Calculus I · Unit 3A · lesson
Improper Integrals with Unbounded Integrands
Learn Improper Integrals with Unbounded Integrands through clear explanation, worked examples, visual reasoning, checks, and connected integral-calculus practice.
Section overview
Numerical and improper integrationWhat this section is building
Learn Improper Integrals with Unbounded Integrands through clear explanation, worked examples, visual reasoning, checks, and connected integral-calculus practice.
Numerical rules replace a curve with simple local shapes; improper integrals replace a forbidden endpoint or infinite interval with a limit.
Choose the rule and partition, estimate scale and sign, or write the correct defining limit before evaluating.
Treating an approximation as exact, using Simpson's Rule with an invalid partition, or substituting infinity as though it were a number.
Learning objectives
Split at singularities, use one-sided limits, and require every piece to converge.
Improper Integrals with Unbounded Integrands
An infinite spike may or may not have finite accumulation
When an integrand becomes unbounded at an endpoint or inside the interval, the ordinary definite-integral notation hides a limit. Replace the problematic boundary by a nearby finite value and approach it from the correct side. If the singularity lies inside the interval, split the integral there and require both one-sided improper integrals to converge.
A graph can look alarming without determining the answer. A very tall but sufficiently narrow spike may have finite area, while a milder-looking singularity may diverge. The decision comes from the limiting calculation. Never cancel divergences from opposite sides unless a different concept, such as a principal value, has been explicitly introduced; ordinary improper-integral convergence requires each piece to exist independently.
If is unbounded at , define
provided both one-sided integrals converge.
Integrable vertical blow-up
The function becomes unbounded at , but the area remains finite.
Never evaluate across a singularity with a single antiderivative subtraction. Split the interval first. Cancellation between two divergent sides does not make the ordinary improper integral converge.
u3a-improper-sing-01Does converge?
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Show hint
The exponent corresponds to a p-integral near zero with .
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