Improper Integral: Infinite Interval, Finite Area
Infinity is not an endpoint you plug into an integral. First stop at a finite boundary B. The area from 1 to B is 1−1/B. Then let B move farther and farther right. The interval keeps expanding, but the missing tail 1/B keeps shrinking.
Move the idea
Try it: push the finite boundary B outward
Change one parameter and watch what changes with it.
Infinite width does not force infinite area. What matters is how quickly the function decays: here the remaining area after B is exactly 1/B, and that remainder tends to zero.
Start finite
For B>1, integrate only from 1 to B. This is an ordinary definite integral: ∫₁ᴮ x⁻²dx=[−1/x]₁ᴮ=1−1/B.
Then take the limit
The improper integral is defined by lim(B→∞)(1−1/B). Since 1/B→0, the partial areas approach 1.
The decay rate is the reason
An infinite interval alone says nothing about convergence. For powers 1/xᵖ on [1,∞), the p-test converges when p>1 and diverges when p≤1.
Make the idea reusable.
Area so far versus area remaining
At B=2 the accumulated area is 1/2; at B=5 it is 4/5; at B=20 it is 19/20. The remaining tail is respectively 1/2, 1/5 and 1/20. The picture becomes intuitive when you track the remainder, not just the growing interval.
Why ∫₁^∞1/x dx behaves differently
For 1/x the finite integral is ln B, and ln B grows without bound. The curves both approach zero, but 1/x² decays fast enough for its total tail area to converge while 1/x does not.
Improper does not mean invalid
An improper integral is handled through a limit because the interval or integrand is unbounded. It may converge to a finite value or diverge; the limit decides which.