f is continuously differentiable and $f'(0) \neq 0$ and $f(0) = 0$. show that $\int_0^1\frac{1}{f(\sqrt{x})}$ converges a)f is continuously differentiable and $f'(0) \neq 0$ and $f(0) = 0$. show that $$\int_0^1\frac{1}{f(\sqrt{x})}$$ converges
b) now f is continuously differentiable twice and $f(0) = f'(0) = 0$ but $f''(0) \neq 0$ prove that $$\int_0^1\frac{1}{f(\sqrt{x})}$$ diverges
Hi all, This question has to do with Improper integrals convergence. I thought maybe that because we know that the derivative in 0 is not 0 but the value is, the function is monotonous. Then I thought maybe to try and derive the integral but I'm not sure how. A solution would be appreciated.
 A: I'm going to assume that $f$ does not go to 0 on any other point of the interval, so we just have to worry of what happens near 0. On both cases use substitution $u = \sqrt{x}$ then the integral we are interested in becomes
$
2\int_{0}^{1} \frac{u}{f\left(u\right)} du.
$
Now, for item a, you get
$
\lim_{u \to 0} \frac{f\left(u\right)}{u} = f'\left(0\right) \neq 0
$
So you can find a lower bound for $\left|\frac{f\left(u\right)}{u}\right|$ in a neighboorhood of $0$, which gives you an upper bound for $\left|\frac{u}{f\left(u\right)}\right|$ and from this you can conclude.
For item b, you get using L'Hopital
$
\lim_{u \to 0} \frac{f\left(u\right)}{u^{2}} = \lim_{u \to 0} \frac{f'\left(u\right)}{2u} = f''\left(0\right) \neq 0
$
Now we have an upper bound (because the limit exists) and an lower bound, but for $\left|\frac{f\left(u\right)}{u^{2}}\right|$. I am going to assume this is positive, the negative case is analogous, so there are $\epsilon,c,C >0$ such that
$
c > \frac{f\left(u\right)}{u^{2}} > C 
\ \forall u \in \left[0,\epsilon\right],
$
hence
$
2\int_{0}^{1} \frac{u}{f\left(u\right)}du > 
2\int_{0}^{\epsilon} \frac{u}{f\left(u\right)}du > 
2\int_{0}^{\epsilon} \frac{u}{cu^{2}}  du = 2\int_{0}^{\epsilon} \frac{du}{cu} = +\infty
$
