proof of Gauss’ digamma theorem

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

$$\mathrm{\Gamma }(x+n)=(x+n-1)(x+n-2)\mathrm{\cdots }x\mathrm{\Gamma }(x)$$

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

$$\psi \left(\frac{p}{q}\right)+\gamma =\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{n+1}-\frac{q}{p+nq}\right)=\underset{t\to 1^{-}}{lim}\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{n+1}-\frac{q}{p+nq}\right)t^{p+nq}$$

using Abel’s limit theorem.

Now,

$$\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{n+1}-\frac{q}{p+nq}\right)t^{p+nq}=\sum _{n=0}^{\mathrm{\infty }}\frac{t^{p+nq}}{n+1}-\sum _{n=0}^{\mathrm{\infty }}\frac{q{t}^{p+nq}}{p+nq}=t^{p-q}\sum _{n=0}^{\mathrm{\infty }}\frac{t^{(n+1)q}}{n+1}-q\sum _{n=0}^{\mathrm{\infty }}\frac{t^{p+nq}}{p+nq}$$

Since

$$-\ln (1-t)=\sum _{n=1}^{\mathrm{\infty }}\frac{t^n}{n}$$

the first term is

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

Using the algorithm for extracting every $q^{\mathrm{th}}$ term of a series (http://planetmath.org/ExtractingEveryNthTermOfASeries), the second term is

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

and therefore

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

$=-t^{p-q}\ln {\displaystyle \frac{1-t^q}{1-t}}-(t^{p-1}-1)\ln (1-t)+{\displaystyle \sum _{n=1}^{q-1}}{\omega }^{-np}\ln (1-{\omega }^{n}t)$

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

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

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

$$\mathrm{\Re }(\ln (1-{\omega }^n))=\ln {|1-{\omega }^n|}^{1/2}=\ln {|{(1-\cos \frac{2\pi n}{q})}^2+{\sin }^2\frac{2\pi n}{q}|}^{1/2}=\frac{1}{2}\ln (2-2\cos \frac{2\pi n}{q})$$

and so

$$\psi \left(\frac{p}{q}\right)+\psi \left(\frac{q-p}{q}\right)=-2\gamma -2\ln q+\sum _{n=1}^{q-1}\cos \left(\frac{2\pi np}{q}\right)\ln (2-2\cos \frac{2\pi n}{q})$$(1)

But

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

by the Euler reflection formula and thus

$$\psi \left(\frac{p}{q}\right)-\psi \left(\frac{q-p}{q}\right)=-\pi \cot \frac{\pi p}{q}$$(2)

Add equations (1) and (2) to get

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

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

where the last equality holds since

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