# Suppose $f(0)=0$, $f^{\prime}(0)$ exists, and $f$ is positive on $\mathbb{R}_+$. Prove $\int_0^b\frac{dx}{f(x)}$ diverges, where $b>0$ is arbitrary.

The proof that my TA gave was that, since $f$ is differentiable at 0, it is somehow asymptotically related to some polynomial $x^p$ for $p\geq1$. Then since the integral of $x^{-p}$ diverges over any interval [0,b], so must the integral of $f^{-1}$ (since $f<x^p\implies f^{-1}>x^{-p}$). I don't actually understand precisely what it means to be asymptotically related to some polynomial, so I tried to go by it by first principles.

My best guess is that, since $\lim_{h\rightarrow 0}\frac{f(h)}{h}$ converges then $f$ converges faster than the identity $function$ so $\frac{1}{f}$ diverges faster than $\frac{1}{x}$ and since the integral of $\frac{1}{x}$ diverges on the interval then the integral of $f$ diverges on that interval.

However, that is not a rigorous proof at all.

• $b$ is just some positive number?
– Moya
Commented Feb 25, 2016 at 2:52
• Yes it's just an arbitrary $b>0$. I will edit question to reflect that. Commented Feb 25, 2016 at 2:53

The statement that $f'(0)$ "exists" is normally taken to mean it is a real number -- thus finite rather than either $\pm\infty$. If $f'(0)$ were negative, then $f(x)$ itself would be negative for $x$ sufficiently close to $0$, so $f'(0)$ must be $\ge 0$. Let $a=f'(0)+1<\infty$. Recall that $$f'(0) = \lim_{x\,\downarrow\,0} \frac{f(x)-f(0)}{x-0}.$$ So for $x$ close enough to $0$, we have $$\frac{f(x)-f(0)}{x-0} < a.$$ Hence $f(x) < ax$. Therefore $\dfrac 1 {f(x)} \ge \dfrac 1 {ax}$. Hence $$\int_0^b \frac{dx}{f(x)} \ge \int_0^b \frac{dx}{ax} = \infty.$$