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For a related question, I need to know the $n$th integral of $\ln(x)$ and the fractional derivative of $\ln(x)$.

A break down of how fractional derivatives may be found on the Wikipedia.

In particular, I need to calculate $\frac{d^{1/2}}{dx^{1/2}}\ln(x)$ and $\frac{d^{-n}}{dx^{-n}}\ln(x)$ where that is the $n$th integral of $\ln(x)$.

The fractional derivative in this scenario is given by:

$$\frac{d^{1/2}}{dx^{1/2}}\ln(x)=\lim_{h\to0}\frac{(-1)^{1/2}}{h^{1/2}}\sum_{0\le m<\infty}\frac{\Gamma(1.5)}{\Gamma(m+1)\Gamma(1.5-m+1)}\ln(x+mh)$$

It is rather difficult to take the limit from my skills, so I was hoping someone could solve it. (I do accept power series answers or anything the cannot be written in easy closed form)

Secondly, I have attempted to find the $n$th integral of $\ln(x)$ and this is what I found

$$\frac{d^{-n}}{dx^{-n}}\ln(x)=\frac{x^n[\ln(x)-\sum_{i=1}^{n}\frac1i]}{\Gamma(n+1)}$$

Two problems about this formula, I'm unsure if it works fully, and I need it to work for non whole $n$.

Thanks for your time and efforts.

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  • $\begingroup$ You may want to use $\displaystyle \sum_{i=1}^{n} \frac1i= \int_{y=0}^1\frac{1-y^n}{1-y}\,dy $ or some other known relation for your extension into fractional $n$ $\endgroup$ – Henry Feb 27 '16 at 14:02
  • $\begingroup$ @Henry Thanks. ${}$ $\endgroup$ – Simply Beautiful Art Feb 27 '16 at 14:03
  • $\begingroup$ Your $n^{\text{th}}$ indefinite integral could also add an arbitrary $n-1^{\text{th}}$ degree polynomial of $x$ (corresponding to the constant of integration) and this might raise issues for fractional $n$ $\endgroup$ – Henry Feb 27 '16 at 14:05
  • $\begingroup$ @Henry Yes, I wish to try and ignore the constant of integration as much as possible for simplicity's sake. $\endgroup$ – Simply Beautiful Art Feb 27 '16 at 14:08
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If the fraction derivative for $0<\alpha<1$ is defined through: $$ D^{\alpha} f(x)=\frac{1}{\Gamma(1-\alpha)}\cdot\frac{d}{dx}\int_{0}^{x}\frac{f(t)}{(x-t)^\alpha}\,dt \tag{1}$$ we have: $$ D^{\frac{1}{2}}\log x = \frac{1}{\sqrt{\pi x}}\left(\log x+\log 4\right).\tag{2}$$ The $n$-th antiderivative (for $n\in\mathbb{N}$) is indeed $$ D^{-n}\log x = \frac{x^n}{n!}\left(\log x-H_n\right)\tag{3} $$ and that is straightforward to prove through integration by parts and induction.

Notice that $(2)$ and $(3)$ agree as expected since $H_{-1/2}=-\log 4$.

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