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Why does $\int_{0}^{\infty} \frac{\ln(x^2)}{x^2}$ converge?

I've tried using the substitution $u = ln(x^2)$, but it doens't seem to lead anywhere...

Any help would be appreciated!

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    $\begingroup$ It does not converge, there is bad trouble near $0$. $\endgroup$ – André Nicolas Mar 25 '16 at 5:22
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For $ 0 < x < 1$, we have the inequality $1-x < -\ln x.$

This follows from

$$-\ln x = \int_x^1 t^{-1}\, dt > \frac{1-x}{1}.$$

Hence,

$$\frac{- \ln x^2}{x^2} > \frac{1-x^2}{x^2} = x^{-2} - 1.$$

Therefore, we have as $c \to 0$

$$-\int_c^1\frac{ \ln x^2}{x^2} \, dx > \int_c^1 (x^{-2} - 1) \, dx = c^{-1} - 2 + c \to \infty,$$

and the integral diverges.

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  • $\begingroup$ @John B: You're welcome. $\endgroup$ – RRL Mar 25 '16 at 6:15

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