Closed form for n-th anti-derivative of $\log x$ Is it possible to write a closed-form expression with free variables $x, n$ representing the n-th anti-derivative of $\log x$?
 A: Related problems: (I), (II). Here is a unified formula for the $n$th derivative and the $n$th anti-derivative of real order of $\ln(x)$ in terms of the Meijer G function

$$ G^{1, 2}_{2, 2}\left(x-1\, \Big\vert\,^{1-n, 1-n}_{1-n, 0}\right). $$

The above formula gives
a) derivatives of real order if $n>0$,
b) anti-derivatives of real order if $n<0$.
One can have the above formula in terms of the hypergeometric formula

$$ {\frac { \left( x-1 \right) ^{1-n}{\mbox{$_2$F$_1$}(1,1;\,2-n;\,1-x)}
}{\Gamma  \left( 2-n \right) }},$$

with some restrictions on $n$.
A: An intuitive way to get the formula from the answer of Vladimir Reshetnikov without induction:
Recall that $\displaystyle\ln x=\lim_{s\rightarrow 0}\frac{x^s-1}{s}$. But the $n$th antiderivative of $x^s-1$ is very easy to compute:
$$\frac{1}{s}\underbrace{\int\ldots\int}_{n\;\text{times}}(x^s-1)dx=\frac{x^{s+n}}{s(s+1)\ldots(s+n)}-\frac{x^n}{s\cdot n!}=_{s\rightarrow0}\frac{x^n}{n!}\left(\ln x-\sum_{k=1}^n\frac{1}{k}\right).$$
A: $$\log^{(-n)}x=\frac{x^n}{n!}(\log x-H_n),$$
where $H_n$ is the harmonic number: $H_n=\sum_{k=1}^n k^{-1}=\,$$\gamma$$\,+\,$$\psi$$(n+1)=\gamma+\frac{\Gamma'(n+1)}{n!}$
Proof: By induction.
