I wonder if there any established formula for fractional derivative of a 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

and I want to compare it with the other expressions, and, possibly, equate them to derive further results.

My derivation is as follows.

First of all there is a known formula that is only valid for integer $n$:

$\psi_n(z)=(-1)^{(n+1)}n!\zeta(n+1,z)$ (can be seen here: http://mathworld.wolfram.com/PolygammaFunction.html, formula 12).

Espinoza and Moll mention in their article on the definition of balanced polygamma that they were unable to find a generalization for this formula despite any attempts.

If to try to directly expand the formula to non-integer values, the resulting function will be non-real. If to replace the $(-1)^n$ with a cosine, the resulting polygamma generalization will be undefined at negative integer values. Red line on this graph: alt text

This is highly undesirable because many formulas for integrals and discrete integrals use negapolygamma at negative integer order. If we take a traditional or balanced polygamma, multiplying it by cosine or another simple fiunction will not give the zeta function. Even more the resulting function will not be discrete-analytic. So this approach is also undesirable.

Following this a thought came to my mind that the alternating sign naturally arises if we differentiate a function of negative argument.

Since $\psi(1-x)=\psi(x) + \pi\cot\pi x$ we can rewrite the previously cited formula in the following form:

$(\psi(x) + \pi\cot\pi x)^{(n)}=-n! \zeta(n+1,1-x)$

By taking different definitions of fractional derivative of cotangent one can arrive at different generalizations of polygamma.

The formula also works in the opposite way: by taking a given generalization of polygamma one can arrive at different expressions for fractional derivative of cotangent.


$(\pi\cot\pi x)^{(n)}=-n! \zeta(n+1,1-x)-\psi^{(n)}(x)$

If we assume the polygamma being the balanced generalization, the formula

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


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.

Below is the graphic of function $2\cot x \csc x^2$ which is the second derivative of cotangent and the 1.9999th derivative of cotangent following the above formula. The graphs seem to coincide.

I wonder whether there are known other generalizations of fractional derivative of cotangent and which corresponding generalizations of polygamma will arise if to insert those in the previous formula.

alt tex


I found a document by Mauro Bologna from Chile, "Short Introduction to Fractional Calculus" (PDF link), which shows this on p.52:

    enter image description here
where $\cos_\mu$ and $\sin_\mu$ are "generalized" sine and cosine functions (p.50 of that document).

(I post this without penetrating the mysteries of this topic.)


I am not in conditions of giving an answer to your question. We have several questions involved in the computation. I presume that you are considering that the function is defined in Z. If it is the case, it can be expanded in series with infinite order one poles. Considering it as a complex variable function, we should use the generalised Cauchy integral that creates several problems because it needs an amalytical function over one half plane. So, let us consider that we work in R. We have a series of functions of the type 1/(t-k.pi). This means that we have to compute the derivative of t^(-1) and its translated. Here we must be careful, because most results found in books give the derivative of t^(-1) but they omit that it is valid only for t>0. If t is defined over R, we cannot use a causal fractional derivative, we have to use a two-sided derivative or a Riesz potential. In this case the a order derivative is, aside a possible error, because I'm doing fast computations, |t|^(-a-1).sgn(t)/[2gamma(-a) sin(a.pi/2)]. sgn(t) is the signum function.

I wish I could help. My mail is mdo@fct.unl.pt


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.