2023-01-gamma-properties.md

title	description	date
the gamma extensions	The Gamma function, $\Gamma$, is at the heart of important results in mathematics including the [Riemann hypothesis][riemann-hypothesis]. It is defined as follows: ```math \begin{aligned} \Gamma \;\colon \CC &\to \CC \\ s &\mapsto \int\limits_0^\infty t^{s-1} e^{-t} \, \d t \end{aligned} ``` What does it represent, and why is it touted as an [analytic continuation][analytic-continuation] of the factorial function? [riemann-hypothesis]: /math/2023-02-riemann-zeta-properties [analytic-continuation]: https://en.wikipedia.org/wiki/Analytic_continuation	2023-12-29 15:00:00 -0800

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

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

What makes $\Gamma$ so interesting?

Definition

The Gamma function is defined as follows:

$$\begin{aligned} \Gamma \;\colon \CC &\to \CC \\ \\\ \Gamma(s) &= \int\limits_0^\infty t^{s-1} e^{-t} \, \d t \end{aligned}$$

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 $\Gamma(s)$ 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, $\Gamma$ has two general properties, outlined below.

Special Case

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

$$\begin{aligned} \Gamma(1) &= \int\limits_0^\infty t^{1-1} e^{-t} \, \d t \\\ &= \int\limits_0^\infty e^{-t} \, \d t \;\; \text{(since $t^0 = 1$)} \\\ &= \biggl [ -e^{-t} \biggr ]_0^\infty \\\ &= 0 - (-1) \\\ &= 1 \end{aligned}$$

General Case: Positive Integers

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

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

$$\begin{aligned} \Gamma(n+1) &= \int\limits_0^\infty t^{n} e^{-t} \, \d t \\\ \end{aligned}$$

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

$$\begin{aligned} \int u \, \d v = uv - \int v \, \d u \end{aligned}$$

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

$$\begin{aligned} \d u &= \frac{\d}{\d t} t^n = n t^{n-1} \\\ v &= \int e^{-t} \, \d t = -e^{-t} \\ \\\ \mathrm{Thus,} \\ \\\ \Gamma(n+1) &= \int\limits_0^\infty t^{n} e^{-t} \, \d t \\\ &= \int\limits_0^\infty \underbrace{t^n}_{u} \cdot \underbrace{e^{-t}}_{\d v} \, \d t \\\ &= \underbrace{\biggl [ -t^n e^{-t} \biggr ]_0^\infty}_{uv} - \int\limits_0^\infty \underbrace{nt^{n-1}}_{v} \cdot \underbrace{\left(- e^{-t}\right)}_{\d u} \, \d t\\ \\\ &= (0 - 0) - \int\limits_0^\infty -nt^{n-1} e^{-t} \, \d t \\\ &= n \int\limits_0^\infty t^{n-1} e^{-t} \, \d t \\ \\\ &= n \cdot \Gamma(n) \\\ \end{aligned}$$

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

$$\begin{aligned} \Gamma(n+1) &= n \Gamma(n) \\\ \Gamma(n) &= \frac{1}{n} \Gamma(n+1) \end{aligned}$$

Proof by Induction

So far, we have seen the following results:

$$\begin{aligned} \Gamma(1) &= 1 \equiv 0! \\\ \Gamma(n+1) &= n \Gamma(n) \; \; \text{for $n \in \NN_{>1}$} \end{aligned}$$

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

$$\begin{aligned} \Gamma(n+1) &= n \cdot \Gamma(n) \\\ &= n \cdot (n-1) ! \\\ &= n! \end{aligned}$$

::alert $\Gamma$ 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 $\Gamma$ to the positive integers, we discovered that:

$$\begin{aligned} \Gamma(n) &= (n-1)! \; \; \text{for $n \in \NN$} \end{aligned}$$

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

$$\begin{aligned} \Gamma \left( \frac{3}{2} \right) &= \Gamma(\frac{1}{2} + 1) \\\ &= \frac{1}{2} \cdot \Gamma(\frac{1}{2}) \\\ &= \frac{1}{2} \int\limits_0^\infty t^{-\frac{1}{2}} e^{-t} \, \d t \\\ \end{aligned}$$

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

$$\begin{aligned} \Gamma\left(\frac{3}{2}\right) &= \frac{1}{2} \int\limits_0^\infty t^{-\frac{1}{2}} e^{-x^2} \cdot 2t^\frac{1}{2} \, \d x \\\ &= \int\limits_0^\infty e^{-x^2} \, \d x \\\ \end{aligned}$$

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

$$\begin{aligned} I(a) &= \int\limits_a^a e^{-x^2} \, \d x \\ \\\ I(\infty) &= \int\limits_\infty^\infty e^{-x^2} \, \d x \\ \\\ I^2(\infty) &= \left( \int\limits_{-\infty}^\infty e^{-x^2} \, \d x \right) \cdot \left( \int\limits_{-\infty}^\infty e^{-y^2} \, \d y \right) \\\ &= \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty e^{-(x^2 + y^2)} \, \d x \, \d y \\\ &= \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty e^{-r^2} \, r\, \d r \, \d \theta \\\ &= \int\limits_0^{2\pi} \int\limits_0^\infty e^{-r^2} \, r\, \d r \, \d \theta \\\ &= 2\pi \int\limits_0^\infty e^{-r^2} \, r\, \d r \\\ &= 2\pi \int\limits_{-\infty}^0 \frac{1}{2} e^{u} \, \d u \;\; \text{(substituting $u = -r^2$)} \\\ &= \pi \biggl [ e^u \biggr ]_{-\infty}^0 \\\ &= \pi (e^0 - e^{-\infty}) \\\ &= \pi (1 - 0) \\ &= \pi \end{aligned}$$

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

$$\begin{aligned} \Gamma\left(\frac{3}{2}\right) &= \frac{1}{2} \int\limits_{-\infty}^\infty e^{-x^2} \, \d x \\\ &= \frac{1}{2} \cdot \sqrt{I^2(\infty)} \\\ &= \frac{\sqrt{\pi}}{2} \\\ \end{aligned}$$

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

$$\begin{aligned} \Gamma\left(\frac{1}{2}\right) &= \sqrt{\pi} \\\ \Gamma\left(\frac{3}{2}\right) &= \frac{1}{2} \sqrt{\pi} \\\ \Gamma\left(\frac{5}{2}\right) &= \frac{3}{4} \sqrt{\pi} \\\ \Gamma\left(\frac{7}{2}\right) &= \frac{15}{8} \sqrt{\pi} \\\ \Gamma\left(\frac{9}{2}\right) &= \frac{105}{16} \sqrt{\pi} \\\ \vdots \end{aligned}$$

In general,

$$\begin{aligned} \Gamma\parens{\frac{1}{n}} \sim n - \gamma \end{aligned}$$

where $\gamma$ 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

$\Gamma$ 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

$$\displaystyle \Gamma(s+1) = s \Gamma(s)$$

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

$$\begin{aligned} \Gamma(i) &= \parens{-1 + i}! \approx -0.1549 - 0.4980i \end{aligned}$$

Various values of $\Gamma$ are tabulated here.

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