# Indefinite Integral of $\ln(x)^{\ln(x)^{\ln(x)^{.^{.^{.^{\ln(x)}}}}}}$ for an $n$-number of $\ln(x)$'s with respect to $x$

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

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

• With one $\ln$ this is well known. Now do it with two $\ln$s before asking us the general question. – GEdgar Dec 30 '17 at 18:49
• I am aware of how to approach those but not this one. @GEdgar – Fred Wieser Dec 30 '17 at 18:54
• 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? – Βασίλης Μάρκος Dec 30 '17 at 18:58
• yes that is correct @ΒασίληςΜάρκος – Fred Wieser Dec 30 '17 at 18:59
• 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. – Jean Marie Dec 30 '17 at 19:20

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...