Evaluating Improper Integrals with Infinite Limits

You can't plug infinity into an antiderivative, so you don't try. Instead you replace the infinite bound with a letter, say b, evaluate the definite integral from a to b using the usual antiderivative, and then ask what happens as b races off toward infinity. If that limit settles on a finite number, the integral converges and the infinite region really does have finite area. If the limit grows without bound, the integral diverges.

This is why a shrinking curve is not enough to guarantee finite area. What matters is the rate of shrinking. For 1/x², the antiderivative is -1/x, and as b goes to infinity that term dies out fast, leaving a finite value. For 1/x, the antiderivative is ln(x), and ln(b) grows forever even though 1/x itself is sinking to 0. Same destination for the function, totally different outcome for the area, because ln grows slower than any power but still grows without bound.

This split is exactly what controls convergence of the family 1/x^p: when p>1 the integral converges, when p≤1 it diverges, with p=1 sitting right on the borderline. That threshold is not a coincidence. It is the same test that decides whether the p-series sum of 1/n^p converges, since the integral and the series rise and fall together for a positive, decreasing function. Improper integrals here are the bridge between the integration techniques already built and the series convergence tests coming next.