Here $\psi(z)$ is digamma function, $\Gamma(z)$ is gamma function. $$\psi(z)=\frac{{\Gamma}'(z)}{\Gamma(z)},$$ For positive integers $m$ and $k$ (with $m < k$), the digamma function may be expressed in terms of elementary functions as: $$\psi\left(\frac{m}{k}\right)=-\gamma-\ln(2k)-\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right)+2\sum^{[(k-1)/2]}_{n=1}\cos\left(\frac{2\pi nm}{k}\right)\ln\left(\sin \left(\frac{n\pi}{k}\right)\right). $$ How to prove it ?
How to prove Gauss's Digamma Theorem?
5
$\begingroup$
calculus
number-theory
special-functions
1 Answers
3
You can look at this, and the references therein.
Added: In fact, a quick Google search gives several references for the proof. Also, if the math does not render well, the Planetmath team suggests to switch the view style to HTML with pictures (you can choose at the bottom of the page).
-
0@M Turgeon Thank you very much! I think it's helpful, but I can't find a simple proof. – 2012-05-22
-
0@DaoyiPeng What would be a simple proof for you? – 2012-05-22
-
0The proof is ill formatted! – 2012-05-22
-
0@PeterTamaroff Well, I sent a comment so that someone check and fix it. Meanwhile, it is still possible to look at the source file and figure out what is not being processed. – 2012-05-23
-
0@M Turgeon I can understand it's proof,Thank you again!en,I am a Chinese student,My English is poor,I will do my best to understand your words. – 2012-05-23
-
0@PeterTamaroff The team at Planetmath suggests to switch from jsMath to HTML with pictures (at the bottom of the page, you can choose the view style). – 2012-05-28
-
1@MTurgeon Thanks! – 2012-05-28