Appendix A: Special functions

\[ \require{physics} \require{braket} \]

\[ \newcommand{\dl}[1]{{\hspace{#1mu}\mathrm d}} \newcommand{\me}{{\mathrm e}} \]

\[ \newcommand{\Exp}{\operatorname{E}} \newcommand{\Var}{\operatorname{Var}} \newcommand{\Mode}{\operatorname{mode}} \]

\[ \newcommand{\pdfbinom}{{\tt binom}} \newcommand{\pdfbeta}{{\tt beta}} \newcommand{\pdfpois}{{\tt poisson}} \newcommand{\pdfgamma}{{\tt gamma}} \newcommand{\pdfnormal}{{\tt norm}} \newcommand{\pdfexp}{{\tt expon}} \]

\[ \newcommand{\distbinom}{\operatorname{B}} \newcommand{\distbeta}{\operatorname{Beta}} \newcommand{\distgamma}{\operatorname{Gamma}} \newcommand{\distexp}{\operatorname{Exp}} \newcommand{\distpois}{\operatorname{Poisson}} \newcommand{\distnormal}{\operatorname{\mathcal N}} \]

A.1 Gamma functions

Definition A.1 (Gamma function) Let \(z\) be any complex number that \(\mathfrak R(z)>0\). Then \[ \Gamma(z)=\int_0^{\infty}t^{z-1}\me^{-t}\dl3t. \tag{A.1}\]

Theorem A.1 \[ \Gamma(z+1)=z\Gamma(z). \tag{A.2}\]

Proof. \[ \begin{split} \Gamma(z+1)&=\int_0^{\infty}t^{z+1-1}\me^{-t}\dl3t=-\int_0^{\infty}t^{z+1-1}\dl3\me^{-t}\\ &=-t\me^{-t}\biggr\rvert_0^{\infty}+\int_0^{\infty}\me^{-t}\dl3t^{z}=\int_0^{\infty}zt^{z-1}\me^{-t}\dl3t=z\Gamma(z). \end{split} \]

A.2 Beta functions

Definition A.2 (Beta function) Let \(z_1\), \(z_2\) be two complex numbers that \(\mathfrak R(z_1),\mathfrak R(z_2)>0\). Then \[ B(z_1,z_2)=\int_0^1t^{z_1-1}(1-t)^{z_2-1}\dl3t. \tag{A.3}\]

Theorem A.2 (Relations between Beta functions and Gamma functions) \[ \Gamma(\alpha)\Gamma(\beta)=\Gamma(\alpha+\beta)B(\alpha, \beta). \tag{A.4}\]

Proof. Use the following trick to change a product of two integrals into a double integral. \[ \begin{align} \Gamma(\alpha)\Gamma(\beta)&=\int_0^{\infty}u^{\alpha-1}\me^{-u}\dl3u\int_0^{\infty}v^{\beta-1}\me^{-v}\dl3v\\ &=\int_0^{\infty}\int_0^{\infty}u^{\alpha-1}v^{\beta-1}\me^{-u-v}\dl3u\dl3v. \end{align} \]

Set \(u=st\), \(v=s-st\). Then \(s=u+v\), \(t=\dfrac{u}{u+v}\), and \(\abs{\dfrac{\partial(u,v)}{\partial(s,t)}}=\abs{\mqty[t&s\\1-t&-s]}=s\). Then \[\begin{split} \Gamma(\alpha)\Gamma(\beta)&=\int_0^{\infty}\int_0^{\infty}u^{\alpha-1}v^{\beta-1}\me^{-u-v}\dl3u\dl3v\\ &=\int_{v=0}^{v=\infty}\int_{u=0}^{u=\infty}(st)^{\alpha-1}(s(1-t))^{\beta-1}\me^{-s}s\dl3s\dl3t\\ &=\int_{t=0}^{t=1}\int_{s=0}^{s=\infty}s^{\alpha+\beta-1}t^{\alpha-1}(1-t)^{\beta-1}\me^{-s}\dl3s\dl3t\\ &=\int_{0}^{\infty}s^{\alpha+\beta-1}\me^{-s}\dl3s\int_{0}^{1}t^{\alpha-1}(1-t)^{\beta-1}\dl3t\\ &=\Gamma(\alpha+\beta)B(\alpha,\beta). \end{split}\]