# Particular Values of Gamma Function

1. Mar 14, 2013

### GreenPrint

Hi,

I would like to know why is it that
gamma(1/2) = squareroot(pi)

The other values as well. I don't see why out of all values that the gamma function converges to operations of pi of all things. Is there some kind of relationship to the gamma function and the trigonometric function. I don't see how else one would know that it converges exactly to the square root of pi... One could plug it into a calculator and would get that it's about the square root pi due to the fact that the calculator would round off to so many digits but how does one know it's' exactly the square root of pi? There has to be some other way to know this as I don't see how else this can be known.

Thanks.

2. Mar 14, 2013

### CompuChip

Assuming you are using $\Gamma(z) = \int t^{z-1} e^{-t} \, dt$ you can substitute $x = \sqrt{t}$ to reduce $\Gamma(\tfrac12)$ to a Gaussian integral.

3. Mar 14, 2013

### LCKurtz

$$\Gamma(x) =\int_0^\infty t^{x-1}e^{-t}\, dt$$ so$$\Gamma(\frac 1 2) =\int_0^\infty t^{-\frac 1 2}e^{-t}\, dt$$Let $t = u^2,\, dt=2udu$:$$=2\int_0^\infty e^{-u^2}\, du$$Are you familiar with how to evaluate that integral?

 Apparently CompuChip types faster than I do.

Last edited: Mar 14, 2013
4. Mar 14, 2013

### CompuChip

Yeah, well, I had less $\LaTeX$ code to type so it's not really fair.

5. Mar 14, 2013

### GreenPrint

I do not know how to evaluate those integrals. I however was able to look up the definition of the error function and got the square root of pi.

How then do you evaluate the gamma function for any given input x? I'm assuming you have to get it into a form of the error function one way or the other but I can't seem to figure it out.

thanks for all of your help.

6. Mar 14, 2013

### Ray Vickson

For most values of x the error function has nothing to do with the problem. People have developed sophisticated numerical methods to compute $\Gamma(x)$ for general given values of x.

7. Mar 14, 2013

### SteamKing

Staff Emeritus
8. Mar 15, 2013

### CompuChip

Yes, if you want to integrate exp(-x²) over a finite interval you will need the error function. But if you are interested in
$$\int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi},$$
that result can be proven without defining erf (see e.g. Wikipedia). The trick you need is to square the integral and switch to polar coordinates, once you have seen that proof the occurrence of pi is hopefully no longer a mystery either.

For some other values can calculate the particular value by noting that $\Gamma(z + 1) = z \Gamma(z)$ (proof by partial integration), which immediately gives you, for example, that $\Gamma(\tfrac32) = \tfrac12 \sqrt{\pi}$ as well as $\Gamma(n + 1) = n!$ (by induction and performing the trivial integral $\Gamma(1) = \int_0^\infty e^{-t} \, dt = 1$).

Last edited: Mar 15, 2013