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

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?

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.
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: $$\forall\varepsilon>0:\quad\lim_{x\to\infty}\frac{\log x}{x^\varepsilon}=0.$$
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.