Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
add comment

1 Answer 1

up vote 4 down vote accepted

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).

share|improve this answer
    
@M Turgeon Thank you very much! I think it's helpful, but I can't find a simple proof. –  Daoyi Peng May 22 '12 at 4:21
    
@DaoyiPeng What would be a simple proof for you? –  M Turgeon May 22 '12 at 12:29
    
The proof is ill formatted! –  Pedro Tamaroff May 22 '12 at 23:21
    
@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. –  M Turgeon May 23 '12 at 0:27
1  
@MTurgeon Thanks! –  Pedro Tamaroff May 28 '12 at 20:19
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.