Tetration and $f(x) = \exp(\int_1^x \ln(f(t)) dt)$

Question

Let $g(x)$ satisfy $g(1) = 1 , g'(1) = 1 , g(1+x) = \exp(g(x))$

Now it is clear that $g'(5) = g(5)g(4)g(3)g(2)$

This invites to think of the function $f(x)$ which is defined similarly and might or might not behave alot like some $g(x)$.

$f(x)$ satisfies for $x>1$:

$$f(1) = f'(1) = 1$$

$$f(x) = \exp(\int_1^x \ln(f(t)) dt)$$

What are the solutions for $f(x)$ ?

Do we have uniqueness ?

Answer 1

No differentiable function $f$ can satisfy $f(1)=f'(1)=1$ along with $$f(x) = \exp\left(\int^x_1 \ln(f(t))dt\right).$$

Indeed, assuming that $f$ is differentiable, and satisfies the functional equation, taking the logarithm of the equation, we have $$\ln(f(x)) = \int^x_1 \log(f(t))dt.$$ Let $F(x) = \ln(f(x))$, so $$F(x) = \int^x_1F(t)dt.$$ By the fundamental theorem of calculus, $$F'(x) = F(x),$$ so $$F(x) = F(1)e^{x-1}.$$ Now if $f(1) = 1$, then $F(1) = 0$ which would mean $F$ is identically zero, so $f$ is identically $1$, and cannot satisfy $f'(1)=1$.