If $f(x)$ is continuous for $f(x)\geq a$ and $\lim_{x\to \infty} x^\mu f(x)=A$, then prove the following
if $\mu>1$, $\int_a^\infty f dx $ converges absolutely
if $\mu=1$ and $A\neq 0$,$\int_a^\infty f dx $ diverges
Considering $\int_1^\infty \frac{dx}{x^2}$ and $\int_1^\infty \frac{dx}{x \ln{x}}$ state what will be the concusion in second case if $A=0$.
As $$\lim_{x\to \infty} x^\mu f(x)=A$$
$$\implies \lim_{x\to \infty} x^\mu |f(x)|=|A|$$ $$\implies |A|-\epsilon <x^\mu |f(x)|<|A|+\epsilon, \text{for} M>0, \forall x\geq A$$ $$\implies |f(x)|<(|A|+\epsilon)x^{-\mu} , \text{for} M>0, \forall x\geq A$$
As $$\int_M^\infty \frac{dx}{x^\mu}$$ converges if $\mu>1$, then by comparison test,
$$\int_M^\infty |f(x)|dx$$ converges.
Please help me.
For the second question. Assume that $A<0$, then for large $M>0$, $f(x)<0$ for all $x\geq M$. Assume further that $M>0$ is such that $xf(x)-A<-A/2$, then $f(x)<(A/2)x^{-1}$ and we have $(A/2)\displaystyle\int_{M}^{\infty}x^{-1}dx=-\infty$, so $\displaystyle\int_{M}^{\infty}f(x)dx=-\infty$.
For the third part, $\lim_{x\rightarrow\infty}x\cdot\dfrac{1}{x^{2}}=\lim_{x\rightarrow\infty}\dfrac{1}{x}=0$ but $\displaystyle\int_{1}^{\infty}\dfrac{1}{x^{2}}dx=1<\infty$, also that $\lim_{x\rightarrow\infty}x\cdot\dfrac{1}{x\log x}=\lim_{x\rightarrow\infty}\dfrac{1}{\log x}=0$ but $\displaystyle\int_{2}^{\infty}\dfrac{1}{x\log x}dx=\log(\log x)\bigg|_{x=2}^{\infty}=\infty$.