Is this improper integral $\int_e^\infty \frac{dt}{t^a \log^b (t)}$ convergent? $\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.
