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Gamma Function

Let h⁢(x)=xα-1⁢exp⁡(-x) for 0<x<∞. The Gamma function determines how the integral of this function over the range (0,∞) varies with α.

Γ⁢(α)=∫0∞xα-1⁢exp⁡(-x)⁢dx,

for α>0. The Gamma function can be evaluated in R by the command gamma(a):

> gamma(0.5)        # Calculate the gamma function at 0.5
[1] 1.772454
> x_seq(1,10)       # Let x = (1,2,3,4,5,6,7,8,9,10)
> x
 [1]  1  2  3  4  5  6  7  8  9 10
> gamma(x)          # Calculate the gamma function at these values
 [1]  1    1    2    6    24   120   720   5040  40320 36288

Properties:

  1. 1.

    Recurrence relation (see below): Γ⁢(α+1)=α⁢Γ⁢(α) for α>0.

  2. 2.

    Positive integers (see notes): Γ⁢(1)=1, Γ⁢(α)=(α-1)! for α a positive integer.

  3. 3.

    Special (see below): Γ⁢(1/2)=π.

  4. 4.

    Limits (see below): Γ⁢(α)→∞ as α→0 or α→∞.

Proofs:

Proof that Γ⁢(α+1)=(α)⁢Γ⁢(α) for α>1.

Γ⁢(α+1) =∫0∞tα⁢exp⁡(-t)⁢dt
=[tα⁢(-1)⁢exp⁡(-t)]0∞+∫0∞α⁢tα-1⁢exp⁡(-t)⁢dt
=0-0+α⁢Γ⁢(α)

for α>0.

Proof that Γ⁢(1/2)=π.

In Math113 you saw that

I=∫-∞∞e-x2⁢dx=π.

Substituting x=t1/2, so d⁢x=t-1/2/2⁢d⁢t, we have, by symmetry,

I=2⁢∫0∞e-x2⁢dx=2⁢∫0∞e-t⁢t-1/2/2⁢dt=∫0∞t1/2-1⁢e-t⁢dt=Γ⁢(1/2).

Proof that Γ⁢(α)→∞ as α→0.

Γ⁢(α)>∫01xα-1⁢e-x⁢dx>∫01xα-1⁢e-1⁢dx=e-1⁢[xαα]01=e-1⁢α-1→∞

as α→0.

Proof that Γ⁢(α)→∞ as α→∞.

Γ⁢(α)>∫2∞xα-1⁢e-x⁢dx>∫2∞2α-1⁢e-x⁢dx=2α-1⁢[-e-x]2∞=2α-1⁢e-2→∞

as α→∞.