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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.

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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.

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  • $\begingroup$ Ah, so $D^\mu\ln(x)=\frac{x^{-\mu}\left(\ln(x)-\gamma-\psi(1-\mu)\right)}{\Gamma(1-\mu)}$... ok, thanks. This was helpful. Can different methods to computing fractional differintegrals result in different results? $\endgroup$ – Simply Beautiful Art Jun 1 '16 at 20:45
  • $\begingroup$ Wow, I just graphed that, and it turns out that the result is the same. Thank you. desmos.com/calculator/hk9h7zsu5b $\endgroup$ – Simply Beautiful Art Jun 1 '16 at 20:51

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