Convergence of an improper integral,depending on parameters

Question

I have to show how the convergence of $$\int_2^\infty \frac {1}{x^\alpha (\ln x)^\beta} \mathrm{ d}x $$ depends on parameters $$\alpha,\beta\gt0$$ And considering the case $\alpha\gt1$,my textbook says $x\geq2$ implies $\ln x\geq \ln 2$ and hence $$\frac {1}{x^\alpha (\ln x)^\beta}\leq\frac {1}{x^\alpha (\ln2)^\beta}$$ $$\forall x\geq2$$ And it is easy to see that $\frac {1}{x^\alpha (\ln2)^\beta}$ converges,so does $\frac {1}{x^\alpha (\ln x)^\beta}$,by Comparison Theorem. But $$\frac {1}{x^\alpha (\ln x)^\beta}\leq\frac {1}{x^\alpha (\ln2)^\beta}$$ Implies $$x^\alpha(\ln x)^\beta\geq x^\alpha(\ln2)^\beta$$ $\forall x\geq2$ and $\alpha\gt1$. But I struggle to see why $$\frac {1}{x^\alpha (\ln x)^\beta}\leq\frac {1}{x^\alpha (\ln2)^\beta}$$ and therefore $$x^\alpha(\ln x)^\beta\geq x^\alpha(\ln2)^\beta$$ doesn't hold for $0\lt\alpha\leq1$

Answer 1

Your inequality does hold, but you cannot use the comparison theorem anymore, since $\int_2^{\infty}{\frac{1}{x^{\alpha}\ln(2)^{\beta}}} = \infty$.