14.4: Proofs of (some) properties

Last updated
Save as PDF

Page ID: 6560

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Proofs of (some) properties of \(\Gamma\)

Property 1. This is clear since the integral converges absolutely for \(\text{Re} (z) > 0\).

Property 2. We know (see the Laplace table) \(\mathcal{L} (t^n; s) = \dfrac{n!}{s^{n + 1}}\). Setting \(s = 1\) and using the corollary to the claim above we get

\[\Gamma (n + 1) = \mathcal{L} (t^n; 1) = n!. \nonumber \]

(We could also prove this formula directly from the integral definition of of \(\Gamma (z)\).)

Property 3. We could do this relatively easily using integration by parts, but let’s continue using the Laplace transform. Let \(f(t) = t^z\). We know

\[\mathcal{L} (f, s) = \dfrac{\Gamma (z + 1)}{s^{z + 1}} \nonumber \]

Now assume \(\text{Re} (z) > 0\), so \(f(0) = 0\). Then \(f' = zt^{z - 1}\) and we can compute \(\mathcal{L} (f';s)\) two ways.

\[\mathcal{L} (f';s) = \mathcal{L} (zt^{z - 1}; s) = \dfrac{z \Gamma (z)}{s^z} \nonumber \]

\[\mathcal{L} (f';s) = s \mathcal{L} (t^{z}; s) = \dfrac{\Gamma (z + 1)}{s^z} \nonumber \]

Comparing these two equations we get property 3 for \(\text{Re} (z) > 0\).

Property 4. We’ll need the following notation for regions in the plane.

\[\begin{array} {l} {B_0 = \{\text{Re} (z) > 0\}} \\ {B_1 = \{\text{Re} (z) > -1\} - \{0\}} \\ {B_2 = \{\text{Re} (z) > -2\} - \{0, -1\}} \\ {B_n = \{\text{Re} (z) > -n\} - \{0, -1, \ ..., -n + 1\}} \end{array} \nonumber \]

So far we know that \(\Gamma (z)\) is defined and analytic on \(B_0\). Our strategy is to use Property 3 to analytically continue \(\Gamma\) from \(B_0\) to \(B_n\). Along the way we will compute the residues at 0 and the negative integers.

Rewrite Property 3 as

\[\Gamma (z) = \dfrac{\Gamma (z + 1)}{z} \nonumber \]

The right side of this equation is analytic on \(B_1\). Since it agrees with \(\Gamma (z)\) on \(B_0\) it represents an analytic continuation from \(B_0\) to \(B_1\). We easily compute

\[\text{Res} (\Gamma, 0) = \lim_{z \to 0} z \Gamma (z) = \Gamma (1) = 1. \nonumber \]

Similarly, Equation 14.5.6 can be expressed as \(\Gamma (z + 1) = \dfrac{\Gamma (z + 2)}{z + 1}\). So,

\[\Gamma (z) = \dfrac{\Gamma (z + 1)}{z} = \dfrac{\Gamma (z + 2)}{(z + 1) z} \nonumber \]

The right side of this equation is analytic on \(B_2\). Since it agrees with \(\Gamma\) on \(B_0\) it is an analytic continuation to \(B_2\). The residue at −1 is

\[\text{Res} (\Gamma, -1) = \lim_{z \to -1} (z + 1) \Gamma (z) = \dfrac{\Gamma (1)}{-1} = -1. \nonumber \]

We can iterate this procedure as far as we want

\[\Gamma (z) = \dfrac{\Gamma (z + m + 1)}{(z + m) (z + m - 1) + \ ... + (z + 1)z} \nonumber \]

The right side of this equation is analytic on \(B_{m + 1}\). Since it agrees with \(\Gamma\) on \(B_0\) it is an analytic continuation to \(B_{m + 1}\). The residue at \(-m\) is

\[\text{Res} (\Gamma, -m) = \lim_{z \to -m} (z + m) \Gamma (z) = \dfrac{\Gamma (1)}{(-1)(-2)\ ... (-m)} = \dfrac{(-1)^m}{m!}. \nonumber \]

We’ll leave the proofs of Properties 5-8 to another class!