4.08c - Improper integrals

4.08c - Improper integrals

Use limits to evaluate integrals with an infinite range or an undefined integrand. You will decide whether a finite value exists, handle singularities at either endpoint, and split an integral when the difficulty lies inside its interval.

A limit decides whether an integral exists

Can a region stretch infinitely far to the right but still have a finite area?

For y=1/x2y = 1/x^2, the area from x=1x = 1 to x=10x = 10 is 9/109/10; to x=100x = 100 it is 99/10099/100. Extending the region adds area, but the total approaches 11. Infinite width alone does not decide the answer.

You already know

For an ordinary definite integral, find an antiderivative FF and calculate F(b)−F(a)F(b)-F(a). A limit describes what an expression approaches as its variable approaches an endpoint or grows without bound.

The usual endpoint substitution needs care when an endpoint is infinite, or when the integrand is undefined at a point of the integration interval. Instead, first integrate over a safe finite interval, then examine its limit.

Improper integral

An improper integral has an infinite range of integration, or an integrand that is undefined at a point in the integration interval, including an endpoint. It converges if the required one-sided limits all exist as finite real numbers. Otherwise it diverges and has no finite value.

For example, ∫1∞x−2 dx\int_1^\infty x^{-2}\,dx has an infinite range, while ∫01x−1/2 dx\int_0^1 x^{-1/2}\,dx has a finite range but an undefined integrand at 00. Both are improper and both converge. The ordinary integral ∫12x−2 dx\int_1^2 x^{-2}\,dx is neither of these: its integrand is continuous throughout a finite interval.

[DIAGRAM: asset_name: Two infinite tails behave differently; asset_slug: h245_improper_tails; caption: Both curves tend to zero, but only the area under 1 over x squared from 1 to infinity is finite; description: Side-by-side plots from x equals one to eight, showing y equals one over x squared in blue and y equals one over x in lavender. The shaded finite portions have labels giving their integrals to R, one minus one over R and log R. Arrows show that both regions continue beyond the frame.]
Both curves tend to zero, but only the area under 1 over x squared from 1 to infinity is finite

Both curves in the picture tend to 00, but ∫1∞1/x dx\int_1^\infty 1/x\,dx diverges. A decreasing height is not enough: the accumulated area must approach a finite number.

Key idea

An improper integral is a limit of ordinary integrals; a finite antiderivative expression is not enough until the limit has been checked.

Use that definition to distinguish the source of the difficulty.

Infinite ranges of integration

Replace an infinite endpoint by a finite variable, integrate, and then let that variable tend in the required direction.

An infinite upper limit

If ff is integrable on every finite interval [a,R][a,R], define

∫a∞f(x) dx=lim⁡R→∞∫aRf(x) dx\int_a^\infty f(x)\,dx = \lim_{R\to\infty}\int_a^R f(x)\,dx

only when that limit is finite. For an infinite lower endpoint, use lim⁡L→−∞∫Lbf(x) dx\lim_{L\to-\infty}\int_L^b f(x)\,dx.

Not in the formula booklet: learn it.

The finite variable makes every step of the fundamental theorem of calculus legitimate. The limit is a separate, final operation.

Worked example 1: a finite area

Question. Determine the exact value of ∫1∞x−2 dx\int_1^\infty x^{-2}\,dx.

Plan. Integrate to a finite upper limit RR, then let RR increase without bound.

∫1Rx−2 dx=[−x−1]1Rpower rule=1−1Rupper minus lowerlim⁡R→∞(1−1R)=11R→0\begin{aligned} \int_1^R x^{-2}\,dx &= [-x^{-1}]_1^R && \text{power rule} \\ &= 1-\frac1R && \text{upper minus lower} \\ \lim_{R\to\infty}\left(1-\frac1R\right) &= 1 && \frac1R\to0 \end{aligned}

Why this step? Infinity is not a number to substitute into −1/x-1/x; the limit tells you what the finite-area expression approaches.

Answer. The integral converges and has exact value 11.

Check. Up to x=10x = 10, the area is 0.90.9, and the remaining area is 0.10.1; together they give 11.

Change only the exponent, from −2-2 to −1-1. The curve still tends to zero, so the limit must decide whether its area behaves differently.

Worked example 2: a divergent area

Question. Determine whether ∫1∞x−1 dx\int_1^\infty x^{-1}\,dx converges.

Plan. Find the area to RR and examine its limit.

∫1R1x dx=[ln⁡x]1Rx>0=ln⁡Rln⁡1=0ln⁡R→∞R→∞\begin{aligned} \int_1^R \frac1x\,dx &= [\ln x]_1^R && x > 0 \\ &= \ln R && \ln1 = 0 \\ \ln R &\to\infty && R\to\infty \end{aligned}

Why this step? The antiderivative grows without bound, even though the integrand tends to zero.

Answer. The integral diverges; it has no finite value.

Check. Each interval from RR to 2R2R contributes ln⁡(2R)−ln⁡R=ln⁡2\ln(2R)-\ln R = \ln2. Extending through more such intervals keeps adding the same positive amount.

For a range infinite in both directions, choose any ordinary point cc and split there: ∫−∞∞f(x) dx\int_{-\infty}^{\infty} f(x)\,dx exists only if ∫−∞cf(x) dx\int_{-\infty}^{c} f(x)\,dx and ∫c∞f(x) dx\int_c^{\infty} f(x)\,dx both converge separately. For example, ∫−∞0ex dx=lim⁡L→−∞(1−eL)=1\int_{-\infty}^0 e^x\,dx = \lim_{L\to-\infty}(1-e^L) = 1. Do not tie the two distant endpoints together to manufacture cancellation.

Watch out

Wrong: 1/∞=01/\infty = 0, treating infinity as a substituted value. Right: lim⁡R→∞1/R=0\lim_{R\to\infty}1/R = 0. In detailed reasoning, show the finite endpoint and the limit explicitly.

The same three-stage calculation applies to other functions: finite integral, limit, conclusion about convergence.

Undefined integrands at endpoints

When an integrand is undefined at an endpoint, approach that endpoint from inside the interval.

At a lower endpoint aa, replace it by a+εa+\varepsilon and take ε→0+\varepsilon\to0^+. At an upper endpoint bb, replace it by b−εb-\varepsilon and take the same positive limit. The direction matters because every temporary integration interval must lie in the integrand's real domain.

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