5
$\begingroup$

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 ?

$\endgroup$
3
$\begingroup$

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).

$\endgroup$
7
  • $\begingroup$ @M Turgeon Thank you very much! I think it's helpful, but I can't find a simple proof. $\endgroup$ – Daoyi Peng May 22 '12 at 4:21
  • $\begingroup$ @DaoyiPeng What would be a simple proof for you? $\endgroup$ – M Turgeon May 22 '12 at 12:29
  • $\begingroup$ The proof is ill formatted! $\endgroup$ – Pedro Tamaroff May 22 '12 at 23:21
  • $\begingroup$ @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. $\endgroup$ – M Turgeon May 23 '12 at 0:27
  • 1
    $\begingroup$ @MTurgeon Thanks! $\endgroup$ – Pedro Tamaroff May 28 '12 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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