Fractional Gamma Functions

Earlier we explored how the Gamma function works. One of the more useful formulas for generalizing integral forms was found to be:

G(x + 1) = x G (x)


This will also be found very useful in working with fractional Gamma functions, as I will show in this article. Most solutions of fractional G (x) entail already knowing at least one basic form, usually obtained from a special integral.

For example, working with most fractionals we make use of the basic integral that generates:  G (½)  . This was actually the problem No. 3 in the last set. For those who consulted the link provided, they would have found:

G (½)  =   ò^¥ _o  [t^-1/2 exp(-t)]dt

The resulting integral yields:   G (½) = Öp

 
Now let's see how to obtain G (-½):

From the basic Gamma function formula (letting x = -½) :

G (-½) = G (-½ + 1) = -½ G (-½) Or:


G (-½) = -2G (½) = -2 Öp 

Another application, decimals - which are merely another form of fraction:

Say you wish to obtain G (-0.30)

(In this case, one is assumed to know the basic Gamma function G (1.70) = 0.90864)

In this case, from the Gamma function formula given earlier:

G (-0.30) = G (1 - 0.30) = -0.30 G (-0.30)

G (0.70)/ (-0.30) = G (-0.30)

But: G (0.70) = G (1.70)/ 0.70 = (0.90864)/ 0.70 = 1.29805


So: G (-0.30) = G (0.70)/ -0.30 = 1.29805/ -0.30 = -4.32683


Fractional sequences can also come into play, e.g.

Find: G (n + ½):    G (n + ½) = (n- ½) G (n – ½)

 
Since: we use x = (n – ½) in: G (x + 1) = xG (x)

 
G ([n – ½] + 1) = (n – ½) G (n – ½) 

  Þ G (n + ½) = (n – ½) G (n – ½)

And we can go further, focusing on treating the right hand side:

(n – ½) G (n – ½) = (n – ½) (n – 3/2) G (n –3/2)

= (n - ½)(n - 3/2) . .. . .3/2·½·G (½)


But, G (½) =  Öp, so:

 
G (n + ½) = (2n -1)/2 · (2n -3)/2 . . .. 3/2· ½· (p)^1/2

  Further factoring and additional algebra yields:


G (n + ½) = (2n - 1)! (p)^1/2 / 2ⁿ n! 

This is left as an exercise for the ambitious reader!


Problems for the Math Maven:

1) Find: G (3/2)

2) Find: G (-0.70)

3) Use the form: G (n + ½) = (2n - 1)! (p)^1/2 / 2ⁿ n!

and show it is equal to the value for G (3/2)  you obtained in (1)

4) Hence, or otherwise, compute: G (5/2)