The * fractional* gamma function is an offshoot of how the regular gamma function works. One of the more useful formulas for the latter, introduced in several earlier posts e.g.

is:

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) .

G (½) = ò

The resulting integral yields: Öp

^{¥ }_{o}[t^{-}^{1/2}exp(-t)] dtThe resulting integral yields: 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 form of the fractional gamma function appears as

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

Then 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 =

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

Fractional

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)

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

But, G (½) = Öp, so:

G (n + ½) = (2n -1)/2

Further factoring and additional algebra yields:

G (n + ½) = (2n - 1)! (p)

This is left as an exercise for the ambitious reader!

1) Find: G (3/2)

2) Find: G (-0.70)

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

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

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

G (-½) = -2G (½) = -2 Öp

Another form of the fractional gamma function appears as

*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

Then 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)

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}/ 2^{n}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}/ 2^{n}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)

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