# proof of Gauss’ digamma theorem

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

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 (http://planetmath.org/ExtractingEveryNthTermOfASeries), the second term is

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

and therefore

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

Let $t\to 1^{-}$ to get

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

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

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

and so

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

But

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

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

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

where the last equality holds since

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

## References

Title proof of Gauss’ digamma theorem ProofOfGaussDigammaTheorem 2013-03-22 16:24:07 2013-03-22 16:24:07 rm50 (10146) rm50 (10146) 5 rm50 (10146) Proof msc 30D30 msc 33B15