1
$\begingroup$

It's a well know fact that $$d(x^a)=ax^{a-1}dx\; \text{ for }\;a\in \Bbb{R}$$ It's a less well-known, but easily provable, fact that this generalizes to $$d^n(x^a)=\frac{a!}{(a-n)!}x^{a-n}\; \text{ for }\;n\in \Bbb{N}$$ Using fractional calculus, we can generalize this to $$D^n(x^a)=\frac{\Gamma(a+1)}{\Gamma(a-n+1)}x^{a-n}\;\text{ for }\;n\in \Bbb{R}^+$$ (for $\Bbb{R}^-$, and thus antiderivatives, and thus indefinite integrals, we need the slightly different $(-1)^{-n}\frac{\Gamma(n-a)}{\Gamma(-a)}x^{(a+n)}$).

There are just two problems: $\int(x^{-1})dx=\ln(x)+C$, which isn't a power function and stays that way under further (integral) integration, and $d(x^0)=0(x^{-1})$, which is technically a power function, but stops further (integral) differentiation cold with that 0.

Now, these wouldn't normally be problems (in fact, there are very good reasons the integral of the reciprocal function is the natural logarithm, not the least of which because the derivative of the constant function is 0). The problem isn't that the inconsistency is there - it's that I don't know how it behaves.

Are the fractional integrals of the reciprocal function logarithmic or power? (or both? or neither?) If both, where is the line drawn? Or is it "both" in the sense of multi-valued? What about the fractional derivatives of the constant function? Are they identically zero or do they wiggle around between the integral values?

And what about the square root? It's first derivative is on the order of the inverse square-root, while its three-halves derivative is on the order of the reciprocal, implying that the half-integral of the reciprocal should be on the order of the inverse square-root, but it apparently isn't.

And what about other non-integer powers going through this zone? How do they avoid getting zeroed by the constant function or getting swept up in the logarithm? Do the fractional "differintegrals" of the power functions depend on where in the chain you start?

Edit: It would seem, in some naive sense, that it is the case that fractional derivatives depend on where in the chain you start, if only because constants of integration are themselves integrated in subsequent reintegrations. So, better question: Do the constants of differintegration, which must be normalized by initial conditions, allow the differintegration chain to pass through the interval $a \in [-1, 0]$? And if so, how?

$\endgroup$
2
  • $\begingroup$ The important bit is in bold. The other questions can be considered background. Or extra credit, if you're feeling ambitious. $\endgroup$
    – No Name
    Commented Mar 19, 2018 at 15:37
  • $\begingroup$ The important bit is still in bold. The old important bit(s) are in italics. The rest of my previous comment still stands. $\endgroup$
    – No Name
    Commented Mar 22, 2018 at 6:23

1 Answer 1

1
$\begingroup$

Introduction

You've already figured out how to find the Fractional Derivative and Fractional Integral for all $x^{a}$ where $a \in \left] -1,\, 0 \right]$. The only fragment that is still missing is the special case $a = -1$.

You also already underlined what the problem is, namely that the antiderivative of $x^{-1} = \frac{1}{x}$ is given by the Natural Logarithm $\ln\left( \cdot \right)$, but that contradicts your formula. Since the Natural Logarithm $\ln\left( \cdot \right)$ isn't a monomial, so your formula won't work here either. Then let's just look at the Natural Logarithm $\ln\left( \cdot \right)$ itself.

In the following i'll use $\operatorname{D_{x}^{\alpha}}$ as $\dfrac{\operatorname{d}^{\alpha}}{\operatorname{d}x^{\alpha}}$.

Note As is usually the case with Fractional Derivatives in Fractional Calculus, there is also not the Fractional Derivative. They differ depending on the method of derivation used and the type of Differential Operator. But i'll ignore that here. But I'll ignore that here for the sake of simplicity and just cover a few special cases.

Calculation

Via Euler Method

With the Generalized Hypergeometric Function $\operatorname{_{1}F_{0}}$

There is a useful relation between $z^{a}$ and the Generalized Hypergeometric Function (I will call it GHF in the Following) $\operatorname{_{p}\tilde{F}_{q}}$ given by: $$ \begin{align*} \operatorname{_{1}F_{0}}\left( a;\, \cdot;\, z \right) = \left( 1 - z \right)^{-a} \implies \operatorname{_{1}F_{0}}\left( -a;\, \cdot;\, -z + 1 \right) = z^{a}\\ \end{align*} $$

This GHF has given an already well-researched Fractional Derivation called Fractional Integro-Differentiation Formula (with respect to $z$) by: $$ \begin{align*} \operatorname{D_{z}^{\alpha}}\left[ \operatorname{_{1}F_{0}}\left( a;\, \cdot;\, z \right) \right] &= z^{-\alpha} \cdot \operatorname{_{2}\tilde{F}_{1}}\left( 1,\, a;\, 1 - a;\, z \right)\\ \end{align*} $$ where $\operatorname{_{p}\tilde{F}_{q}}$ is the Generalized Regularized Hypergeometric Function defined by $\operatorname{_{p}\tilde{F}_{q}} = \frac{\operatorname{_{p}F_{q}}}{\Gamma}$ where $\Gamma\left( z \right)$ the Complete Gamma Function is. In particular, $\operatorname{_{2}\tilde{F}_{1}}$ is the Regularized Gaussian / Ordinary Hypergeometric Function.

So with the linear differential-rules we get: $$ \begin{align*} \operatorname{_{1}F_{0}}\left( -a;\, \cdot;\, -z + 1 \right) &= z^{a}\\ \operatorname{D_{z}^{\alpha}}\left[ \operatorname{_{1}F_{0}}\left( -a;\, \cdot;\, -z + 1 \right) \right] &= \left( -z + 1 \right)^{-\alpha} \cdot \operatorname{_{2}\tilde{F}_{1}}\left( 1,\, -a;\, 1 + a;\, -z + 1 \right)\\ \end{align*} $$

So $$\fbox{$ \begin{align*} \operatorname{D_{z}^{\alpha}}\left[ z^{a} \right] &= \left( -z + 1 \right)^{-\alpha} \cdot \operatorname{_{2}\tilde{F}_{1}}\left( 1,\, -a;\, 1 + a;\, -z + 1 \right)\\ \end{align*} $}$$

With the $\ln\left( \cdot \right)$

Leonhard Euler came up with a nice 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, which I always affectionately call the Euler method. If we try that with the Natural Logarithm $\ln\left( \cdot \right)$ we probably only get a special case of Faà di Bruno's Formula in most cases.

But it is possible to find a generalized formula for the integrals The following generalization for $n \in \mathbb{N}$ is not from me. I got it form here from another user. Respect to cyclochaotic for that. He proved it too. $$ \begin{align*} \operatorname{D_{z}^{0}}\left[ \ln\left( z \right) \right] &= \ln\left( z \right)\\ \operatorname{D_{z}^{-1}}\left[ \ln\left( z \right) \right] &= z \cdot \left( \ln\left( z \right) - 1 \right)\\ \operatorname{D_{z}^{-2}}\left[ \ln\left( z \right) \right] &= z^{2} \cdot \left( \frac{1}{2} \cdot \ln\left( z \right) - \frac{3}{4} \right)\\ \operatorname{D_{z}^{-3}}\left[ \ln\left( z \right) \right] &= z^{3} \cdot \left( \frac{1}{6} \cdot \ln\left( z \right) - \frac{11}{36} \right)\\ \operatorname{D_{z}^{-4}}\left[ \ln\left( z \right) \right] &= z^{4} \cdot \left( \frac{1}{24} \cdot \ln\left( z \right) - \frac{25}{288} \right)\\ \operatorname{D_{z}^{-5}}\left[ \ln\left( z \right) \right] &= z^{5} \cdot \left( \frac{1}{120} \cdot \ln\left( z \right) - \frac{137}{7200} \right)\\ &\cdots\\ \operatorname{D_{z}^{-n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( n! \right)^{2}} \cdot \left( n! \cdot \ln\left( z \right) + \cos\left( n \cdot \pi \right) \cdot S_{n + 1}^{\left( 2 \right)} \right)\\ \end{align*} $$ where $S_{\cdot}^{\left( \cdot \right)} = s\left( \cdot,\, \cdot \right)$ is a Stirling Number Of The First Kind.

$S_{z}^{\left( 2 \right)}$ is a realy good special case of Stirling Numbers Of The First Kind, for it has a useful relation to the Harmonic Numbers $H_{n}$ given by $S_{n}^{\left( 2 \right)} = \left( -1 \right)^{n} \cdot \left( n - 1 \right)! \cdot H_{n - 1}$, which is so useful because of its relation to the Digamma Function $\psi_{0}\left( \cdot \right)$ given by $H_{n} = \gamma + \psi_{0}\left( n + 1 \right)$ where $\gamma$ si the Euler-Mascheroni Constant. With this we can generalize the term and get $H_{n - 1} = \gamma + \psi_{0}\left( n \right)$ $$ \begin{align*} \operatorname{D_{z}^{-n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( n! \right)^{2}} \cdot \left( n! \cdot \ln\left( z \right) + \cos\left( n \cdot \pi \right) \cdot S_{n + 1}^{\left( 2 \right)} \right)\\ \operatorname{D_{z}^{-n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( n! \right)^{2}} \cdot \left( n! \cdot \ln\left( z \right) + \cos\left( n \cdot \pi \right) \cdot \left( -1 \right)^{n} \cdot \left( n - 1 \right)! \cdot H_{n - 1} \right)\\ \operatorname{D_{z}^{-n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( n! \right)^{2}} \cdot \left( n! \cdot \ln\left( z \right) + \cos\left( n \cdot \pi \right) \cdot \left( -1 \right)^{n} \cdot \left( n - 1 \right)! \cdot \left( \gamma + \psi_{0}\left( n \right) \right) \right)\\ \operatorname{D_{z}^{-n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( \Gamma\left( n + 1 \right) \right)^{2}} \cdot \left( \Gamma\left( n + 1 \right) \cdot \ln\left( z \right) + \cos\left( n \cdot \pi \right) \cdot \left( -1 \right)^{n} \cdot \Gamma\left( n \right) \cdot \left( \gamma + \psi_{0}\left( n \right) \right) \right)\\ \operatorname{D_{z}^{-n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( \Gamma\left( n + 1 \right) \right)^{2}} \cdot \left( \Gamma\left( n + 1 \right) \cdot \ln\left( z \right) + \left( -1 \right)^{n} \cdot \left( -1 \right)^{n} \cdot \Gamma\left( n \right) \cdot \left( \gamma + \psi_{0}\left( n \right) \right) \right)\\ \operatorname{D_{z}^{-n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( \Gamma\left( n + 1 \right) \right)^{2}} \cdot \left( \Gamma\left( n + 1 \right) \cdot \ln\left( z \right) + \Gamma\left( n \right) \cdot \left( \gamma + \psi_{0}\left( n \right) \right) \right)\\ \end{align*} $$ where $\Gamma\left( \cdot \right)$ is the Complete Gamma Function.

But the last of these equations is also defined for non-natural $n$. This allows us to generalize $$ \begin{align*} \operatorname{D_{z}^{-\alpha}}\left[ \ln\left( z \right) \right] &= \frac{z^{\alpha}}{\left( \Gamma\left( \alpha + 1 \right) \right)^{2}} \cdot \left( \Gamma\left( \alpha + 1 \right) \cdot \ln\left( z \right) + \Gamma\left( \alpha \right) \cdot \left( \gamma + \psi_{0}\left( \alpha \right) \right) \right)\\ \end{align*} $$

But we shouldn't forget the constants that come with indefinite integration (assuming $\alpha \in \mathbb{R}^{+}$ or $\alpha > 0$) $$\fbox{$ \begin{align*} \operatorname{D_{z}^{-\alpha}}\left[ \ln\left( z \right) \right] &= \frac{z^{\alpha}}{\left( \Gamma\left( \alpha + 1 \right) \right)^{2}} \cdot \left( \Gamma\left( \alpha + 1 \right) \cdot \ln\left( z \right) + \Gamma\left( \alpha \right) \cdot \left( \gamma + \psi_{0}\left( \alpha \right) \right) \right) + \sum\limits_{k = 0}^{\left\lfloor \alpha \right\rfloor}\left[ c_{k} \cdot x^{k - 1} \right]\\ \end{align*} $}$$ where all $c_{k}$ are some constants and $\left\lfloor \cdot \right\rfloor$ is the Floor Function.

Unfortunately, this function has singularitys at non-positiv integers $\alpha$.

Now using $\operatorname{D_{z}^{m}}\left[ \operatorname{D_{z}^{n}} \right] = \operatorname{D_{z}^{m + n}}$ aka $\operatorname{D_{z}^{-\alpha}}\left[ \ln\left( z \right) \right] = \operatorname{D_{z}^{-\alpha}}\left[ \operatorname{D_{z}^{-1}}\left[ \frac{1}{z} \right] \right] = \operatorname{D_{z}^{-\alpha - 1}}\left[ \frac{1}{z} \right]$ gives us $$\fbox{$ \begin{align*} \operatorname{D_{z}^{\alpha}}\left[ z^{-1} \right] &= \begin{cases} \frac{z^{-\alpha - 1}}{\left( \Gamma\left( -\alpha \right) \right)^{2}} \cdot \left( \Gamma\left( -\alpha \right) \cdot \ln\left( z \right) + \Gamma\left( -\alpha - 1 \right) \cdot \left( \gamma + \psi_{0}\left( -\alpha \right) \right) \right) + \sum\limits_{k = 0}^{\left\lfloor -\alpha - 1 \right\rfloor}\left[ c_{k} \cdot x^{k - 1} \right],\, &\text{for}\, \alpha \leq -1 \wedge a = -1\\ \left( -1 \right)^{\alpha} \cdot \Gamma\left( 1 + \alpha \right) \cdot x^{-1 - \alpha},\, &\alpha -1\\ \end{cases} \end{align*} $}$$

This gives us the general solution $$\fbox{$\fbox{$ \begin{align*} \operatorname{D_{z}^{\alpha}}\left[ z^{a} \right] &= \begin{cases} \begin{cases} \frac{z^{-\alpha - 1}}{\left( \Gamma\left( -\alpha \right) \right)^{2}} \cdot \left( \Gamma\left( -\alpha \right) \cdot \ln\left( z \right) + \Gamma\left( -\alpha - 1 \right) \cdot \left( \gamma + \psi_{0}\left( -\alpha \right) \right) \right) + \sum\limits_{k = 0}^{\left\lfloor -\alpha - 1 \right\rfloor}\left[ c_{k} \cdot x^{k - 1} \right],\, &\text{for}\, \alpha \leq -1\\ \left( -1 \right)^{\alpha} \cdot \Gamma\left( 1 + \alpha \right) \cdot x^{-1 - \alpha},\, &\text{for}\, \alpha > -1\\ \end{cases},\, &\text{for}\, a = -1\\ \left( -1 \right)^{\alpha} \cdot \frac{\Gamma\left( a + \alpha \right)}{\Gamma\left( a \right)} \cdot \left( x + b \right)^{-a - \alpha},\, &\text{for}\, -1 > a \leq 1\\ \end{cases} \end{align*} $}$}$$

Calculation Via Series Expansion

Another well-known and frequently used method to differentiate a function fractionally is simply to differentiate the series expansion of this function with addends from functions that are easier to differentiate fractionally. The Natural Logarithm $\ln\left( \cdot \right)$ has a wonderfully simple series expansion (Taylor Series around $z = 1$) given by $\ln\left( 1 + z \right) = \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot z^{k} \right]$ (Mercator Series) aka $\ln\left( z \right) = \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \left( z - 1 \right)^{k} \right]$.

So we'll get: $$ \begin{align*} \operatorname{D_{z}^{-\alpha}}\left[ \ln\left( z \right) \right] &= \operatorname{D_{z}^{-\alpha}}\left[ \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \left( z - 1 \right)^{k} \right] \right]\\ \operatorname{D_{z}^{-\alpha}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \operatorname{D_{z}^{-\alpha}}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \left( z - 1 \right)^{k} \right] \right]\\ \operatorname{D_{z}^{-\alpha}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \operatorname{D_{z}^{-\alpha}}\left[ \left( z - 1 \right)^{k} \right] \right]\\ \operatorname{D_{z}^{-\alpha}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \operatorname{D_{z}^{-\alpha}}\left[ \left( z - 1 \right)^{k} \right] \right]\\ \end{align*} $$

Via using $\operatorname{D_{x}^{\alpha}}\left[ \left( x - 1 \right)^{m} \right] = \frac{\Gamma\left( m + 1 \right)}{\Gamma\left( m - \alpha + 1 \right)} \cdot \left( x - 1 \right)^{m - \alpha}$ (for $m > 0$) where $\Gamma\left( \cdot \right)$ is the Complete Gamma Function we'll get: You can read a step-by-step derivation of the formula used here. $$ \begin{align*} \operatorname{D_{z}^{-\alpha}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \operatorname{D_{z}^{-\alpha}}\left[ \left( z - 1 \right)^{k} \right] \right]\\ \operatorname{D_{z}^{-\alpha}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k + \alpha + 1 \right)} \cdot \left( z - 1 \right)^{k + \alpha} \right]\\ \end{align*} $$

But we shouldn't forget the constants that come with indefinite integration (assuming $\alpha \in \mathbb{R}^{+}$ or $\alpha > 0$): $$\fbox{$ \begin{align*} \operatorname{D_{z}^{\alpha}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot \left( z - 1 \right)^{k - \alpha} \right] + \sum\limits_{k = 0}^{\left\lfloor -\alpha \right\rfloor}\left[ c_{k} \cdot x^{k - 1} \right]\\ \end{align*} $}$$ where all $c_{k}$ are some constants and $\left\lfloor \cdot \right\rfloor$ is the Floor Function.

But this series expansion converges very very slowly.

This function is defined for $\alpha = -1$. In fact, it gives: $$ \begin{align*} \operatorname{D_{z}^{1}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k + -1 + 1 \right)} \cdot \left( z - 1 \right)^{k + -1} \right]\\ \operatorname{D_{z}^{1}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k \right)} \cdot \left( z - 1 \right)^{k - 1} \right]\\ \operatorname{D_{z}^{1}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot k \cdot \left( z - 1 \right)^{k - 1} \right]\\ \operatorname{D_{z}^{1}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \left( -1 \right)^{k + 1} \cdot \left( z - 1 \right)^{k - 1} \right]\\ \operatorname{D_{z}^{1}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \left( -1 \right)^{k - 1} \cdot \left( z - 1 \right)^{k - 1} \right]\\ \operatorname{D_{z}^{1}}\left[ \ln\left( z \right) \right] &= \sum\limits_{n = 0}^{\infty}\left[ \left( -1 \right)^{n} \cdot \left( z - 1 \right)^{n} \right]\\ \operatorname{D_{z}^{1}}\left[ \ln\left( z \right) \right] &= \frac{1}{z}\\ \end{align*} $$

Wolfram|Alpha gives the same Series Expansion. So:

$$ \operatorname{D_{z}^{1}}\left[ \ln\left( z \right) \right] = \frac{1}{z} $$

This means that this formula is universally valid. So with this we get: $$\fbox{$ \begin{align*} \operatorname{D_{z}^{\alpha}}\left[ z^{a} \right] &= \begin{cases} \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha \right)} \cdot \left( z - 1 \right)^{k - \alpha - 1} \right] + \sum\limits_{k = 0}^{\left\lfloor -\alpha - 1 \right\rfloor}\left[ c_{k} \cdot x^{k - 1} \right],\, &\text{for}\, \alpha \leq -1 = a\\ \left( -1 \right)^{\alpha} \cdot \frac{\Gamma\left( a + \alpha \right)}{\Gamma\left( a \right)} \cdot \left( x + b \right)^{-a - \alpha},\, &\text{otherwise}\\ \end{cases} \end{align*} $}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .