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I have to check if $\int_{0}^\infty \mathrm 1/(x\ln(x)^2)\,\mathrm dx $ is convergent or divergent.

My approach was to integrate the function , hence : $\int_{0}^\infty \mathrm 1/(x\ln(x)^2)\,\mathrm dx=-\lim_{x \to \infty} 1/\ln(x)+ \lim_{x \to 0} 1/\ln(x)=0 $

Still my book says that it is divergent. Maybe the $\infty$ sign of the integral means to check for $+\infty$ and $-\infty $ or i just overlooked something. Any help would be appreciated.

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    $\begingroup$ You have a problematic point at $x=1$... $\endgroup$ – David Mitra Jul 9 '13 at 15:12
  • $\begingroup$ $\ln(1)$ is zero. $\endgroup$ – OR. Jul 9 '13 at 15:12
  • $\begingroup$ @DavidMitra thanks $\endgroup$ – sigmatau Jul 9 '13 at 15:13
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Look at the following improper integral: $$ \int_1^2f(x)dx $$ Certainly, $\lim_{x\to 1^+}(x-1)f(x)=+\infty$ so the Comparison test admits the series is divergent.

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    $\begingroup$ $x \to +1^+ = +\infty^+\quad \ddot\smile$ $\endgroup$ – Namaste Jul 10 '13 at 0:09
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Let's change the variables:

$$y=\ln {x} \\ dy =\frac{dx}{x}$$

Now the integral becomes:

$$I=\int_0^\infty \frac{dx}{x(\ln{x})^2}=\int_{-\infty}^\infty \frac{dy}{y^2}.$$

Note that the integrand is an even function, ergo(the main goal is to avoid the singularity):

$$I=2\int_0^{\infty} \frac{dy}{y^2}=-\frac{2}{x}\large{|_0^{\infty}}=\infty $$

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