Define each infinite or singular contribution by a limit

Try

The region under 1/x² from x=1 onward extends forever. Does an infinitely long region necessarily have infinite area? Compare its progressively smaller tails. Then compare 1/x, whose values also approach zero. A vanishing integrand alone does not determine whether its integral converges.

Predict

Predict the difference between approaching an infinite endpoint and approaching a singularity inside a finite interval. Both require limits, but an internal singularity splits the problem into two independent one-sided integrals. Opposite signed infinities cannot be canceled to manufacture ordinary convergence.

Experiment

First integrate over finite intervals, where the ordinary fundamental theorem applies.

Upper endpoint R	∫₁ᴿx⁻²dx	∫₁ᴿx⁻¹dx
2	1/2	ln 2
10	9/10	ln 10
100	99/100	ln 100

The first formula is 1−1/R and approaches 1. The second is ln R and grows without bound. Sketch both positive decreasing curves and mark that their similar pointwise endpoint limits conceal different accumulated totals.

Understand

An improper integral ∫ₐ^∞f is defined as the limit of ∫ₐᴿf as R approaches infinity, if the limit is finite. An integrable singular endpoint is treated by moving a finite cutoff toward it from within the interval. For ∫₀¹x^(−1/2)dx, compute from ε to 1 to get 2−2√ε, then let ε approach zero from above. The result is 2 despite the unbounded integrand.

For the tail family ∫₁^∞x^(−p)dx, if p≠1 the truncated value is (R^(1−p)−1)/(1−p). It converges exactly when p>1, giving 1/(p−1). At p=1, the logarithm diverges. Near zero the condition reverses: ∫₀¹x^(−p)dx converges for p<1 and diverges for p≥1. The location of the problematic endpoint determines which power dominates the limit.

An internal singularity requires a split. For ∫₋₁¹1/x dx, the negative and positive sides diverge separately. Symmetric truncation gives zero, called a Cauchy principal value, but this is not the ordinary improper integral. Different relationships between the two cutoff rates can produce different apparent cancellations, which explains why ordinary convergence requires each side to converge on its own.

For ∫₀²1/(x−1)²dx, the primitive −1/(x−1) might tempt an endpoint difference of −2. That negative result contradicts the nonnegative integrand and signals invalid use of FTC across the pole at 1. Splitting into limits shows both sides grow to positive infinity. A finite symbolic endpoint expression is not evidence that the improper integral exists.

Comparison can establish convergence without a closed-form value. On x≥1, 0≤1/(x²+1)≤1/x², so its improper integral converges and is at most 1. Comparison requires pointwise bounds and nonnegative functions for this form of the test. The bound does not provide the exact value, and it cannot be inferred from a few numerical samples alone.

Apply

Locate every singularity and infinite endpoint before integrating. Introduce a separate cutoff for each problematic side, evaluate finite integrals, then take limits. In practice explanations, distinguish convergence, divergence and a bound on a convergent value. Do not combine undefined quantities under the label “infinity minus infinity.”

Continue

The final lesson estimates integrals numerically. A numerical routine returning a finite number does not override divergence or guarantee accuracy when regularity assumptions fail.