convergence test : $\int_{0}^\infty \mathrm 1/(x\ln(x)^2)\,\mathrm dx $

Question

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.

You have a problematic point at $x=1$... – David Mitra Jul 09 '13 at 15:12 — David Mitra, Jul 09 '13 at 15:12

score 2 · Accepted Answer · answered Jul 09 '13 at 15:31

2

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.

answered Jul 09 '13 at 15:31

Mikasa

67,374

1

$x \to +1^+ = +\infty^+\quad \ddot\smile$ – amWhy Jul 10 '13 at 00:09

score 1 · Answer 2 · answered Jul 09 '13 at 16:54

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):

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

convergence test : $\int_{0}^\infty \mathrm 1/(x\ln(x)^2)\,\mathrm dx $

2 Answers2