I was trying to prove for what values of p eq.1 converges or diverges, they didn't give the proof for eq.1 but for eq.2 a proof was given and when I was done with the proof for eq.1 I noticed that for the second one they specified from which-side they approached zero (see eq.4) but I didn't specify from which side I approached infinity (see eq.3). I didn't do that because without graphing I usually can't tell which side to use and this is general and not a specific exercise or example, anyways, could proof eq.3 be considered incomplete if I don't specify from which side I approached infinity?
First of all, for eq.1, we see that, by the fundamental theorem of calculus, that $$\int_a^{\infty}x^{-p}dx=\lim_{c\to\infty}\frac{c^{-p+1}}{(-p+1)}-\frac{a^{-p+1}}{(-p+1)}$$Except the special case $p=1$, we have $\lim_{c\to \infty}\ln(c)-\ln(a)$, which diverges.
But for most values of $p$, this will diverge based on $$\lim_{c\to\infty}c^{-p+1}=\pm\infty?$$
It hangs on whether or not $p\le1$. In those cases, it will diverge.
For eq. 2, we use the same method:$$\int_0^ax^{-p}dx=\frac{0^{-p+1}}{(-p+1)}-\frac{a^{-p+1}}{(-p+1)}$$What determines if this diverges is based on $$0^{-p+1}$$If $p\le1$, then this is undefined and the integral will diverge.
For eq.3 and 4, I cannot help you.