Example of a function $f(x)$ such that the integral of $f(x^2)$ converges but the integral of $f(x)$ diverges?

Question

Does anybody know an example of a function $f(x)$ such that the integral from $1$ to infinity of $f(x^2)$ converges but the integral of $f(x)$ from $1$ to infinity diverges? Thanks!

Answer 1

Take $f(x)=\frac1x$ then

$$\int_1^\infty f(x)dx$$ is divergent and $$\int_1^\infty f(x^2)dx$$ converges.

Answer 2

Easiest example I can think of is, $f(x) = x^{-1}$, $\int_1^\infty x^{-1}dx$ diverges but $\int_1^\infty x^{-2}dx =1$

Answer 3

The integral $$\int_1^{\infty} x^p \,dx$$ converges (to $\frac{1}{p + 1}$) iff $p < -1$, so the integral $f(x) := x^p$ diverges and the integral of $f(x^2) := x^{2p}$ converges iff $-1 \leq p < -\frac{1}{2}$.