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I would like to find a differentiable $f:[1,\infty) \to \mathbb{R}$ satisfying:

$$f(x) = \ln(x)f'(x)$$.

My work:

We have $f(1) = 0$ and $f(e)=f^k(e)$. A Taylor series centered at $x_0=e$ gives us the following polynomial: $$f(e)+f(e)(x-e)+f(e)\frac{1}{2}(x-e)^2 \dots$$ $$=f(e) \sum_{k=1}^\infty \frac{(x-e)^k}{k!}$$ although I don't see how this might help me, and I also am not sure how else to approach this problem. Does the condition $f(e)=f^k(e)$ force $f(x)$ to be $e^x$, which would contradict $f(x)=\ln(x)f'(x)$, and thus no $f$ exists? I wouldn't think so, although I know that $Ae^x$ is the unique solution to $f(x)=f'(x)$ from this post.

doraemonpaul
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Descartes Before the Horse
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    If $$\frac{1}{\ln(x)}=\frac{f'(x)}{f(x)}$$ then $$ \int \frac{1}{\ln(x)}=\int \frac{f'(x)}{f(x)}=\ln(f(x)) $$ The problem is that the LHS does not have an expression in elementary terms. – Reveillark Dec 02 '19 at 04:55
  • @Reveillark The question does not ask for an elementary function $f$, just a differentiable $f$. One should perhaps focus on the existence. – WhatsUp Dec 02 '19 at 05:12
  • @Reveillark But shouldn't you take care of the pole at $x = 1$? – WhatsUp Dec 02 '19 at 05:16
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    @WhatsUp Hmmm yeah you might be right, you run into trouble there. I think there might be something to salvage in my original comment, so I'll leave it up. The constantly $0$ function definitely works though :P – Reveillark Dec 02 '19 at 05:23
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    @Reveillark Alright... you win with zero... – WhatsUp Dec 02 '19 at 05:24
  • @WhatsUp I do not think that one can really take care of the pole: the solution to the differential equation is $ce^{\text{li}(x)}$ which has a cuspid at $x=1$ for all the $c\neq 0$. –  Dec 02 '19 at 06:49
  • How do you draw $f(e)=f^k(e)$, which seems to be a wrong identity ? –  Dec 02 '19 at 09:35

2 Answers2

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As Caffeine commented $$\frac{f'(x)}{f(x)}=\frac1{\log(x)}\implies f(x)=C e^{\text{li}(x)}=C e^{\text{Ei}(\log(x))}$$

If you look for a series expansion of $e^{\text{li}(x)}$ around $x=e$, it is given by $$e^{\text{Ei}(1)}\left(1+(x-e)+\frac{(e-1) }{2 e}(x-e)^2+\frac{e^2-3e+3}{6 e^3}(x-e)^3+ O\left((x-e)^4\right)\right)$$ where $e^{\text{Ei}(1)}\approx 6.65333$

Claude Leibovici
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As written, the equation has a nasty singularity at $x=1$ and this is reflected in the solution with the integral logarithm (see Claude's answer).

Around $x=1$ we can write $\log x\approx x-1$, leading to the solution $f(x)\approx|x-1|$. Hence we substitute $f(x)=(x-1)g(x)$, giving us

$$\frac{g'(x)}{g(x)}=1-\frac{\log x}{x-1}.$$

With the kind help of WA,

$$\log g(x)+c=x+\text{Li}_2(1-x)=x+\sum_{k=1}^\infty\frac{(1-x)^k}{k^2}.$$

($\text{Li}_2$ is the dilogarithm function, which is quite smooth.) The given series converges in $[0,2]$. Values elsewhere can be obtained by these identities: https://en.wikipedia.org/wiki/Spence%27s_function#Identities.