Little theorems for improper integrals

Question

We know that integral $ \int_1^{\infty} f(x) d x $ converges. What can we say about the following integrals: $$ \int_1^{\infty} f^3(x) d x, \\ \int_1^{\infty} \frac {|f(x)|}{x^2} d x. $$ What can I do to proof it? Don't even have any ideas.

Answer 1

Nothing can be said. Concerning the first integral, take $f(x) = 1/x^2$ for all $ x >1$ and $g(x) = 1/(x-1)^{1/3}$ for $x \in (1,2)$ and $g(x)=0$ for $x \ge 2$. Then both

$$ \int_1^\infty f(x)dx $$ and $$ \int_1^\infty g(x)dx$$ are convergent, but $$\int_1^\infty f^3(x)dx $$ is convergent, while $$\int_1^\infty g^3(x)dx $$ is divergent. Concerning the second integral, tale $f$ as above and now take $g(x) = x^2 \sin(x-1)/(x-1)$ for all $x >1$. Again $\int_1^\infty f(x)dx$ and $\int_1^\infty g(x)dx$ are convergent, but $$ \int_1^\infty \frac{|f(x)|}{x^2}dx$$ is convergent, while $$ \int_1^\infty \frac{|g(x)|}{x^2}dx$$ is divergent.