How to prove The Gauss Digamma theorem

Let $p/q$ be a rational number with $0<p<q$ then
$$\psi\left(\frac{p}{q}\right) = -\gamma-\log(2q)-\frac{\pi}{2}\cot\left(\frac{p}{q}\pi\right)+2\sum_{k=1}^{q/2-1} \cos\left(\frac{2\pi pk}{q}\right)\log\left[ \sin \left(\frac{\pi k}{q}\right) \right]$$

If some how it can make it simpler you can take the case $p=1$.

I feel I should use the following formulas

$$\sum_{r=1}^m \psi \left( \frac{r}{m}\right) = -m(\ln(m)+\gamma)$$

$$\sum_{r=1}^m \psi \left( \frac{r}{m}\right) \mathrm{exp}{\frac{2\pi rk i}{m}}= m\log\left(1-\mathrm{exp} \frac{2k\pi i}{m}\right)$$

Maybe a Fourier series expansion is my best bet but it is not my best category.


I am aware of How to prove Gauss's Digamma Theorem? but seems like that question didn't recieve that much attenstion also the supposed-to-be accepted answer provides a link that doesn't work anymore. I think it is better to have our sloution here for future refrences.

  • $\begingroup$ Another quite simple proof follows from a contour integration. $\endgroup$ – Machinato Feb 15 at 12:13

From PlanetMath:

We know that $$\Gamma(x+n) =(x+n-1)(x+n-2)\cdots x\Gamma(x)$$ By the partial fraction decomposition associated with the $\psi $ function, $$\psi ({\frac {p}{q}}) +\gamma = \sum_{n=0}^{\infty}(\frac {1}{n+1}-\frac {q}{p+nq}) =\lim_{t \to 1^{-}} \sum_{n=0}^{\infty}(\frac {1}{n+1}-\frac {q}{p+nq})t^{p+nq} $$ using Abel's limit theorem. Now, $$\sum_{n=0}^{\infty}(\frac {1}{n+1} -\frac {q}{p+nq})t^{p+nq} = \sum_{n=0}^{\infty}\frac {t^{p+nq}}{n+1} -\frac {qt^{p+nq}}{p+nq} =t^{p-q}\sum_{n=0}^{\infty} \frac {t^{(n+1)q}}{n+1}-q\sum_{n=0}^{\infty} \frac {t^{p+nq}}{p+nq}...(1) $$

Since $-\ln (1-t) =\sum_{n=1}^{\infty}t^n $, the first term is $-t^{-p-q}\ln (1-t^q) $. Using the algorithm for extracting every qth term of the series the second term is $\sum_{n=0}^{q-1} \omega^{-np}\ln (1-\omega^nt)$ and therefore we can write $(1) $ as equal to:$$ -t^{-p-q}\ln \frac {1-t^q}{1-t} -(t^{p-1}-1)\ln (1-t) +\sum_{n=0}^{q-1} \omega^{-np}\ln(1-\omega ^nt)$$

Let $t \to 1^- $ to get $$\psi (\frac {p}{q}) =-\gamma-\ln q +\sum_{n=1}^{q-1}\omega^{-np}\ln (1-\omega^n) $$ Replacing $p $ By $q-p $ and adding gives us, $$\psi (\frac {p}{q}) +\psi (\frac {q-p}{q}) = -2\gamma-2\ln q +2\sum_{n=1}^{q-1}\cos (\frac {2\pi np}{q })\ln (1-\omega^n) $$ The LHS is real, so it is equal to the real part of the RHS. But, $$\mathbb R(\ln (1-\omega^n)) = \ln|1-\omega^n|^{1/2} =\ln|(1-\cos (\frac {2\pi n}{q}))^2 +\sin \frac {2\pi n}{q} ^2|^{1/2} = \frac {1}{2}\ln (2-2\cos \frac {2\pi n}{q}) $$

And so, $$\psi(\frac {p}{q}) +\psi( \frac{q-p}{q}) = -2\gamma-2\ln q+\sum_{n=1}^{q-1} \cos \frac {2\pi np}{q}\ln (2-2\cos \frac{2\pi n}{q})...(2) $$ 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, $$\psi (\frac {p}{q})-\psi (\frac {q-p}{q}) =-\pi \cot \frac {\pi p}{q}...(3) $$

Now add $(2) $ and $(3) $ to get, $$\psi (\frac {p}{q}) =-\gamma-\frac {\pi}{2}\cot \frac {\pi p}{q} -\ln q +\sum_{n=1}^{q-1}\cos \frac {2\pi np}{q}\ln (\sin \frac {2\pi n}{q}) $$ where we have used the fact that $\ln (2-2\cos 2x) =2\ln (2\sin x) $. Hope it helps.


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