1
$\begingroup$

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.

  • 0
    I just added the definition of Balanced Polygamma to the above formula.2012-04-17

1 Answers 1

1

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.

  • 1
    There is the [expression](http://functions.wolfram.com/ElementaryFunctions/Cot/20/02/0003/) $(-i)^{n+1}2^n (iu-1)\sum_{k=0}^n \left(-\frac{iu+1}{2}\right)^k k!\left[{n}\atop{k}\right]$ Familiar? I don't quite know myself...2012-04-17