(H.A.Priestley Introduction to integration 18.4.3)
Let $f(x) = \frac{x sin(x)}{1+x^2}$ on $\mathbb R$. We claim that f is not integrable. Note that $f$ behaves like $g(x)=\frac{xsin(x)}{x^2}$ for large $|x|$ and that $g$ is not integrable. Since $1+x^2 \le 2x^2$ for large $|x|$ we have $$ \frac{|x sin(x)|}{1+x^2} \ge |\frac{sin(x)}{2x}|$$ Thus arguing by contradiction we see that $f$ cannot be integrable.
Now I don't understand the exact form of the contradiction here. The comparaison theorem is:
(H.A.Priestley Introduction to integration 18.4.3)
Let $f$ be continuous except at the point of a finite set $S$, possibly empty. Assume that for each $a \in S \bigcup \{-\infty , +\infty\} $ there exists a puncture neighbourhood $U_a$ of $a$ and a function $g_a$ integrable on $U_a$ such that $f(x) = O(g_a(x))$ as $x \to a$. Then f is integrable.
So in the example above, what is the contrapose? it seems to say :
$f$=O(g) and $g$ not integrable, therefore $f$ not integrable ?
but surely that is not the contrapose of the theorem I have quoted ?