I was reading this question Showing this to $1 \leq p < \infty$ , here on this site.
But I did not understand this part (in the third solution mentioned there given by @mechanodroid):
" Now notice that for large enough $t_0 \ge 1$ we have $t \ge t_0 \implies \ln t \le \sqrt[2p]{t}$ so the above integral can be bounded by $$\int_1^{t_0}\frac{(\ln t)^p}{t^2}\,dt + \int_{t_0}^{+\infty}\frac{dt}{t^{3/2}} < +\infty$$ which is finite since $\frac{(\ln t)^p}{t^2}$ is bounded on $[1,t_0]$ and $t \mapsto \frac1{t^{3/2}}$ is integrable on $[1, +\infty)$."
Could anyone clarify this part for me, please?
2- Also for the second solution given by @Awegan, I can see a typo in $p$ and $q$, the powers of $L$, am I correct?
3- Also, is there an alternative proof (detailed) for the question given in the link I mentioned above?
$$\int_1^{t_0}\frac{(\ln t)^p}{t^2}\,dt + \int_{t_0}^{+\infty}\frac{dt}{t^{3/2}} < +\infty$$
Since the integrand of the left hand integral is bounded, then $$\int_1^{t_0}\frac{(\ln t)^p}{t^2}dt\leq M(t_0-1),$$ where $M$ is the bound (see here). Clearly this is finite.
For the right hand integral, just evaluate it, e.g.
$$\int_{t_0}^\infty\frac{dt}{t^{3/2}}dt=\left[-\frac{2}{\sqrt{t}}\right]_{t_0}^{+\infty}=\frac{2}{\sqrt{t_0}},$$ which is also finite.
The sum of two finite numbers gives a finite number, so by definition of $L^p$ spaces $f\in L^p(0,1]$.
In answer to your comment, here's a plot of $\log t$ (in blue) and $t^{1/(2p)}$ for $p=1.4$, for example. As you can see, $t_0\approx 35$ in this case, after which the inequality holds true for $t\geq t_0$. This always happens for some $t_0$ value.