Gamma Extensions

December 29, 2023

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The Gamma function, $\Gamma$ , is at the heart of important results in mathematics including the Riemann hypothesis.

It is defined as follows:

\begin{align} \Gamma \;\colon \C &\to \C~\nonumber \\ s &\mapsto \int\limits_0^\infty t^{s-1} e^{-t} \, \d t~\label{1} \end{align}

What makes $\eqref{1}$ so interesting?

Basic Properties

These are:

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.

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{align} \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{align}

General Case: Positive Integers

Take $n$ to be an arbitrary positive integer, i.e. $n \in \N$ 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{align*} \Gamma(n+1) &= \int\limits_0^\infty t^{n} e^{-t} \, \d t \\ \end{align*}

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

\begin{align*} \int u \, \d v = uv - \int v \, \d u \end{align*}

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

\begin{align*} \d u &= \frac{\d}{\d t} t^n = n t^{n-1} \\ v &= \int e^{-t} \, \d t = -e^{-t} \\ \end{align*}

Thus,

\begin{align*} \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{align*}

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

\begin{align*} \Gamma(n+1) &= n \Gamma(n) \\ \Gamma(n) &= \frac{1}{n} \Gamma(n+1) \end{align*}

Proof by Induction

So far, we have seen the following results:

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

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{align*} \Gamma(n+1) &= n \cdot \Gamma(n) \\ &= n \cdot (n-1) ! \\ &= n! \end{align*}

info

$\Gamma$ 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{align*} \Gamma(n) &= (n-1)! \; \; \text{for $n \in \N$} \end{align*}

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

\begin{align*} \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{align*}

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

\begin{align*} \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{align*}

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

\begin{align*} 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{align*}

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

\begin{align*} \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{align*}

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

\begin{align*} \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{align*}

In general,

\begin{align*} \Gamma\parens{\frac{1}{n}} \sim n - \gamma \end{align*}

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{align*} \Gamma(i) &= \parens{-1 + i}! \approx -0.1549 - 0.4980i \end{align*}

Various values of $\Gamma$ are tabulated here.

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

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