Improper Integrals and Their Convergence
Give meaning to an integral with an infinite limit or an unbounded integrand by evaluating a proper integral and taking a limit, then decide whether the result is finite.
What a learner can do afterwards
- Rewrite an infinite-range integral as a limit and evaluate it
- Decide convergence for 1/x^p on [1, ∞) and say where the cut-off sits
- Handle an integrand that blows up inside the interval by splitting it
1 · Read
An improper integral stretches past comfort: either a bound is infinite or the integrand blows up inside. The fix is always a limit. Integrate out to a finite t, or stop short of the bad point, then let t creep toward it. A finite limit means converge; otherwise it diverges.
Compare 1/x and 1/x^2 from 1 onward. The integral of 1/x is ln t, which grows without bound, so it diverges. The integral of 1/x^2 is 1 - 1/t, which settles at 1, so it converges to 1. Infinity is a direction you approach, never a number you plug in.
The model case is 1/x^p from 1 onward: it converges exactly when p beats 1. A blow-up inside works the same way by splitting: for 1/sqrt(x) from 0 to 1, stop short at a small a, integrate, and let a approach 0. The result converges to 2.
Always rewrite first, evaluate second, and take the limit last. An integral across two infinities must be split, for example at 0, and both halves must converge for the whole to converge.
Rewrite with a limit, evaluate, and let the limit decide converge or diverge.
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8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.