Function that converges to infinity slow enough and is yet integratable

Question

I'm looking for a continuously differentiable and positive function $f(x)$ for which

1. $\lim_{x \to 0} f(x) = \infty$
2. $\lim_{x \to 0} \frac{d}{dx} \frac{1}{f(x)} \neq \infty$
3. $\int_0^1 f(x) dx < \infty$

I tried several functions:

• $1/x^a$ with $0 < a < 1$ violates 2.: The derivative of the inverse is $a x^{a -1}$ and goes to $\infty$
• $1/x^b$ with $b \geq 1$ violates 3. (the integral won't converge)
• $\log(x)$ violates 2.

I have the impression that such a function might not exist. 1. and 2. basically ask for a function that converges "slowly enough" to 0. Because the function then converges too slowly, the integral does not converge.

Does that make sense? Or is there a functional form I missed?

Answer 1

If my calculations are right, $f=e^{x^{-a}}$ works for $a\in\left( 0,\,1\right)$. Then $f^{-1}$ has $x$-derivative $ay^{a+1}e^{-y^a}$ with $y:=x^{-1}$, so the finite $y\to\infty$ limit required in your second riterion is in fact $0$. Since $f$ ranges from $1$ to $e$ over the last criterion's integration range, that integral should be finite too.

Answer 2

Consider the map :

$$f:(0,+\infty)\to\mathbb{R},\,x\mapsto\frac{1}{\sqrt x}+\sin\left(\frac 1x\right)$$

$f$ is infinitely differentiable

$\int_0^1f(x)\,dx$ exists because $x\mapsto\frac 1{\sqrt x}$ is integrable and $x\mapsto\sin\left(\frac{1}{x}\right)$ is locally integrable on $(0,1]$ and bounded.

$(1/f)'$ doesn't have any limit as $x\to0^+$