How to integrate $\int \ln(x+\ln(x+\ln(x+...))) dx$?

Question

How to integrate: $$\int \ln(x+\ln(x+\ln(x+...))) dx$$

I am trying to integrate this expression by taking the above expression as $y$.

Therefore, $y=\ln(x+\ln(x+\ln(x+...)))$.

Now from this we can conclude that $y=\ln(x+y)$.

Therefore, $e^{y}=(x+y)$.

Now if I was asked to differentiate this expression then my method would be $e^{y}y'=1+y'$. Therefore, $y'(e^{y}-1)=1$. Therefore, $y'=\frac{1}{e^{y}-1}$, where $y=\ln(x+\ln(x+\ln(x+...)))$.

But I can't integrate this expression. Please help me out with this question.

Answer 1

$$y:{=\ln(x+\ln(x+\ln(x+...)))\\\Rightarrow y=\ln(x+y)\\\Rightarrow e^y-y=x\\\Rightarrow dx=(e^y-1)dy}$$

Without making things compliacted we are calculating for $x\ge1$ branch,

$${\quad\,\int \ln(x+\ln(x+\ln(x+...))) dx\\ =\int y(e^y-1)dy\\ =\color{#027cc4}{\int ye^ydy} -\int ydy \\ =\color{#027cc4}{y\int e^ydy -\int{\left[\,{\frac{d}{dy}(y)\int e^ydy}\,\right]\,dy}}-\int ydy\quad\color{#027cc4}{\text{(IBP)}}\\ =ye^y-e^y-\frac{y^2}2+c\\ =e^y(y-1)-\frac{y^2}2+c}$$

If we write the answer in terms of $x$,

$$[x+\ln(x+\ln(x+...)))][\ln(x+\ln(x+...)))-1]-\frac{[\ln(x+\ln(x+...)))]^2}2+c$$

Answer 2

first $x\geq 1$ or the logarithm would be complex $f_0(x) :=x \ \ , f_n(x):=\ln(x+f_{n-1}(x)) $ then the question is equivalent to find $\int \lim\limits_{n \to \infty }f_n(x)dx$ before doing anything we must know if the limit exist or not (if not since the sequence is monotone then it must be divergent which mean we cannot integrate it )

by the definition of logarithm if $a>b \ ,c>0$ then $\log_c(a)> \log_c(b)$ since $f_1(x) = \ln(x)$ and $f_2(x)= \ln(x+\ln(x))$ $\forall x\geq 1$ $f_2(x)>f_1(x)$ now assume for some $k \in \mathbb{N} \ \ f_{k+1}(x) > f_{k}(x)$ then $f_{k+2}(x)- f_{k+1}(x) =\ln(x+f_{k+1}) - \ln(x+f_{k}(x))>0$ so the sequence is monotone by induction then it has a limit if it was bounded (by Monotone convergence theorem) note that $\sqrt x \geq \ln(x) \ \forall x\ge1$ so define $g_{0}(x) =x ,g_{n+1}(x)=\sqrt{x+g_n(x)}$ if $g_n(x)< \frac{1+ \sqrt{1+4x}}{2}$ then $g_{n+1}(z)< \frac{1+ \sqrt{1+4x}}{2}$ which mean that the sequence $g_n(x)$ is bounded and since $g_n(x)> f(n) $ that also mean that the sequence is bounded above then the limit exist for all $x\in [1, \infty)$

now for the easy part since the limit exist lets call $\lim\limits_{n \to \infty} f_n(x):=$ $f(x)$ conclude that $e^{f(x)}=x+f(x)$ let $e^{f(x)}-f(x)=x$ then $dx= (e^{f(x)}-1) df(x)$ so that $$\int f(x)dx= \int(e^{f(x)}-1) f(x) df(x)=f(x)e^{f(x)}- e^{f(x)} -\frac{(f(x))^2}{2} +C $$

since $e^{f(x)}=x+f(x)$ $$= \frac{f(x)^2}2+xf(x)-x-f(x) +C$$

Answer 3

$$y= \ln(x+\ln(x+\ln(x+...))) $$ From the ideas given by O M and leonbloy $$e^y-y=x$$ Solving for $y$ involves a special function : $$y=-W(-e^{-x})-x$$ $W$ is the LambertW function. https://en.wikipedia.org/wiki/Lambert_W_function

The LambertW function is multivalued. The two real branches are noted $W_{0}$ and $W_{-1}$

In the present case we consider only the real branches which implies $x\geq 1$.

Given $x\geq 1$ the iterative numetical calculus of a finite series $y= \ln(x+\ln(x+\ln(x+...))) $ is possible. This leads to approximate values of $y\geq 0$ of the limit value for infinite series calculated with the Lambert $W_{-1}$ function.

The branch with Lambert $W_0$ for $y\leq 0$ can't be computed with the above series. So we don't consider it as an effective solution of the problem.

Then considering the real positive branch alone, the integration leads to : $$\int y(x)\:dx =\int \big(-W_{-1}(-e^{-x})-x \big)\:dx $$ $$\boxed{\int y(x)\:dx=\frac12 \big(W_{-1}(-e^{-x})\big)^2+W_{-1}(-e^{-x})-\frac12x^2+C}$$ This is consistent with the pie's result when his function $f(x)$ is expressed in term of Lambert W function.