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How does one prove the polygamma reflection formula:

$$\psi^{(n)}(1-z)+(-1)^{n+1}\psi^{(n)}(z)=(-1)^n \pi \frac{d^n}{d z^n} \cot \pi z $$

Do we have to invoke the power of contour integration and kernels? I searched the web but the proof was to be found nowhere.

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You just need to prove the reflection formula: $$ \psi(1-z)-\psi(z)=\pi\cot(\pi z)\tag{1}$$ then differentiate it multiple times. In order to prove $(1)$, let's start from the Weierstrass product for the $\Gamma$ function: $$\Gamma(t+1) = e^{-\gamma t}\prod_{n=1}^{+\infty}\left(1+\frac{t}{n}\right)^{-1}e^{\frac{t}{n}}\tag{2}$$ leading to: $$ \Gamma(z)\,\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}\tag{3} $$ (the reflection formula for the $\Gamma$ function), then consider the logarithmic derivative of $(3)$: $$\psi(z)-\psi(1-z) = \frac{d}{dz}\,\log\sin(\pi z)\tag{4} $$ and $(1)$ is proved.

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    $\begingroup$ Wow... Jack... super! I did not think that this formula is SOOOO easy to prove... I thought we would invoke contour integration or apply many acrobatics in order to prove that.. Thank you. $\endgroup$ – Tolaso Apr 3 '15 at 16:32

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