Prove $\int_0^1 \frac{\log x}{x^2} \cos x$ is convergent

Question

I am trying to prove $\int_0^1 \frac{\log x}{x^2} \cos x$ is convergent. I have tried various approaches but none of those work.

For example using by parts to change the integral.

For example using the fact that $\cos x = 1 - \frac{x^2}{2} + \operatorname{o}(x^3)$ to establish that:

for all $\varepsilon > 0$ we can find $\delta > 0$ such that for all $x \in (0, \delta)$ we have $0 < \cos x < 1 - \frac{x^2}{2} + \varepsilon x^3$

and then try to show the Cauchy criterion, i.e.

for all $\varepsilon > 0$ we can find $\delta > 0$ and $\epsilon_1 > 0$ such that for all $a, a' \in (0, \delta)$ we have $|\int_a^{a'} \frac{\log x}{x^2}(1 - \frac{x}{2} + \varepsilon_1x^3)| < \epsilon$

I think the main problem here is that I don't know how to bound $\cos x$. Thanks for your help!

Answer 1

The only problem occurs at $x=0$, and at $x=0$, $\cos x$ is nowhere near $0$ so you can more or less just ignore it.

You probably know that $$\int_0^1 \frac{-1}{x^2}dx$$ does not exist, and using the fact that if $\log x < -1$ for all $x<\frac1e$, you can see that $$\frac{\log x}{x^2} < \frac{-1}{x^2}$$

which should be enough to prove that the integral does not converge.

Answer 2

This proof is based on the one given by 5xum but with details filled in.

Since $$\lim_{x \to 0^+} \cos x = 1$$

We can find $\delta_1 > 0$ such that for all $x \in (0, \delta_1)$

$$\cos x > \frac{1}{2}$$

On the other hand, for any $x \in (0, \frac{1}{3})$

$$-\log x > 1$$

Now take $\delta = \min\{\delta_1, \frac{1}{3}\}$.

Then

$$\int_0^1 -\frac{\log x}{x^2} \cos x = \int_0^\delta -\frac{\log x}{x^2} \cos x + \int_\delta^1 -\frac{\log x}{x^2} \cos x$$

Since $-\frac{\log x}{x^2} \cos x \ge 0$ for all $x \in (0, 1]$, it suffices to show $\int_0^\delta -\frac{\log x}{x^2} \cos x$ diverges.

Now for all $x \in (0, \delta)$

$$\cos x > \frac{1}{2}$$ $$\implies -\frac{\log{x}}{x^2} \cos x \ge -\frac{1}{2}\frac{\log{x}}{x^2} \ge 0$$

Finally, since $-\frac{1}{2}\frac{\log{x}}{x^2}$ diverges by direct computation. Using comparison test, we must have $\int_0^\delta -\frac{\log x}{x^2} \cos x$ being divergent. So we are done.

Answer 3

To 5xum As I know, improper integral bounded by divergent improper integral does not imply divergence. But improper integral bounded by convergent improper integral implies convergence.