Closed form for n-th anti-derivative of $\log x$

Question

Is it possible to write a closed-form expression with free variables $x, n$ representing the n-th anti-derivative of $\log x$?

http://www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1692-02.pdf — Jose Garcia, May 18 '13 at 22:45
Basically a duplicate of: http://math.stackexchange.com/q/13591/27624 — Argon, May 18 '13 at 22:59
@Argon: Even though I cast a close vote, I now believe it is slightly different, as the $H_n$ term becomes important for $\log x$, but not for $\frac{1}{x}$. — Aryabhata, May 19 '13 at 05:54

Vladimir Reshetnikov · Accepted Answer · 2013-05-19T03:21:08.450

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$$\log^{(-n)}x=\frac{x^n}{n!}(\log x-H_n),$$ 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.

edited May 19 '13 at 03:21

answered May 18 '13 at 22:44

Vladimir Reshetnikov

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I am not sure if it would be a proper notation if I wrote $(d/dx)^{-n}\log x=...$ – Vladimir Reshetnikov May 18 '13 at 22:52
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Writing $\log^{(-n)}(x)$ looks a little better : link – Raymond Manzoni May 18 '13 at 22:57
Oldham and Spanier in their "The Fractional Calculus" use $\frac{d^{-n}}{dx^{-n}}$ – Argon May 18 '13 at 23:00
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I think your $H_n$ is defined incorrectly. – GEdgar May 19 '13 at 02:53
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@GEdgar Thanks, it was totally wrong! I kept in mind the sum of an initial segment of the harmonic series, but for some reason wrote the divergent sum of the whole series. – Vladimir Reshetnikov May 19 '13 at 03:27
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You can add an arbitrary $(n-1)^{\text{th}}$ degree polynomial... – Aryabhata May 19 '13 at 05:54

Mhenni Benghorbal · Answer 2 · 2020-03-12T21:06:45.640

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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 $n$.

edited Mar 12 '20 at 21:06

answered May 23 '13 at 07:02

Mhenni Benghorbal

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(+1) its so beautiful to find closed form using Meijer G function but how to prove it ? – mnsh Aug 14 '13 at 11:39

Start wearing purple · Answer 3 · 2013-05-23T15:32:54.307

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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: $$\frac{1}{s}\underbrace{\int\ldots\int}_{n\;\text{times}}(x^s-1)dx=\frac{x^{s+n}}{s(s+1)\ldots(s+n)}-\frac{x^n}{s\cdot n!}=_{s\rightarrow0}\frac{x^n}{n!}\left(\ln x-\sum_{k=1}^n\frac{1}{k}\right).$$

edited May 23 '13 at 15:32

answered May 23 '13 at 07:35

Start wearing purple

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There is a typo, you are missing a $\frac{1}{s}$ in $\ln x = \lim\dots$. Don't you need some justification for taking the limit inside though? – Aryabhata May 23 '13 at 15:26
@Aryabhata: Thank you! I've edited it. Of course this is not a rigorous derivation, I just tried to make the formula of V.R. less misterious, as usually the hardest part in the proof by induction is to guess the form of the answer... – Start wearing purple May 23 '13 at 15:32

Closed form for n-th anti-derivative of $\log x$

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