ProofOfGaussDigammaTheorem 2006-11-12 17:11:01 2006-11-13 08:09:44 Proof Gauss' digamma theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \textbf{Proof.} The proof follows the argument given in \cite{bib:andrews}, which in turn derives from that given in \cite{bib:jensen}. The first formula is the logarithmic derivative of \[\Gamma(x+n)=(x+n-1)(x+n-2)\cdots x\Gamma(x)\] By the partial fraction decomposition satisfied by the $\psi$ function, \[\psi\left(\frac{p}{q}\right)+\gamma=\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{q}{p+nq}\right)= \lim_{t\to 1^-}\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{q}{p+nq}\right)t^{p+nq}\] using Abel's limit theorem. Now, \[\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{q}{p+nq}\right)t^{p+nq}= \sum_{n=0}^{\infty}\frac{t^{p+nq}}{n+1}\ -\ \sum_{n=0}^{\infty}\frac{qt^{p+nq}}{p+nq}= t^{p-q}\sum_{n=0}^{\infty}\frac{t^{(n+1)q}}{n+1}\ -\ q\sum_{n=0}^{\infty}\frac{t^{p+nq}}{p+nq}\] Since \[-\ln(1-t)=\sum_{n=1}^{\infty} \frac{t^n}{n}\] the first term is \[-t^{p-q}\ln(1-t^q)\] Using the algorithm for \PMlinkname{extracting every $q^\mathrm{th}$ term of a series}{ExtractingEveryNthTermOfASeries}, the second term is \[\sum_{n=0}^{q-1}\omega^{-np}\ln(1-\omega^n t)\] and therefore \begin{align*}\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{q}{p+nq}\right)t^{p+nq}&=-t^{p-q}\ln(1-t^q)+\sum_{n=0}^{q-1}\omega^{-np}\ln(1-\omega^n t)\\ &=-t^{p-q}\ln\frac{1-t^q}{1-t}-(t^{p-1}-1)\ln(1-t)+\sum_{n=1}^{q-1}\omega^{-np}\ln(1-\omega^n t) \end{align*} Let $t\to 1^-$ to get \[\psi\left(\frac{p}{q}\right)=-\gamma-\ln q+\sum_{n=1}^{q-1}\omega^{-np}\ln(1-\omega^n)\] Replace $p$ by $q-p$ and add the two expressions to obtain \begin{equation*}\psi\left(\frac{p}{q}\right)+\psi\left(\frac{q-p}{q}\right)=-2\gamma-2\ln q+2\sum_{n=1}^{q-1}\cos\left(\frac{2\pi n p}{q}\right)\ln(1-\omega^n) \end{equation*} The left side is real, so it is equal to the real part of the right side. But \[\Re(\ln(1-\omega^n))=\ln\lvert 1-\omega^n\rvert^{1/2}=\ln\left\lvert \left(1-\cos\frac{2\pi n}{q}\right)^2+\sin^2\frac{2\pi n}{q}\right\rvert^{1/2}=\frac{1}{2}\ln\left(2-2\cos\frac{2\pi n}{q}\right) \] and so \begin{equation}\label{eqn:one}\psi\left(\frac{p}{q}\right)+\psi\left(\frac{q-p}{q}\right)=-2\gamma-2\ln q+\sum_{n=1}^{q-1}\cos\left(\frac{2\pi n p}{q}\right)\ln(2-2\cos\frac{2\pi n}{q})\end{equation} But \[\psi(x)-\psi(1-x)=\frac{d}{dx}\ln(\Gamma(x)\Gamma(1-x))=-\pi\cot\pi x\] by the Euler reflection formula and thus \begin{equation}\label{eqn:two}\psi\left(\frac{p}{q}\right)-\psi\left(\frac{q-p}{q}\right)=-\pi\cot \frac{\pi p}{q}\end{equation} Add equations (\ref{eqn:one}) and (\ref{eqn:two}) to get \begin{align*}\psi\left(\frac{p}{q}\right)&=-\gamma-\frac{\pi}{2}\cot\frac{\pi p}{q}-\ln q+\frac{1}{2}\sum_{n=1}^{q-1}\cos\frac{2\pi n p}{q}\ln\left(2-2\cos\frac{2\pi n}{q}\right)\\ &=-\gamma-\frac{\pi}{2}\cot\frac{\pi p}{q}-\ln q+\sum_{n=1}^{q-1}\cos\frac{2\pi n p}{q}\ln\left(2\sin\frac{\pi n}{q}\right) \end{align*} where the last equality holds since \[\ln(2-2\cos (2\theta))=\ln(2-2(1-2\sin^2\theta)=\ln(4\sin^2\theta)=2\ln(2\sin\theta)\] \begin{thebibliography}{10} \bibitem{bib:andrews} G.E.~Andrews, R.~Askey, R.~Roy, \emph{Special Functions}, Cambridge University Press, 2001. \bibitem{bib:jensen} J.L.~Jensen [1915-1916], An elementary exposition of the theory of the gamma function, \emph{Ann. Math.} \textbf{17}, 124-166. \end{thebibliography} \include{Footer}