$\int_0^{1/e} (1/(x (\ln x)^2))^t dx$ diverges for $t>1$

Question

Consider the function $g:(0,\infty)\to \Bbb R$ defined by $g(x)=1/(x (\ln x)^2)$. I am trying to show that $\int_0^{1/e} g(x)^tdx$ converges for $0<t\leq 1$ and diverges for $t>1$. For $t=1$, I showed that the integral is finite by direct computation with making a change of variables $u=\ln x$. For $t<1$, I showed that the integral is finite by comparison, using that $(\ln x)^2 \geq 1$ for $0<x \leq 1/e$. But I have no idea for $t>1$.

Answer 1

Assume $t>1$.

if we put $u=\frac 1x$ the integral to study becomes

$$\int_e^{+\infty}\frac{u^{t-2}du}{(\ln(u))^{2t}}$$

observe that $2-t<1$.

take $\gamma$ such that

$$2-t<\gamma<1$$ then

$$\lim_{u\to+\infty}u^\gamma \frac{u^{t-2}}{(ln(u))^{2t}}=+\infty$$ thus for great enough $u$, we have

$$u^\gamma \frac{u^{t-2}}{(\ln(u))^{2t}}>100$$

and

$$\frac{u^{t-2}}{(\ln(u)^{2t}}>\frac{100}{u^\gamma}$$

now use comparison test with $ \gamma<1$.