Is the integral of a function that goes to infinity in an open interval also infinity? Consider a continuously differentiable function $f$ in the interval $[a,b]$, such that $f$ is positive in $[a,b)$ and $f(b) = 0$. Then is it always true that 
$$\lim_{x \rightarrow b}\int_a^x \frac{1}{f(t)} dt = +\infty$$
Clearly, if we relax the condition that $f$ be continuously differentiable, the conjecture doesn't hold; simply take the function $f(x) = x^{\frac{2}{3}}$ for $a=-1$ and $b=0$.
However, I haven't been able to prove one way or the other the fact for continuously differentiable functions. I'd be much obliged if someone could give me a hint or two in the right direction.
 A: If we assume that $f'(b)$ exists, then $f(b-h)\approx -hf'(b)$ so your integral is similar to $\lim_{h\to0^+}\int_h^1\frac{\mathrm dx}{x}$. You can make this more precise by noticing that  for any $c>-f'(b)\ge0$ we have $f(b-h)<ch$ for sufficiently small $h$ (say, for $h<h_0$), hence for $x>b-h_0$,
$$ \int_a^x\frac{\mathrm dt}{f(t)} =\int_a^{b-h_0}\frac{\mathrm dt}{f(t)}+\int_{b-x}^{h_0}\frac{\mathrm dt}{f(b-t)}>\int_a^{b-h_0}\frac{\mathrm dt}{f(t)}+\frac1c\int_{b-x}^{h_0}\frac{\mathrm dt}{t}$$
where the first summand is a constant and the second diverges to $\infty$ as $x\to b$.
A: By the mean value theorem, for all $x \in [a,b]$ there exists $\xi$ between $x$ and $b$ such that
$$f(x) = f(b) - f'(\xi)(b-x)= -f'(\xi)(b-x).$$
Since $f$ is continuously differentiable, $f'(x)$ is bounded on $[a,b]$ with $|f'(x)| < M$.
Therefore, $f(x) =|f(x)| < M(b-x)$ and
$$\frac{1}{f(x)}> \frac{1}{M(b-x)}.$$
Hence, the integral of $1/f$ diverges.
A: Try functions which are flat towards in the vicinity of your zero. An explicit example would be $f(x) = x^3$ on the interval $I = [0,1]$.
