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I have to determine using the Comparison theorem whether the following integral converges or diverges:

$$\int_0^\infty \frac{\ln{x}}{x^2} dx $$

Its kinda obvious that this integral is improper because the upper limit goes to infinity and it is asymptotic at $x=0$. Then i split it in two parts:

$$\int_0^\infty \frac{\ln{x}}{x^2} dx = \int_0^1 \frac{\ln{x}}{x^2} dx + \int_1^\infty \frac{\ln{x}}{x^2} dx$$

Of course, the first part has negative sign and the second, positive sign.

Then, is possible to determine with Comparison Theorem the nature of this integral? In that case, which are the functions to compare?

Also, is it possible that the negative area could be canceled by the positive area?

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For the upper integral, note that $\ln(x) < \sqrt{x}$ for all $x > 1$.

For the lower integral, note that $\int_0^1 x^{-2}dx$ already diverges, and $\ln(x)$ just makes it diverge even harder.

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Similar to eyeballfrog's answer, since for any $x \geq 1$, $\ln{(x)} \leq x^\frac 1e$ $$ \int_1^p \frac{\ln{(x)}}{x^2}\,dx \leq \int_1^p\frac {x^\frac 1e}{x^2}\,dx=\frac{e \left(p-p^{\frac{1}{e}}\right)}{(e-1) p} \lt \frac e {e-1}$$

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If the integral converges then $\int_{\epsilon}^{2\epsilon} \frac {\ln \, x } {x^{2}}\, dx$ must tend to $0$ as ${\epsilon} \to 0$. However, $\int_{\epsilon}^{2\epsilon} \frac {\ln \, x } {x^{2}}\, dx \leq \ln\, (2 \epsilon) \int_{\epsilon}^{2\epsilon} \frac 1 {x^{2}}\, dx=\ln\, (2 \epsilon)[\frac 1 { \epsilon}-\frac 1 {2 \epsilon} ]\to -\infty$ .

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