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proof of Gauss' digamma theorem
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(Proof)
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Proof. The proof follows the argument given in [1], which in turn derives from that given in [2].
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 extracting every $q^\mathrm{th}$ term of a series, the second term is $$ \sum_{n=0}^{q-1}\omega^{-np}\ln(1-\omega^n t $$ and therefore
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 $$ \psi(x)-\psi(1-x)=\frac{d}{dx}\ln(\Gamma(x)\Gamma(1-x))=-\pi\cot\pi $$ 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 (![[*]](http://images.planetmath.org:8080/cache/objects/8550/js//usr/share/latex2html/icons/crossref.png "[*]") ) and () to get
where the last equality holds since $$ \ln(2-2\cos (2\theta))=\ln(2-2(1-2\sin^2\theta)=\ln(4\sin^2\theta)=2\ln(2\sin\theta $$
- 1
- G.E. Andrews, R. Askey, R. Roy, Special Functions, Cambridge University Press, 2001.
- 2
- J.L. Jensen [1915-1916], An elementary exposition of the theory of the gamma function, Ann. Math. 17, 124-166.
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Cross-references: equality, equations, Euler reflection formula, right, real part, real, side, expressions, algorithm, term, Abel's limit theorem, function, decomposition, partial fraction, logarithmic derivative, formula, argument, proof
This is version 2 of proof of Gauss' digamma theorem, born on 2006-11-12, modified 2006-11-13.
Object id is 8550, canonical name is ProofOfGaussDigammaTheorem.
Accessed 1389 times total.
Classification:
| AMS MSC: | 33B15 (Special functions :: Elementary classical functions :: Gamma, beta and polygamma functions) | | | 30D30 (Functions of a complex variable :: Entire and meromorphic functions, and related topics :: Meromorphic functions, general theory) |
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Pending Errata and Addenda
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