Evaluating $\psi^{(1/2)}(x)$ of the extended polygamma function

Question

The polygamma function is generally given by

$$\psi^{(n)}(x)=\frac{d^{n+1}}{dx^{n+1}}\ln(\Gamma(x)),~n\in\mathbb N_{\ge0}$$

where $\Gamma$ is the gamma function. This can be extended to negative integers by letting

$$\psi^{(n-1)}(x)=\int_a^x\psi^{(n)}(x)$$

Unfortunately, there is no fixed $a$ such that the above holds true for any $n$, but it does capture the general idea.

Using fractional calculus, one can extend the polygamma function to complex values. Here, I could have

$$\psi^{(z)}_u(x)=\frac1{\Gamma(1-\{a\}-bi)}\frac{d^{\lceil a\rceil+2}}{dx^{\lceil a\rceil+2}}\int_u^x\frac{\ln(\Gamma(t))}{(x-t)^{\{a\}+bi}}~\mathrm dt$$

where $z=a+bi$, $a>-1$, $\lceil a\rceil$ is the ceiling function, and $\{a\}$ is the fractional part of $a$. $u$ could be any constant such that the integral converges. For $z\in\mathbb R^+$, the above expression reduces down to the normal polygamma function regardless of $u$ (at least for $x>0$).

For $a<-1$,

$$\psi^{(z)}_u(x)=\frac1{\Gamma(z)}\int_u^x(x-t)^{z-1}\ln(\Gamma(t))~\mathrm dt$$

---

Particularly of interest, I'm trying to find the closed form of $\psi^{(1/2)}_u(x)$.

$$\psi^{(1/2)}_u(x)=\frac1{\sqrt\pi}\frac{d^3}{dx^3}\int_u^x\frac{\ln(\Gamma(t))}{\sqrt{x-t}}~\mathrm dt$$

By applying Leibniz rule, this reduces down to

$$\frac{d}{dx}\int_u^x\frac{\ln(\Gamma(t))}{\sqrt{x-t}}~\mathrm dt=\lim_{y\to x}\left[\frac{\ln(\Gamma(y))}{\sqrt{x-y}}-\frac12\int_u^y\frac{\ln(\Gamma(t))}{(x-t)^{3/2}}~\mathrm dt\right]$$

However, I can't figure out how to simplify this, if possible. Trying to attack the original integral, then differentiating the result, I'm tempted to use this series expansion, but integrating term by term of that expression is quite nasty and the resulting series looks quite doubtful from anywhere near a closed form.

Also unsure if this is any help, but if we appropriately define $\sqrt z=|z|^{1/2}e^{i\theta/2}$, where $\theta\in[0,2\pi)$, one could attempt a keyhole contour of radius $u$ centered at $x$ to deduce that (hopefully this is correct)

$$\int_u^x\frac{\ln(\Gamma(t))}{\sqrt{x-t}}~\mathrm dt=\frac12\int_\pi^{3\pi}\frac{\Gamma(x+ue^{i\theta})}{\sqrt{-ue^{i\theta}}}~d\theta$$

But again, doesn't seem much helpful

Answer 1

What about applying a fractional derivative to each term of the Kummer-Malmsten Fourier series for $\log\Gamma$? It is not difficult to apply $D^{\alpha}$ (fractional derivative with respect to $x$) to the integral representation

$$ \psi(x)=\int_0^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}\right)\, dt,$$ too.

Answer 2

Introduction

As is usually the case with Fractional Derivatives, there is also not "the Fractional Derivatives" for the Polygamma Function $\psi^{\left( \alpha \right)}\left( \cdot \right) = \psi_{\alpha}\left( \cdot \right) = \psi\left( \alpha,\, \cdot \right)$ (i'll use the notation $\psi^{\left( \alpha \right)}\left( \cdot \right)$). They differ depending on the method of derivation used and the type of Differential operator.

Manual Calculation of the Fractional Derivative

Calculation Via Series Expansion

The Polygamma Function $\psi^{\left( \alpha \right)}\left( z \right)$ has a usefull series expansion for $\alpha > 0$ ($\alpha \in \mathbb{R}^{+}$) given by $$ \begin{align*} \psi^{\left( \alpha \right)}\left( z \right) &= \left( -1 \right)^{\alpha + 1} \cdot \alpha! \cdot \sum\limits_{k = 0}^{\infty}\left[ \left( z + k \right)^{-\alpha - 1} \right]\\ &\text{or}\\ \psi^{\left( \alpha \right)}\left( z \right) &= \left( -1 \right)^{\alpha + 1} \cdot \Gamma\left( \alpha + 1 \right) \cdot \sum\limits_{k = 0}^{\infty}\left[ \left( z + k \right)^{-\alpha - 1} \right]\ \end{align*} $$ where $\Gamma\left( \cdot \right)$ ist the Complte Gamma Function.

If we insert $\alpha = 0.5$ we get: $$ \begin{align*} \psi^{\left( 0.5 \right)}\left( z \right) &= \left( -1 \right)^{0.5 + 1} \cdot \Gamma\left( 0.5 + 1 \right) \cdot \sum\limits_{k = 0}^{\infty}\left[ \left( z + k \right)^{-0.5 - 1} \right]\\ \psi^{\left( 0.5 \right)}\left( z \right) &= \left( -1 \right)^{1.5} \cdot \Gamma\left( 1.5 \right) \cdot \sum\limits_{k = 0}^{\infty}\left[ \left( z + k \right)^{-1.5} \right]\\ \end{align*} $$

The whole thing can be simplified with the Hurwitz Zeta Function $\zeta\left( \alpha,\, z \right)$: $$\fbox{$ \begin{align*} \psi^{\left( 0.5 \right)}\left( z \right) &= \left( -1 \right)^{1.5} \cdot \Gamma\left( 1.5 \right) \cdot \zeta\left( 1.5,\, z \right)\\ \psi^{\left( 0.5 \right)}\left( z \right) &= \pm i \cdot \Gamma\left( 1.5 \right) \cdot \zeta\left( 1.5,\, z \right)\\ \end{align*} $} \tag{1}$$

But $\left( -1 \right)^{1.5} = \left( -1 \right)^{\frac{3}{2}}$ has a different solution depending on the branch. The main branch gives $\left( -1 \right)^{1.5} = i$ and the other branch $\left( -1 \right)^{1.5} = -i$ where $i^{2} = -1$. This would give you utible possible Polygamma Functions.

Plot for $\left( -1 \right)^{1.5} = i$:

Plot for $\left( -1 \right)^{1.5} = -i$:

"Euler Method"

Euler had come up with his own method for some Fractional Derivatives (see Euler's Approach). He has sorted the derivatives of the function by order and is trying to find a pattern in it. He then tried to transfer this pattern from integer orders to fractional ones.

But here is a problem. Without knowledge of special functions, one cannot come up with a general formula for a Derivative of the $\alpha$th order. We would only come up with a special case of Faà di Bruno's Formula.

However, if we write the derivatives in terms of certain special functions, such as the Hurwitz Zeta Function $\zeta\left( \alpha,\, \cdot \right)$ or the Bernoulli Polynomials $B_{\alpha}\left( \cdot \right)$ or the Polylogarithm, we might find such a generalization.

Since I have already carried out a possible derivation of such a Fractional Derivative with the Hurwitz Zeta Function, I will demonstrate it here with the Bernoulli Polynomials. I'll use $\operatorname{D_{x}^{\alpha}}$ given as a differential operator by $\operatorname{D_{x}^{\alpha}} = \frac{\operatorname{d}^{\alpha}}{\operatorname{d}x^{\alpha}}$ for $a \geq 0$, wich is also true for all $\alpha \in \mathbb{R}^{+}$: $$ \begin{align*} \operatorname{D_{x}^{0}}\left[ \psi^{0}\left( z \right) \right] = \psi^{0}\left( z \right) &= +\psi^{0}\left( z \right)\\ \operatorname{D_{x}^{1}}\left[ \psi^{0}\left( z \right) \right] = \psi^{1}\left( z \right) &= -B_{-1}\left( z \right)\\ \operatorname{D_{x}^{2}}\left[ \psi^{0}\left( z \right) \right] = \psi^{2}\left( z \right) &= +B_{-2}\left( z \right)\\ \operatorname{D_{x}^{3}}\left[ \psi^{0}\left( z \right) \right] = \psi^{3}\left( z \right) &= -2 \cdot B_{-3}\left( z \right)\\ \operatorname{D_{x}^{4}}\left[ \psi^{0}\left( z \right) \right] = \psi^{4}\left( z \right) &= +6 \cdot B_{-4}\left( z \right)\\ \operatorname{D_{x}^{5}}\left[ \psi^{0}\left( z \right) \right] = \psi^{5}\left( z \right) &= -24 \cdot B_{-5}\left( z \right)\\ &\cdots\\ \operatorname{D_{x}^{n}}\left[ \psi^{0}\left( z \right) \right] = \psi^{n}\left( z \right) &= \left( -1 \right)^{n} \cdot \left( n - 1 \right)! \cdot B_{-n}\left( z \right),\, \text{for}\, n \in \mathbb{N}\\ \operatorname{D_{x}^{\alpha}}\left[ \psi^{0}\left( z \right) \right] = \psi^{\alpha}\left( z \right) &= \left( -1 \right)^{\alpha} \cdot \Gamma\left( \alpha \right) \cdot B_{-\alpha}\left( z \right),\, \text{for}\, \alpha \in \mathbb{R}^{+}\\ \end{align*} $$ where $\Gamma\left( \cdot \right)$ ist the Complte Gamma Function.

Substituting $\alpha = 0.5$ into the generalized formula we get: $$ \begin{align*} \psi^{0.5}\left( z \right) &= \left( -1 \right)^{\alpha} \cdot \Gamma\left( \alpha \right) \cdot B_{-\alpha}\left( z \right)\\ \psi^{0.5}\left( z \right) &= \left( -1 \right)^{0.5} \cdot \Gamma\left( 0.5 \right) \cdot B_{-0.5}\left( z \right)\\ \psi^{0.5}\left( z \right) &= \sqrt{-1} \cdot \Gamma\left( 0.5 \right) \cdot B_{-0.5}\left( z \right)\\ \psi^{\alpha}\left( z \right) &= \pm i \cdot \Gamma\left( 0.5 \right) \cdot B_{-0.5}\left( z \right)\\ \end{align*} $$ where $i^{2} = -1$ Here, too, as we can see, we get several possible branches and thus several possible functions:

$$\fbox{$ \begin{align*} \psi^{\left( 0.5 \right)}\left( z \right) &= \sqrt{-1} \cdot \Gamma\left( 0.5 \right) \cdot B_{-0.5}\left( z \right)\\ \psi^{\left( 0.5 \right)}\left( z \right) &= \pm i \cdot \Gamma\left( 0.5 \right) \cdot B_{-0.5}\left( z \right)\\ \end{align*} $} \tag{2}$$

Plot for $\left( -1 \right)^{1.5} = i$:

Plot for $\left( -1 \right)^{1.5} = -i$:

Yes. And you guessed it. They are the same as before.

Calculation Via Using Special Differential Operators

Riemann–Liouville Operator

The Riemann–Liouville Fractional Operator, on the other hand, uses the formulas for simplifying mutible integrals into single integral for many other differential operators: $$ \begin{align*} \operatorname{_{a}D^{-v}_{x}}\left[ f\left( x \right) \right] \equiv \frac{1}{\Gamma\left( v \right)} \cdot \int\limits_{a}^{x} f\left( u \right) \cdot \left( x - u \right)^{v - 1} \, \operatorname{d}u \tag{3}\\ \end{align*} $$ where $\Gamma\left( \cdot \right)$ ist the Complte Gamma Function.

This operator has a so-called Convolutional Formula, wich is just spezial case with $a = 0$, but it also takes advantage of the fact that integrating and deriving cancel each other out: $$ \begin{align*}\operatorname{_{0}D^{v}_{x}}\left[ f\left( x \right) \right] \equiv \frac{1}{\Gamma\left( 1 - v \right)} \cdot \operatorname{D^{1}_{x}}\left[ \int\limits_{0}^{x} f\left( u \right) \cdot \left( x - u \right)^{-v}\, \operatorname{d}u \right] \tag{4}\\ \end{align*} $$

So we get: $$ \begin{align*} \psi^{\left( 0.5 \right)}\left( x \right) = \operatorname{_{0}D^{0.5}_{x}}\left[ \psi^{\left( 0 \right)}\left( x \right) \right] &\equiv \frac{1}{\Gamma\left( 0.5 \right)} \cdot \operatorname{D^{1}_{x}}\left[ \int\limits_{0}^{x} \psi^{\left( 0 \right)}\left( u \right) \cdot \left( x - u \right)^{-0.5}\, \operatorname{d}u \right] \tag{5}\\ \end{align*} $$

Using Formula $\left( 5 \right)$ gives you a simplification of your Problem. But I'll leave the solution with this formula to you as an exercise for you.

Mathematical Softwares

Up to now, only a few programs can determine derivatives of fractional order, and those that can can only do so for a few specific functions.

The best-known example of this is probably Wolfram|Alpha: The command FractionalD[f[x], {x, α}] gives the Riemann–Liouville Fractional Derivative of order $\alpha$ of the function $f$ with respect of $x$ ($\operatorname{_{0}D^{\alpha}_{x}}\left( f\left( x \right) \right)$). However, this only works for a few functions so far like some monomials. Wolfram|Alpha also has a small article about the command in which they explain the possibilities that the command opens up (see Wolfram > Language > Reference > FractionalD). There is also the command CaputoD[f, {x, α}], wich gives the Caputo Fractional Differintegral of order $\alpha$ of the function $f$ with respect of $x$ ($\operatorname{_{0}^{C}D^{\alpha}_{x}}\left( f\left( x \right) \right)$). Wolfram|Alpha has also a small article about the command in which they explain the possibilities that the command opens up (see Wolfram > Language > Reference > CaputoD).

Although Wolfram|Alpha cannot fractionally differentiate the Polygamma Function $\psi^{\left( \alpha \right)}\left( z \right)$ (Polygamma[a, z]), it can calculate and plot the Polygamma Function $\psi^{\left( 0.5 \right)}\left( z \right)$.

It uses the following series expansions:

around $x = 0$: $$ \begin{align*} &\psi^{\left( 0.5 \right)}\left( x \right) \approx x^{-0.5} \cdot \left( \frac{0.282095 \cdot \ln\left( x \right) - 0.173123}{x} - 0.325659 + 1.85611 \cdot x - 1.8085 \cdot x^{2} + 1.95403 \cdot x^{3} - 2.13952 \cdot x^{4} + 2.33234 \cdot x^{5} + \mathcal{O}\left( x^{6} \right) \right) \tag{6}\\ \end{align*} $$

around $x = \infty$: $$ \begin{align*} &\psi^{\left( 0.5 \right)}\left( x \right) \approx x^{-0.5} \cdot \left( 0.56419 \cdot \ln\left( x \right) + 0.782133 + \frac{0.141047 \cdot \ln\left( x \right) - 0.345789}{x} + \frac{0.0352618 \cdot \ln\left( x \right) + 0.0601102}{x^2} - \frac{0.016105}{x^{3}} + \frac{-0.00514235 \cdot \ln\left( x \right) - 0.00263636}{x^{4}} + \frac{0.0055425}{x^{5}} + \mathcal{O}\left( \left( \frac{1}{x} \right)^{6} \right) \right) \tag{7}\\ \end{align*} $$

A plot of this definition would look like this: