Consider the following integral: $$\int_{0}^{\infty}\frac{dx}{x^a+x^b}$$
What are the necessary (and sufficing) conditions on $a$ and $b$ for the integral to converge?
It's easy to see that if $a=b$, then the integral is necessarily divergent. But as soon as I start checking for $a\neq b$, I get lost in all the different cases.
WLOG suppose that $b\geqslant a$, then $$ \int_{0}^{\infty }\frac{dx}{x^a+x^b}=\int_{0}^{\infty }\frac{x^{-a}}{1+x^{b-a}}\,d x $$
Now observe that if $a\geqslant 1$ the integral diverges, as $\frac{x^{-a}}{1+x^{b-a}}\geqslant \frac{x^{-a}}2$ for $x\in(0,1]$ and $\int_{0}^1 x^{-a}\,d x=+\infty $. By the other hand, if $a< 1$ then the integral in $[0,1]$ is well-defined, and converges on $[1,\infty )$ if and only if $b>1$. This happens because for $x\geqslant 1$ you can write
$$ \frac{x^{-a}}{1+x^{b-a}}=\frac{x^{b-a}}{1+x^{b-a}}\cdot x^{-b} $$
and $0\leqslant \frac{x^{b-a}}{1+x^{b-a}}\leqslant 1$ for all $x\geqslant 0$. Therefore we conclude that the integral converges if and only if $\max\{ a,b \}>1$ and $\min\{ a,b \}<1$.∎