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Edit : As it get too many downvote, I am about to delete this question. But @insipidintegrator suggest me to try to improve it. Ok, I will try but if it still keep getting more and more downvote. I will delete it.

The full question is

$$f(x) = \int \frac{\mathrm dx}{x\ln(x)\ln(\ln(x))\ln(\ln(\ln(x)))...\overbrace{\ln(...\ln(x)...)}^{100\ \text{copies of}\ln}}$$

$f(x)$ is in the simplified form. How many "ln"s appear in $f(x)$?

It is a challenging integration question that I got from my friend.

My substitution attempt

Let $y = \ln(x), dx=dy/x$

\begin{equation} f(x) = \int \frac{\mathrm dx}{x\ln(x)\ln(\ln(x))\ln(\ln(\ln(x)))...\overbrace{\ln(...\ln(x)...)}^{100\ \text{copies of}\ln}} = \int \frac{\mathrm dy}{y\ln(y)\ln(\ln(y))...\overbrace{\ln(...\ln(y)...)}^{100\ \text{copies of}\ln}} \end{equation}

It seem to get a bit better but I am still don't know what to do next.

I also tried to use "Integration by parts" but I don't see what should be $u$ and $dv$.

This question probably needs an integration formula that I don't know, or it is actually impossible to integrate, but with some technique, it is possible to know how many $\ln$ is in it. Please help me.

Suzu Hirose
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Ro Theory
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    Start with the simpler examples (zero copies of $\log$): what is $\int \frac{dx}{x}$? Next (one copy of $\log$): what is $\int \frac{dx}{x\log(x)}$ ? Next consider $\int \frac{dx}{x\log(x)\log\log(x)}$ (two copies of $\log$)... – Winther Sep 03 '22 at 20:10
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    Hint: prove by induction on $n\ge0$ that$$\frac{d}{dx}\ln^{\circ n}x=\frac{1}{\prod_{k=0}^{n-1}\ln^{\circ k}x}$$where$$f^{\circ 0}(x):=x,,f^{\circ k+1}(x):=f(f^{\circ k}(x)).$$Unfortunately, since $\int\frac{dx}{x}$ already requires some subtlety, your problem has a very rich family of solutions. – J.G. Sep 03 '22 at 20:58
  • @Winther I don't want to solve this question any more too many downvote. Could you tell me why my question is so bad ? – Ro Theory Sep 04 '22 at 05:28
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    Deleting questions counts against you when it comes to question bans. Might be better to keep the question up, no matter what. – Gerry Myerson Sep 04 '22 at 10:26

1 Answers1

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In this answer I use $\log$ rather than $\ln$ to represent the natural logarithm (logarithm to the base $e=2.718...$).

By the chain rule, $$ {d\over dx} \log(\log(x))={d\over dy}\log y\cdot{d\over dx}\log x = {1\over y}\cdot{1\over x}={1\over x\log x} $$ where $y=\log x$. Similarly, let $z=\log(\log x)$ then $$ {d\over dx} \log(\log(\log(x))= {d\over dz}\log z \cdot {d\over dy}\log y \cdot {d\over dx}\log x \\ = {1 \over z}\cdot{1\over y}\cdot{1\over x} ={1\over x\log(x)\log(\log(x))} $$ Hopefully you can see how to extend this to answer your question.

Suzu Hirose
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    So, $f(x) = \overbrace{\ln(...\ln(x)...)}^{101\ copies\ of\ln}+c$ and have total 101 "ln"s. Thank you. – Ro Theory Sep 06 '22 at 10:49