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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 $0\lt\alpha\leq1$

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

Aphelli
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  • Thank you.But,then if we want to show if integral diverges or converges for $\alpha\lt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide? – Turan Nasibli Jan 08 '19 at 08:36
  • Hint: when $x$ is large enough, and $0 < \alpha < 1$, $x^{-\alpha}(\ln{x})^{\beta} \geq x^{-(1+\alpha)/2}$. For $\alpha=1$, you can compute exactly an antiderivative. – Aphelli Jan 08 '19 at 09:38
  • So, if I got it right, we must have $$\frac{1}{x^\frac{\alpha +1}{2}}\leq\frac{1}{x^\alpha (\ln x)^\beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance – Turan Nasibli Jan 08 '19 at 10:47
  • First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $\int_2^{\infty}{f}=\infty$ and $f(x) \leq x^{-\alpha}(\ln{x})^{-\beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-\gamma}$. The conditions then translate to $\gamma \leq 1$, $\gamma \geq \alpha$ (and $\gamma > \alpha$ if $\beta > 0$). So with $\gamma=\frac{1+\alpha}{2}$ we are good in all cases. – Aphelli Jan 08 '19 at 11:00
  • I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $\alpha\gt1$ we picked up a function that held true for $\forall x\geq2$. – Turan Nasibli Jan 08 '19 at 11:06