PlanetMath: proof of Gauss' digamma theorem

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proof of Gauss' digamma theorem

(Proof)

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

$\displaystyle \Gamma(x+n)=(x+n-1)(x+n-2)\cdots x\Gamma(x)$

By the partial fraction decomposition satisfied by the $\psi$ function,

$\displaystyle \psi\left(\frac{p}{q}\right)+\gamma=\sum_{n=0}^{\infty}\left(\fra... ..._{t\to 1^-}\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{q}{p+nq}\right)t^{p+nq}$

using Abel's limit theorem.

Now,

$\displaystyle \sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{q}{p+nq}\right)t^{p+... ...0}^{\infty}\frac{t^{(n+1)q}}{n+1}\ -\ q\sum_{n=0}^{\infty}\frac{t^{p+nq}}{p+nq}$

Since

$\displaystyle -\ln(1-t)=\sum_{n=1}^{\infty} \frac{t^n}{n}$

the first term is

$\displaystyle -t^{p-q}\ln(1-t^q)$

Using the algorithm for extracting every $q^\mathrm{th}$ term of a series, the second term is

$\displaystyle \sum_{n=0}^{q-1}\omega^{-np}\ln(1-\omega^n t)$

and therefore

$\displaystyle \sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{q}{p+nq}\right)t^{p+nq}$	$\displaystyle =-t^{p-q}\ln(1-t^q)+\sum_{n=0}^{q-1}\omega^{-np}\ln(1-\omega^n t)$
	$\displaystyle =-t^{p-q}\ln\frac{1-t^q}{1-t}-(t^{p-1}-1)\ln(1-t)+\sum_{n=1}^{q-1}\omega^{-np}\ln(1-\omega^n t)$

Let $t\to 1^-$ to get

$\displaystyle \psi\left(\frac{p}{q}\right)=-\gamma-\ln q+\sum_{n=1}^{q-1}\omega^{-np}\ln(1-\omega^n)$

Replace

and add the two expressions to obtain

$\displaystyle \psi\left(\frac{p}{q}\right)+\psi\left(\frac{q-p}{q}\right)=-2\gamma-2\ln q+2\sum_{n=1}^{q-1}\cos\left(\frac{2\pi n p}{q}\right)\ln(1-\omega^n)$

The left side is real, so it is equal to the real part of the right side. But

$\displaystyle \Re(\ln(1-\omega^n))=\ln\lvert 1-\omega^n\rvert^{1/2}=\ln\left\lv... ...\pi n}{q}\right\rvert^{1/2}=\frac{1}{2}\ln\left(2-2\cos\frac{2\pi n}{q}\right)$

and so

$\displaystyle \psi\left(\frac{p}{q}\right)+\psi\left(\frac{q-p}{q}\right)=-2\ga... ...\sum_{n=1}^{q-1}\cos\left(\frac{2\pi n p}{q}\right)\ln(2-2\cos\frac{2\pi n}{q})$

(1)

But

$\displaystyle \psi(x)-\psi(1-x)=\frac{d}{dx}\ln(\Gamma(x)\Gamma(1-x))=-\pi\cot\pi x$

by the Euler reflection formula and thus

$\displaystyle \psi\left(\frac{p}{q}\right)-\psi\left(\frac{q-p}{q}\right)=-\pi\cot \frac{\pi p}{q}$

(2)

Add equations (1) and (2) to get

$\displaystyle \psi\left(\frac{p}{q}\right)$	$\displaystyle =-\gamma-\frac{\pi}{2}\cot\frac{\pi p}{q}-\ln q+\frac{1}{2}\sum_{n=1}^{q-1}\cos\frac{2\pi n p}{q}\ln\left(2-2\cos\frac{2\pi n}{q}\right)$
	$\displaystyle =-\gamma-\frac{\pi}{2}\cot\frac{\pi p}{q}-\ln q+\sum_{n=1}^{q-1}\cos\frac{2\pi n p}{q}\ln\left(2\sin\frac{\pi n}{q}\right)$

where the last equality holds since

$\displaystyle \ln(2-2\cos (2\theta))=\ln(2-2(1-2\sin^2\theta)=\ln(4\sin^2\theta)=2\ln(2\sin\theta)$

Bibliography

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, argument

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 803 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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