Gamma function

<math>\frac{d}{dx} \sin x=?\,</math>

This article/section deals with mathematical concepts appropriate for late high school or early college.

The Gamma funciton is defined as

<math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt</math>

Relations and values

Factorial

Using integration by parts,

<math>\Gamma(z) = \int_{0}^{\infty}t^{z-1}d(e^{-t}dt)</math>

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

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

So,

<math>=(z-1) \int_{0}^{\infty}t^{(z-1)-1}e^{-t}dt</math>

<math>=(z-1) \Gamma(z-1).</math>

Thus,

<math> \Gamma (z) = (z-1) \Gamma(z-1).</math>

Using this, and the fact that <math>\Gamma(1)=1</math>, then we can get the factorial function. For <math>n</math> a positive integer,

<math> \Gamma(n) = (n-1)!. </math>

z=1/2

For z=1/2,

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

Substituting <math>x=t^{1/2}</math>,

<math>=\int_{0}^{\infty} (\frac{1}{x}) e^{-x^2} (2xdx )</math>

<math>=2 \int_{0}^{\infty} e^{-x^2}dx </math>

<math>= \int_{-\infty}^{\infty} e^{-x^2}dx </math>

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

Thus,

<math> \Gamma(1/2) = \sqrt\pi. </math>