I have the following function and I have to analyze its integrability in $(0,\infty)$:$\space$
$f(x)=\frac{1}{x^2 (\ln x)^3}$
I'm trying to use this definition:$\space$
f integrable $\leftrightarrow $ $\int_X |f| d\mu$ $\space$
But I don't know what to do... I know that $lnx$ is going tobe negative in $(0,1)$ so I can't say that $f(x)=|f(x)|$... So what do I have to do? I also know that if I think that it is integrable I have to try to bound it with a bigger function. But I have no idea...$\space$
I think that it is finite... but I'm not sure about it... $\space$
Could someone help me to do this exercise?
The function is not integrable. It is enough to show that the integral form $1$ to $\infty$is $\infty$. Note that the function is non-negative on this interval. Put $y =\ln x$. The interval $(1, \infty)$ changes to $(0,\infty)$ since $\ln 1=0$ and $\ln x \to \infty$ as $x \to \infty$. We get $\int_{0} ^{\infty} \frac {e^{-y}} {y^{3}}dy \geq \int_{0} ^{1} \frac {e^{-y}} {y^{3}}dy \geq e^{-1} \int_{0} ^{1} \frac 1 {y^{3}}dy=\infty$