I'd like someone to explain what the teacher did, because I'm not sure I understand.
Basically, the question is for which values of $p$, does the integral $$\int_{1}^{\infty} \frac{dt}{t \log ^p(1+t)}$$ converge.
Here is what the teacher did:
He said that this integral converges, if and only if the integral $$\int_{2}^{\infty}\frac{dt}{t \log ^p(t)}$$
converges.
and this integral converges when $p=2$, since we can use the transform $e^x=t$:
$$\int_{2}^{\infty}\frac{dt}{t \log ^p(t)}=\int_{\log 2}^{\infty}\frac{e^xdx}{e^x \log ^p(e^x)}=\int_{\log 2}^{\infty}\frac{dx}{x^p}$$ which converges for $p \geq 2$.
I agree and understand why the second integral converges when $p \geq 2$, but I don't understand why the first integral converges if and only if the second one converges. i dont see the relation between the two.
First, one makes the substitution $u = t+1$ (and then renames $u$ again to $t$), to write the integral as
$$\int_2^\infty \frac{dt}{(t-1)\log^p t}.$$
Then one sandwiches the integrand
$$\frac{1}{t\log^p t} < \frac{1}{(t-1)\log^p t} \leqslant \frac{2}{t\log^p t}$$
to obtain the equivalence that one integral converges if and only if the other does (which is the case if and only if $p > 1$).