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 ?
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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).
Glorfindel
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M Turgeon
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@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 04:21
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@DaoyiPeng What would be a simple proof for you? – M Turgeon May 22 '12 at 12:29
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The proof is ill formatted! – Pedro May 22 '12 at 23:21
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@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 00:27
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@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. – Daoyi Peng May 23 '12 at 03:13
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@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). – M Turgeon May 28 '12 at 14:14
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1@MTurgeon Thanks! – Pedro May 28 '12 at 20:19