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What is the fractional derivative of the function $\pi \cot (\pi x)$?

I derived the following expression:

$(\pi \cot (\pi q))^{(p)}=-\frac{\zeta'(p+1,q)+(\psi(-p)+\gamma ) \zeta (p+1,q)}{\Gamma (-p)}-\Gamma (p+1) \zeta (p+1,1-q)$

where $\psi$ is digamma, $\zeta$ is Hurwitz zeta, $\zeta'$ is the derivative by first argument

I want to know whether it coincides with other, traditional definitions.

This gives the formula $(\cot (q))^{(p)}=-\frac{\zeta'(p+1,\frac q\pi)+(\psi(-p)+\gamma ) \zeta (p+1,\frac q\pi)}{\pi^{p+1}\Gamma (-p)}-\frac 1{\pi^{p+1}}\Gamma (p+1) \zeta (p+1,1-\frac q\pi)$ for fractional derivative of cotangent.

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You might want to tell us which definition you used to derive the expression; else we won't know which are "other" definitions. – joriki Apr 16 '12 at 11:49
As @joriki alludes to, you neglected to mention what definition you're using. The Riemann-Liouville and Caputo definitions sometimes give the same results, but not always. – J. M. Apr 17 '12 at 1:54
I did not use any particular definition, I derived the formula from completely different considerations (the derivation is only valid for cotangent). – Anixx Apr 17 '12 at 10:10
I presume you've seen this? – J. M. Apr 17 '12 at 10:47
Oh, thanks! Very interesting! – Anixx Apr 17 '12 at 10:49

Just a comment, but too long...

Maybe this ... Write down the derivative, second derivative, third derivative,... of cot(x) to get a sequence of polynomials in cot(x). $$\begin{align} \cot(x) &= \cot(x) \\(d/dx) \cot(x) &= -1-\cot(x)^2 \\(d/dx)^2 \cot(x) &= 2 \cot(x)+2 \cot(x)^3 \\(d/dx)^3 \cot(x) &= -2-8 \cot(x)^2-6 \cot(x)^4 \\(d/dx)^4 \cot(x) &= 16 \cot(x)+40 \cot(x)^3+24 \cot(x)^5 \\(d/dx)^5 \cot(x) &= -16-136 \cot(x)^2-240 \cot(x)^4-120 \cot(x)^6 \\(d/dx)^6 \cot(x) &= 272 \cot(x)+1232 \cot(x)^3+1680 \cot(x)^5+720 \cot(x)^7 \\(d/dx)^7 \cot(x) &= -272-3968 \cot(x)^2-12096 \cot(x)^4-13440 \cot(x)^6-5040 \cot(x)^8 \end{align}$$

RECOGNIZE those polynomials as some previously known orthogonal polynomials. See if there are fractional-index versions for those.

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Did you recognize it as some known polynomials? – Anixx Apr 17 '12 at 15:35
There is the expression $$(-i)^{n+1}2^n (iu-1)\sum_{k=0}^n \left(-\frac{iu+1}{2}\right)^k k!\left[{n}\atop{k}\right]$$ Familiar? I don't quite know myself... – J. M. Apr 17 '12 at 17:19

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