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the gamma extensions

12/29/23

The Gamma function, $Γ$ , is a curious mathematical creature that pops up in many, seemingly unrelated places. Yet it defies naive intuition;

Over the positive integers, it simplifies to the factorial function, shifted by $1$ .
Over the negative integers, it diverges.
At $0$ , it diverges.
It has even more interesting behavior over non-integer fractions and complex numbers.

What makes $Γ$ so interesting?

Definition

The Gamma function is defined as follows:

Γ : C Γ (s) \to C = 0 \int \infty t^{s - 1} e^{- t} d t

It converges for all complex numbers $s$ with positive real part (i.e. $Re (s) > 0$ ). It also converges for complex numbers with negative real part (i.e. $Re (s) < 0$ ) except where $Re (s)$ is a negative integer or $0$ .

When $Re (s)$ is zero or a negative integer and the imaginary part of $s$ is $0$ , then $Γ (s)$ diverges and is undefined.

Over the Positive Integers

Note that the integers are embedded in the complex numbers as the subset obtained by setting the imaginary part to $0$ and restricting the real part accordingly.

Over the positive integers, $Γ$ has two general properties, outlined below.

Special Case

Consider $Γ (n)$ for $n = 1$ . The integral has a surprisingly simple simplification:

Γ (1) = 0 \int \infty t^{1 - 1} e^{- t} d t = 0 \int \infty e^{- t} d t (since t^{0} = 1) = [- e^{- t}]_{0}^{\infty} = 0 - (- 1) = 1

General Case: Positive Integers

Take $n$ to be an arbitrary positive integer, i.e. $n \in N$ and $n > 1$ . To prove that $Γ$ is indeed equivalent to the factorial function (albeit shifted by $1$ ), we need to show that $Γ (n + 1) = n Γ (n)$ .

Revisit the definition of $Γ$ , applied to $n + 1$ :

Γ (n + 1) = 0 \int \infty t^{n} e^{- t} d t

This integral can be simplified using the rule for integration by parts:

\int u d v = uv - \int v d u

Take $u = t^{n}$ and $d v = e^{- t} d t$ ; then:

d u v Thus, Γ (n + 1) = \frac{d}{d t} t^{n} = n t^{n - 1} = \int e^{- t} d t = - e^{- t} = 0 \int \infty t^{n} e^{- t} d t = 0 \int \infty u t^{n} \cdot d v e^{- t} d t = uv [- t^{n} e^{- t}]_{0}^{\infty} - 0 \int \infty v n t^{n - 1} \cdot d u (- e^{- t}) d t = (0 - 0) - 0 \int \infty - n t^{n - 1} e^{- t} d t = n 0 \int \infty t^{n - 1} e^{- t} d t = n \cdot Γ (n)

This is an important identity, usually expressed in any of the following two forms:

Γ (n + 1) Γ (n) = n Γ (n) = \frac{1}{n} Γ (n + 1)

Proof by Induction

So far, we have seen the following results:

Γ (1) Γ (n + 1) = 1 \equiv 0! = n Γ (n) for n \in N_{> 1}

By induction on $n$ , suppose we know that $Γ (n) = (n - 1)!$ (with the base-case being $Γ (1) = 0!$ ). We can compute $Γ (n + 1)$ as follows:

Γ (n + 1) = n \cdot Γ (n) = n \cdot (n - 1)! = n!

info

$Γ$ was not intended to be a generalization of the factorial function. The result was a somewhat accidental one, albeit useful.

Over the Reals

By restricting $Γ$ to the positive integers, we discovered that:

Γ (n) = (n - 1)! for n \in N

But what happens when we plug in a fraction, say $\frac{3}{2}$ ?

Γ (\frac{3}{2}) = Γ (\frac{1}{2} + 1) = \frac{1}{2} \cdot Γ (\frac{1}{2}) = \frac{1}{2} 0 \int \infty t^{- \frac{1}{2}} e^{- t} d t

To simplify the integral, let's try to get rid of the $t^{- \frac{1}{2}}$ term by substituting $t = x^{2}$ :

Γ (\frac{3}{2}) = \frac{1}{2} 0 \int \infty t^{- \frac{1}{2}} e^{- x^{2}} \cdot 2 t^{\frac{1}{2}} d x = 0 \int \infty e^{- x^{2}} d x

This is a familiar integral, known as the Gaussian integral:

I (a) I (\infty) I^{2} (\infty) = a \int a e^{- x^{2}} d x = \infty \int \infty e^{- x^{2}} d x = - \infty \int \infty e^{- x^{2}} d x \cdot - \infty \int \infty e^{- y^{2}} d y = - \infty \int \infty - \infty \int \infty e^{- (x^{2} + y^{2})} d x d y = - \infty \int \infty - \infty \int \infty e^{- r^{2}} r d r d θ = 0 \int 2 π 0 \int \infty e^{- r^{2}} r d r d θ = 2 π 0 \int \infty e^{- r^{2}} r d r = 2 π - \infty \int 0 \frac{1}{2} e^{u} d u (substituting u = - r^{2}) = π [e^{u}]_{- \infty}^{0} = π (e^{0} - e^{- \infty}) = π (1 - 0) = π

Since $Γ (\frac{3}{2})$ restricts the integration to $[0, \infty)$ and the function is even:

Γ (\frac{3}{2}) = \frac{1}{2} - \infty \int \infty e^{- x^{2}} d x = \frac{1}{2} \cdot I^{2} (\infty) = \frac{π}{2}

Other fractions of the form $(n + \frac{1}{2}), n \in N$ have similar results simplifications:

Γ (\frac{1}{2}) Γ (\frac{3}{2}) Γ (\frac{5}{2}) Γ (\frac{7}{2}) Γ (\frac{9}{2}) ⋮ = π = \frac{1}{2} π = \frac{3}{4} π = \frac{15}{8} π = \frac{105}{16} π

In general,

Γ (\frac{1}{n}) \sim n - γ

where $γ$ is the Euler-Mascheroni constant and $\sim$ denotes asymptotic equivalence.

but unfortunately the integral has to be computed or simplified directly for each case.

Over the Complexes

$Γ$ has a meromorphic extension to the complex numbers, with simple poles at the non-positive integers and $0$ . It is defined with the same rules and the relation

Γ (s + 1) = s Γ (s)

holds for all complex numbers $s$ . For example;

Γ (i) = (- 1 + i)! \approx - 0.1549 - 0.4980 i

Various values of $Γ$ are tabulated here.

For some useful applications of $Γ$ in the Riemann zeta function, see this post.

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