Visual walkthrough — Improper integrals — Type I (infinite limits), Type II (discontinuous integrand)

I want to paint under a curve, but the region is broken — either it runs to the right forever, or it shoots straight up at one spot. Either way I can't measure it directly. So I cheat: I build a movable wall at a real position $t$. Between $1$ and my wall the region is normal, so I paint it and get an honest number that depends on $t$. Then I slide the wall away (to infinity for Type I) or creep it toward the spike (to $0$ for Type II) and I just watch the number. If it calms down to a value — finite paint, converges. If it keeps climbing — leaks forever, diverges. When I do the algebra, all the drama collapses into one term $t^{1-p}$, and its fate depends only on the sign of the exponent $1-p$. Positive power of a huge $t$ blows up; negative power fades to nothing — and vice-versa for a tiny $t$. The break-even is exponent zero, i.e. $p=1$, which is exactly where the power rule died and I had to use $\ln$. That $\ln t$ crawls upward forever, so $p=1$ just barely loses — on both the far wall and the spike. That's why the two rules are perfect mirror images with the same lonely loser in the middle.

For ${\int }_1^{\infty }{x}^{-p}dx$, which term decides convergence? ::: The limit of $t^{1-p}$ as $t\to \infty$ — its fate depends on the sign of $1-p$. Why does $p=1$ need separate treatment? ::: The antiderivative $\frac{x^{1-p}}{1-p}$ divides by $1-p=0$; the true antiderivative of $x^{-1}$ is $\ln x$. Value of ${\int }_1^{\infty }{x}^{-2}dx$? ::: $\frac{1}{p-1}=\frac{1}{2-1}=1$. Value of ${\int }_0^1{x}^{-1/2}dx$? ::: $\frac{1}{1-p}=\frac{1}{1-\frac{1}{2}}=2$.