$\int_e^\infty \frac{dt}{t^a \log^b (t)} $
What I've done is that for $t > e$,
$$\int_e^\infty \frac{dt}{t^a \log^b (t)} \le \int_e^\infty \frac{dt}{t^a } $$, which converges for $a > 1$. First, is this correct? But what about the cases where $a = 1$ ans $a < 1$? Can't seem to find a way to prove whether or not it's convergent. Thank you.
Yes it is correct.
The case $a=1$ is $$ \int_e^\infty \frac{\frac{dt}t}{\log^b (t)}=\int_1^\infty \frac{du}{u^b } $$ which converges if and only if $b>1$.
The case $a<1$ gives $$ \int_e^\infty \frac{dt}{t^a\log^b (t)}=\int_1^\infty \frac{e^{(1-a)u}}{u^b }du=\infty $$ since $$ \lim_{u\to \infty}\frac{e^{(1-a)u}}{u^b }=\infty $$ whatever the value of $b$ is.