Fractional anti-derivatives and derivatives of the logarithm

Question

Anti-derivatives and derivatives of the natural logarithm are well defined until we attempt to evaluate the fractional derivative and anti-derivatives.

The background to this problem was that I was trying to evaluate fractional derivatives of $\frac1x$, which is usually given as $D^n\frac1x=\frac{\Gamma(0)}{\Gamma(-n)}x^{-1-n}$, but that is defined only for $n=0$ and nowhere else. You can define it for $n\in\mathbb N$, but this is not the best definition one could have. ($D^n$ is the $n$th derivative with respect to $x$)

I have the general $n$th anti-derivative ($I^n$) of the natural logarithm as

$$I^n\ln(x)=\frac{x^n\left(\ln(x)-\int_0^1\frac{t^n-1}{t-1}dt\right)}{\Gamma(n+1)}$$

Proved by induction: $\frac d{dx}I^n\ln(x)=I^{n-1}\ln(x)$, and holds true for $n=1$.

I have this graphed on Desmos. (I really like that it has an integral feature. And if it is too slow, click the little circles on the right to turn off those functions.)

While this is a great formula, it doesn't really work for derivatives ($D^n=I^{-n}$), or at least, Desmos stops graphing at $n=-0.99$, but before that, it appears as though $\lim_{n\to-1}I^n\ln(x)=\frac1x$

I attempted to evaluate it for $n=-1$

$$\frac1x=D^1\ln(x)=I^{-1}\ln(x)=\lim_{n\to-1}\frac{x^n\left(\ln(x)-\int_0^1\frac{t^n-1}{t-1}dt\right)}{\Gamma(n+1)}$$

If you try to directly substitute, you get an indefinite form, so I attempted to do a limit method instead. I would like to apply L'Hostpital's rule, but I don't quite know how to deal with either the numerator nor the denominator.

I have found that

$$I^{1/2}\ln(x)=\frac{2\sqrt x\left(\ln(x)-2+\ln(4)\right)}{\sqrt\pi}$$

The $2-\ln(4)$ is wolframalpha's evaluation of $\int_0^1\frac{t^{1/2}-1}{t-1}dt$

From here, you can differentiate like normal to get $D^{(2n-1)/2}\ln(x)$, $n\in\mathbb N$.

More importantly, if the limit from above is correct, then my formula works for $I^{-n}\ln(x)$ with $n\in\mathbb N$ and I can finally define what $D^n\frac1x$ is equal to!

So the question:

How can we evaluate the limit: $\lim_{n\to-1}I^n\ln(x)$? What does it equal?

Is the fractional derivative of the natural logarithm and $\frac1x$ already known? I've made posts on this before, and it doesn't appear many people have tackled this problem. Here and Here. Both have received little attention. Attempting to use regular fractional derivative formulas are extremely messy, so I could not actually use them. If someone more experienced can, that'd be great.

Answer 1

The complex order differintegral appears as equation (9) with positive real part of the order (i.e., all derivatives) in Tu, S-T., D-K. Chyan, & H.M. Srivastava, "Certain Operators of the Fractional Calculus and Their Applications Associated with the Logarithmic and Digamma Functions", 1995. Analytic continuation to the rest of the plane appears as (16) there.

Answer 2

Note: As is usually the case with Fractional Derivatives, there is also not "the Fractional Derivative". They differ depending on the method of derivation used and the type of Differential Operator.

Calculation Via "Euler Method"

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. $$ \begin{align*} \operatorname{I_{z}^{0}}\left[ \ln\left( z \right) \right] &= \ln\left( z \right)\\ \operatorname{I_{z}^{1}}\left[ \ln\left( z \right) \right] &= z \cdot \left( \ln\left( z \right) - 1 \right)\\ \operatorname{I_{z}^{2}}\left[ \ln\left( z \right) \right] &= z^{2} \cdot \left( \frac{1}{2} \cdot \ln\left( z \right) - \frac{3}{4} \right)\\ \operatorname{I_{z}^{3}}\left[ \ln\left( z \right) \right] &= z^{3} \cdot \left( \frac{1}{6} \cdot \ln\left( z \right) - \frac{11}{36} \right)\\ \operatorname{I_{z}^{4}}\left[ \ln\left( z \right) \right] &= z^{4} \cdot \left( \frac{1}{120} \cdot \ln\left( z \right) - \frac{25}{288} \right)\\ \operatorname{I_{z}^{5}}\left[ \ln\left( z \right) \right] &= z^{5} \cdot \left( \frac{1}{2} \cdot \ln\left( z \right) - \frac{137}{7200} \right)\\ &\cdots\\ \operatorname{I_{z}^{n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( n! \right)^{2}} \cdot \left( n! \cdot \ln\left( z \right) + \cos\left( n \cdot \pi \right) \cdot S_{n + 1}^{\left( 2 \right)} \right)\\ \end{align*} $$ 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: $H_{n - 1} = \gamma + \psi_{0}\left( n \right)$ $$ \begin{align*} \operatorname{I_{z}^{n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( n! \right)^{2}} \cdot \left( n! \cdot \ln\left( z \right) + \cos\left( n \cdot \pi \right) \cdot S_{n + 1}^{\left( 2 \right)} \right)\\ \operatorname{I_{z}^{n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( n! \right)^{2}} \cdot \left( n! \cdot \ln\left( z \right) + \cos\left( n \cdot \pi \right) \cdot \left( -1 \right)^{n} \cdot \left( n - 1 \right)! \cdot H_{n - 1} \right)\\ \operatorname{I_{z}^{n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( n! \right)^{2}} \cdot \left( n! \cdot \ln\left( z \right) + \cos\left( n \cdot \pi \right) \cdot \left( -1 \right)^{n} \cdot \left( n - 1 \right)! \cdot \left( \gamma + \psi_{0}\left( n \right) \right) \right)\\ \operatorname{I_{z}^{n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( \Gamma\left( n + 1 \right) \right)^{2}} \cdot \left( \Gamma\left( n + 1 \right) \cdot \ln\left( z \right) + \cos\left( n \cdot \pi \right) \cdot \left( -1 \right)^{n} \cdot \Gamma\left( n \right) \cdot \left( \gamma + \psi_{0}\left( n \right) \right) \right)\\ \operatorname{I_{z}^{n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( \Gamma\left( n + 1 \right) \right)^{2}} \cdot \left( \Gamma\left( n + 1 \right) \cdot \ln\left( z \right) + \left( -1 \right)^{n} \cdot \left( -1 \right)^{n} \cdot \Gamma\left( n \right) \cdot \left( \gamma + \psi_{0}\left( n \right) \right) \right)\\ \operatorname{I_{z}^{n}}\left[ \ln\left( z \right) \right] &= \frac{z^{n}}{\left( \Gamma\left( n + 1 \right) \right)^{2}} \cdot \left( \Gamma\left( n + 1 \right) \cdot \ln\left( z \right) + \Gamma\left( n \right) \cdot \left( \gamma + \psi_{0}\left( n \right) \right) \right)\\ \end{align*} $$ 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: $$ \begin{align*} \operatorname{I_{z}^{\alpha}}\left[ \ln\left( z \right) \right] &= \frac{z^{\alpha}}{\left( \Gamma\left( \alpha + 1 \right) \right)^{2}} \cdot \left( \Gamma\left( \alpha + 1 \right) \cdot \ln\left( z \right) + \Gamma\left( \alpha \right) \cdot \left( \gamma + \psi_{0}\left( \alpha \right) \right) \right)\\ \end{align*} $$

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

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 $\ln\left( z \right) = \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \left( z - 1 \right)^{k} \right]$.

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

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. $$ \begin{align*} \operatorname{I_{z}^{\alpha}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \operatorname{D_{z}^{-\alpha}}\left[ \left( z - 1 \right)^{k} \right] \right]\\ \operatorname{I_{z}^{\alpha}}\left[ \ln\left( z \right) \right] &= \sum\limits_{k = 1}^{\infty}\left[ \frac{\left( -1 \right)^{k + 1}}{k} \cdot \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k + \alpha + 1 \right)} \cdot \left( z - 1 \right)^{k + \alpha} \right]\\ \end{align*} $$

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

Wolfram|Alpha gives the same Series Expansion. So:

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