Is it possible to find the maximum of $r\ln\left(\ln\left(1+\frac{1}{r}\right)\right)$ for $r\in(0,1)$?

Question

I was working on Chapter 5, Problem 14 of Evan's Partial Differential Equations 2nd ed and to prove integrability of $\ln \left (\ln \left (1+\dfrac{1}{|x|}\right )\right )$ on the unit $n$-ball, where $n>1$, I used the fact that$$f_n(r):=\begin{cases}r^{n-1}\ln \left (\ln \left (1+\dfrac{1}{r}\right )\right ) & \text{if} & r>0\\ 0 & \text{if} & r=0\end{cases}$$is continuous and thus bounded on $[0,1].$ While this is sufficient, I am curious to know if there is a way to find the maximal value of $f_2.$ I tried setting the derivative equal to $0$ but that does not seem solvable. Note that $|f_n|\leq |f_2|\leq |\ln (\ln (2))|$ which follows from the fact that the maximum of $f_2$ is less than $|\ln (\ln (2))|.$

Answer 1

Maximum clearly exists and is unique in the range $(0,1)$. You could find it by setting derivative to $0$. But I (and wolfram) cannot find a exact form for it. The equation for the maximizer $r^*$ is $$ \ln (\ln(1+1/r))=\frac{1}{(1+r)\ln(1+1/r)} $$ which could be "simplified" if you substitute $t=\ln(1+1/r)$, to $$ t\ln t = 1-e^{-t} $$ Which I don't think have a closed-form solution

Answer 2

As @Binxu Wang 王彬旭 answered, there is no hope to find an explicit expression for the zero of function $$g(t)=t \log(t)+ e^{-t}-1$$ For sure, numerical methods would not have any problem.

Surprising is the fact that (thanks to the $ISC$) the solution is extremely close to $$t_*=2+\sqrt{2}-\sqrt[4]{10}\implies g(t_*)=8.95625\times 10^{-9}$$

This gives $$r_*=\frac{e^{\sqrt[4]{10}}}{e^{2+\sqrt{2}}-e^{\sqrt[4]{10}}}=0.241882\cdots$$ $$f_2(r_*)=\frac{e^{\sqrt[4]{10}} \log \left(2+\sqrt{2}-\sqrt[4]{10}\right)}{e^{2+\sqrt{2}}-e^{\sqrt[4]{10}}}=0.119058\cdots$$ which are exactly the numbers given by Wolfram Alpha.

Edit

Looking more, still thanks to the $ISC$, the solution is such that $$t_1=\frac{\sqrt{2}+\sqrt[4]{6}}{\sqrt[4]{11}} < t <2+\sqrt{2}-\sqrt[4]{10}=t_2$$ and $t_2-t_1=9.21\times 10^{-8}$.

Using $t_1$ instead of $t_2$ as above, the same results.

Update

The general problem of $f_n(r)$ is interesting; it "just" requires to find the inverse of $$\frac{1-e^{-t}}{t \log (t)}=n-1$$

Expanded as a series around $t=1$ and using series reversion $$t=1+x+\frac{3-e }{2 (e-1)}x^2+\frac{11-10 e+2 e^2 }{3 (e-1)^2}x^3+\frac{263-397 e+185 e^2-27 e^3}{24 (e-1)^3}x^4+$$ $$\frac{4381-9279 e+6991 e^2-2229 e^3+256 e^4}{120 (e-1)^4}x^5+O(x^6)$$ where $x=\frac{e-1}{e (n-1)}$.

Truncated at this point, varying $n$, some results $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 2 & 1.6380429 & 1.6359341 \\ 3 & 1.3200040 & 1.3199637 \\ 4 & 1.2130091 & 1.2130053 \\ 5 & 1.1594971 & 1.1594964 \\ 6 & 1.1274324 & 1.1274324 \end{array} \right)$$

Making an expansion to $O\left((t-1)^k\right)$ leads to an inverse series to $O\left (\left (\frac {1} {n-1} \right)^{k+3} \right)$. We can convert the inverse series into a simple $[2,2]$ Padé approximant and obtain $$t=\frac{3+a_1 x+a_2 x^2 } {3+ b_1 x+b_2 x^2} \qquad \text{where} \qquad x=\frac{1}{2 e (5 e-17) (n-1)}$$ $$a_1=495-438 e+87 e^2 \qquad\qquad a_2=15487-26842 e+17520 e^2-4942 e^3+505 e^4$$ $$b_1=393-306 e+57 e^2 \qquad\qquad b_2=7327-9142 e+4452 e^2-994 e^3+85 e^4$$

$$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 2 & 1.6343788 & 1.6359341 \\ 3 & 1.3198883 & 1.3199637 \\ 4 & 1.2129938 & 1.2130053 \\ 5 & 1.1594935 & 1.1594964 \\ 6 & 1.1274312 & 1.1274324 \end{array} \right)$$