Convergence of $\int_2^{\infty}f(x)\,dx$ with a given condition

Question

Let , $f$ be continuous function on $[2,\infty)$ and $\displaystyle\lim_{x\to \infty}x(\log x)^pf(x)=A$ , where $A$ is a non-zero finite number.. Then $\displaystyle\int_2^{\infty}f(x)\,dx$ is

(A) convergent for all $p$.

(B) convergent for $p>1$.

(C) convergent for $0<p<1$.

(D) always convergent.

Attempt :

Let , $g(x)=f(x)x(\log x)^p$ and $h(x)=\frac{1}{x(\log x)^p}$.

I want to apply Able's test for improper integral. So we have to show $g$ is monotone and bounded and $\displaystyle \int_2^{\infty} h(x)\,dx $ is convergent. Now , $\displaystyle \int_2^{\infty} h(x)\,dx $ is convergent for $p>1$. From the given limit it is clear that $g$ is bounded. But I am unable to show $g$ is monotone.

Answer 1

Choose $M$ such that for $x>M$, $\frac{f(x)}{h(x)}$ is between $A/2$ and $2A$. Then write

$$\int_2^\infty f(x) dx = \int_2^M f(x) dx + \int_M^\infty f(x) dx.$$

The first term is finite by the continuity hypothesis. The second term is between $\frac{A}{2} \int_M^\infty h(x) dx$ and $2A \int_M^\infty h(x) dx$ by pointwise comparison. Thus $f$ has the same integrability properties as $h$.

The "test" whose correctness is proven by this argument is called the limit comparison test.

Answer 2

Since $\frac{1}{x\left(\log x\right)^p}$ is not an integrable function over $[2,+\infty)$ for any $p\in (0,1]$, the only reasonable chance is $B)$. So, let we prove $B)$. For any $\varepsilon>0$, there is some $M\in\mathbb{R}^+$ such that $$ \left|x(\log x)^p f(x)-A\right|\leq \varepsilon $$ for any $x\geq M$. Then: $$ \int_{2}^{+\infty}f(x)\,dx = \int_{2}^{M}f(x)\,dx + \int_{M}^{+\infty}\frac{1}{x(\log x)^p}\cdot x(\log x)^p f(x)\,dx $$ is a finite number since $f(x)$ is a continuous function over $[2,M]$, hence an integrable function, and: $$\left|\int_{M}^{+\infty}\frac{1}{x(\log x)^p}\cdot x(\log x)^p f(x)\,dx\right|\leq \left(|A|+\varepsilon\right)\int_{M}^{+\infty}\frac{dx}{x(\log x)^p}=\frac{|A|+\varepsilon}{(p-1)\log(M)^{p-1}}.$$