If I have an improper integral $\displaystyle \int_{a}^{b}f(x)dx$,
and $b$ is the improper extrem, if $f(x)\sim g(x)$ for $x->b^-$, the integrals $\displaystyle \int_{a}^{b}f(x)dx$ and $\displaystyle \int_{a}^{b}g(x)dx$have the same behaviour.
But is it true also if the improper extreme is $a$ and I have that $f(x)\sim g(x)$ for $x->a^+$?
My textbook doesn't say anything about it...
Hint. You may observe that $$ \int_{b}^{a}f(x)dx=-\int_{a}^{b}f(x)dx $$ and, as $x \to b^-$, $\displaystyle f(x)\sim g(x) $ is equivalent to $\displaystyle -f(x)\sim -g(x) $. Then only one extremity may be considered.