Improper integral $\int_1^{+\infty}\frac{\log{x}}{x^{\alpha}}$

Question

Why can I say that

$$\int_1^{+\infty}\frac{\log{x}}{x^{\alpha}}\, dx$$

is convergent surely for $\alpha>1$ as I could forget of $\log{x}$ and I would have

$$\int_1^{+\infty}\frac{1}{x^{\alpha}}\, dx.$$

I have thought that it is due by the fact that whatever power of $x$ goes to infinity faster than $\log{x}$ but I don't know if this is the right remark and above all how formalize this fact for the convergence of the integral. Can you help me?

Answer 1

This is a long the lines of the reasoning in your posting.

If $a>1$, the $a=1+2\varepsilon$ for some $\varepsilon>0$. Since $$ \lim_{x\rightarrow\infty}\frac{\log x}{x^\varepsilon}=0$$ there is $A>0$ large enough so that $\frac{\log x}{x^\varepsilon}<1$. Since $\int^\infty_A\frac{1}{x^{1+\varepsilon}}\,dx$ converges, the conclusion follows.

On the other hand, this is the type of integral that can be evaluated directly by integration by parts.

Answer 2

Not only the integral converges but it can be exactly evaluated. Indeed ntegrating by parts one obtains for $a>1$: $$ \int_1^\infty \frac{\log x}{x^a}dx= \underbrace{\left[-\frac1{a-1}\frac{\log x}{x^{a-1}}\right]_1^\infty}_{=0}+\frac1{a-1}\int_1^\infty\frac{dx}{x^a}=\frac1{(a-1)^2}, $$ where we used the fact: $$ \forall\varepsilon>0:\quad\lim_{x\to\infty}\frac{\log x}{x^\varepsilon}=0. $$

Answer 3

Choose $b >0$ so large such that $\frac{\log(x)}{x^\alpha}$ is decreasing on $[b, \infty)$. This works, just take a look at the derivative. Therefore, by the infinite series test, the integral converges iff $$ \sum_{n = m}^\infty \frac{\log(n)}{n^\alpha} $$ converges where $m:= \lceil b \rceil$. It is obvious that $\displaystyle \int_1^b \frac{\log(x)}{x^\alpha}~\mathrm{d}x$ can be neglected as the integrand is bounded und consequently this is always finite.

Now, according to Cauchy's condensation test, $\displaystyle\sum_{n = m}^\infty \frac{\log(n)}{n^\alpha}$ converges, iff $$ \sum_{n = m}^\infty 2^n \frac{\log(2^n)}{(2^n)^\alpha} = \log(2)\sum_{n = m}^\infty n2^{n(1-\alpha)} $$ converges. If $0 \leq \alpha \leq 1$, this clearly diverges. If $\alpha > 1$, we can use the ratio test to easily find that the series is convergent.