0
$\begingroup$

While I was playing with a CAS I find that makes sense the function $$\psi^{(-k)}(x),$$ for example $\psi^{(-2)}(x)$, where $\psi^{(n)}(x)$ denotes the $n$th derivative of the digamma function, see this MathWorld.

Question 1 (Answered see the comments). Can you explain what is the function $\psi^{(-2)}(x)$? I am asking about what is its definition. Many thanks.

I think that maybe is a notation for the second antiderivative, but I would like to know a definition with rigor about what is previous function, and what is previous notation.

As a puzzle I wondered if it is possible to deduce the convergence of some series involving previous functions, and the Möbius function $\mu(n)$, see the definition of this arithmetic function from this MathWorld.

Question 2. Can you deduce the convergence of $$\sum_{n=1}^\infty\mu(n)\frac{\psi^{(-1)}(n)}{n^3}\tag{1}$$ or $$\sum_{n=1}^\infty\mu(n)\frac{\psi^{(-2)}(n)}{n^3}\,?\tag{2}$$ Many thanks.

Only is required to prove the convergence of some example in previous Question 2, well the first or the second series.

$\endgroup$
  • $\begingroup$ $\psi^{(-2)}(x)$ is no generally known mathematical notation, so if you want to ask about it, you define it, or refer to a definition. Otherwise, you risk your question to be closed for missing context (or not belonging to mathematics at all). $\endgroup$ – Professor Vector Mar 9 '18 at 8:45
  • 1
    $\begingroup$ See for example Adamchik's paper "Polygamma functions of negative order " $\endgroup$ – Raymond Manzoni Mar 9 '18 at 8:54
  • $\begingroup$ Many thanks for the definition @RaymondManzoni $\endgroup$ – user243301 Mar 9 '18 at 9:00
  • $\begingroup$ Many thanks @ProfessorVector $\endgroup$ – user243301 Mar 9 '18 at 9:00
  • 1
    $\begingroup$ You are welcome @user243301! $\endgroup$ – Raymond Manzoni Mar 9 '18 at 9:01
0
$\begingroup$

WolframAlpha's definition: "formula presents (not unique) continuation of the classic definition of $\psi^{(\nu)}(z)$ from positive integer values of $\nu$ to its arbitrary complex values... it is very convenient to use it as basic definition of polygamma function for any arbitrary complex values $\nu$, where symbol $\frac{\partial^\nu\psi(z)}{\partial z^\nu}$ denotes the $\nu$th fractional integro-derivative of $\psi(z)$ with respect to $z$ (which provides the Riemann-Liouville-Hadamand fractional left-sided integro-differentiation with beginning at the point $0$). Such approach was realized in Mathematica."

In essence, it would seem Mathematica's definition relies on taking the fractional derivative of the log gamma function, with $\psi^{(-1)}(z)=\ln\Gamma(z)$. For negative integers, this would give

$$\psi^{(-k)}(z)=\int_0^z\int_0^{z_1}\dots\int_0^{z_{k-2}}\ln\Gamma(z_{k-1})~dz_{k-1}\dots dz_2~dz_1=\frac1{(k-2)!}\int_0^z(z-t)^{k-2}\ln\Gamma(t)~dt$$

One also happens to have the special value of

$$\psi^{(-2)}(z)=\int_0^z\ln\Gamma(t)~dt=\frac z2\ln(2\pi)+\frac{z(1-z)}2+z\ln\Gamma(z)-\ln G(z+1)$$

where $G$ is the Barnes-G function.

Provided Stirling's approximation of the Gamma function, one can quickly see the first series converge absolutely, while the second sum converges conditionally since $\sum\mu(n)\ln n/n$ converges.

$\endgroup$
  • $\begingroup$ Many thanks for your answer. $\endgroup$ – user243301 Aug 14 '18 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy