\(\displaystyle Improper \ Integrals \ are \ loaded \ with \ infinite \ discontinuities.\)
\(\displaystyle Example: \ Use \ the \ formula \ for \ arc \ length \ to \ show \ the \ circumference \ of \ the \ circle\)
\(\displaystyle x^2+y^2 \ = \ 1 \ is \ 2\pi.\)
\(\displaystyle To \ simplify \ the \ work, \ we \ consider \ the \ quarter \ circle, \ y \ = \ \sqrt{1-x^2}, \ 0 \ \le \ x \ \le \ 1.\)
\(\displaystyle s \ = \ \int_{0}^{1}\sqrt{1+(y')^2}dx \ = \ \int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}\)
\(\displaystyle This \ integral \ is \ improper, \ since \ it \ has \ an \ infinite \ discontinuity \ at \ x \ = \ 1.\)
\(\displaystyle Thus, \ s \ = \ \int_{0}^{1}\frac{dx}{\sqrt{1-x^2}} \ = \ \lim_{b\to1^-}\bigg[arcsin(x)\bigg]_{0}^{b} \ = \ \pi/2-0 \ = \ \pi/2.\)
\(\displaystyle Hence, \ 4s \ = \ 2\pi.\)