This is tetration question about finding the indefinite integral. I am not sure where to start so any help would be appreciated.

$$ I= \int \ln(x)^{\ln(x)^{\ln(x)^{\cdot^{\cdot^{\cdot^{\ln(x)}}}}}} dx $$

  • $\begingroup$ With one $\ln$ this is well known. Now do it with two $\ln$s before asking us the general question. $\endgroup$ – GEdgar Dec 30 '17 at 18:49
  • $\begingroup$ I am aware of how to approach those but not this one. @GEdgar $\endgroup$ – Fred Wieser Dec 30 '17 at 18:54
  • $\begingroup$ I assume that $\ln(x)^{\ln(x)}$ is $(\ln x)^{\ln x}$, right? The entire logarithm is the base of the power, not just $x$, isn't it? $\endgroup$ – Βασίλης Μάρκος Dec 30 '17 at 18:58
  • $\begingroup$ yes that is correct @ΒασίληςΜάρκος $\endgroup$ – Fred Wieser Dec 30 '17 at 18:59
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    $\begingroup$ A very personal opinion : tetration issues haven't a wide interest ; you never encounter them in applied mathematics and in "pure mathematics", they are not connected to other branches. $\endgroup$ – Jean Marie Dec 30 '17 at 19:20

These are less of an answer and more of a set of thoughts for a specific case of the problem.

For $n=2$: $$\begin{align*} \int (\ln x)^{\ln x}dx&\overset{x=e^y}{\underset{dx=e^ydy}{=}}\int y^ye^ydy=\tag{$\star$}\\ &=\int e^{y\ln y}e^ydy=\\ &=\int e^{y\ln y+y}dy=\\ &=\int\sum_{k=0}^\infty\frac{(y\ln y+y)^k}{k!}=\\ &=\sum_{k=0}^\infty\frac{1}{k!}\int(y\ln y+y)^kdy=\\ &=\sum_{k=0}^\infty\frac{1}{k!}\int\sum_{n=0}^k\binom{k}{n}y^n\ln^nyy^{k-n}dy=\\ &=\sum_{k=0}^\infty\frac{1}{k!}\int y^k\sum_{n=0}^k\binom{k}{n}\ln^nydy=\\ &=\sum_{k=0}^\infty\frac{1}{k!}\sum_{n=0}^k\binom{k}{n}\int y^k\ln^nydy \end{align*}$$ I introduced power series since I think that integrals of the form $\int x^xdx$ cannot be calculated in terms of "simple" functions.

Now, if one would like to procceed to further calculations, I feel that we should talk about definite integrals. However, one can calculate the integral: $$I_{n,k}=\int y^k\ln^nydy$$ by applying multiple times integration by parts: $$I_{n,k}=\frac{y^{k+1}}{k+1}\ln^ny-\frac{n}{k+1}I_{n-1,k}+c$$

However, when $n>2$, I cannot find something as "straightforward" as the above proccedure...


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