

Gamma function

From Conservapedia

Relations and values

Factorial

z=1/2

Views

Personal tools

Search

Popular Links

Help

World History

Edit Console