## Monday Math 19: The Gamma Function (Part 2/?)

The Gamma Function Part 2: Weierstrass form and Stirling’s formulas
(Part 1 is here)

Let us now define a sequence of functions . We note first that  has simple poles at 0, -1, -2, …, -n. Next,
,
and
.
We also look at the logarithm of  and the derivatives thereof:

Thus:


So  is convex.

Now, let us consider the limit of this sequence, which can be shown to exist via complex analysis:

We thus see from the above that  has poles at 0, -1, -2, -3,…; and the properties of  above tell us:
,
,
And  is convex. As we noted at the end of the previous part, the only logarithmically convex function , x>0, for which  and  is the gamma function. Thus
.

Next, I will remind readers of the Euler-Mascheroni constant, γ, which is defined by the limit:

giving γ=0.57721566…
We note:

Now, suppose we multiply each term in the product by , multiplying out front with the reciprocal terms:
.
Now, we take the limit as , and see that the limit of the bracketed terms in the first exponent is –γ, obtaining:
.
This in known as the Weierstrass form of the gamma function.

Also useful are Stirling’s approximations, which allow one to approximate  for large x, and n! for large n:


A derivation of the second approximation using Laplace’s method can be found here.

### 8 Responses to “Monday Math 19: The Gamma Function (Part 2/?)”

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