# when does a generalized integral diverges

I know that when $$f(0)=0$$,the integral $$I=\int_0^\varepsilon \frac{1}{f(x)}dx$$ may not diverge (like $$f=\sqrt{x}$$.

So I want to know if a function $$f$$ is differentiable in a neighborhood of $$0$$ such that $$f(0)=f’(0)=0$$, can we say that for any $$\varepsilon$$, the integral $$I$$ must diverge.

I don’t know how to prove it ,or give a counterexample.

• What's the relationship between $f(x)$ and $I$? Oct 26 at 14:41
• @5xum sorry! I have rewritten the equation. Oct 26 at 14:42
• @Andrei Sorry! I have rewritten the equation. Oct 26 at 14:43
• Can you assume that $f'$ is continuous on some interval $(0,t)$ or do you specifically want to avoid that?
– ajr
Oct 26 at 14:48

Your conditions are sufficient to prove that $$\lim_{x\to 0}\frac{f(x)}{x} = 0$$

Method 1:

If $$f$$ is differentiable at $$x_0$$, then, by the definition of differentiability, there exists some function $$o$$ such that, for small values of $$h$$, you have

$$f(x_0+h)=f(x_0)+f'(x_0)\cdot h + o(h)$$

and the limit $$\lim_{h\to 0}\frac{o(h)}{h}=0.$$

In your case, $$x_0=0$$ gives you

$$f(x)=0+0\cdot x + o(x)=o(x)$$

so you basically know that

$$\lim_{x\to 0}\frac{f(x)}{x} = 0$$

Method 2:

By another definition of $$f'$$, you have

$$0=f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(h)}{h} = \lim_{h\to0}\frac{f(h)-0}{h} = \lim_{h\to0}\frac{f(h)}{h}$$

In both cases, once you have the limit, it is trivial to show that for values of $$x$$ close to $$0$$, you have $$|f(x)|<|x|$$, or, for positive values of $$x$$, simply $$|f(x)|.