Improper Integrals

Some integrals run to infinity, and some integrands blow up mid-interval. Improper integrals handle both the same way: compute a proper integral up to a movable stopping point, then take a limit. If the limit is finite, the integral converges; otherwise it diverges.

By the end you'll be able to rewrite an integral over an infinite interval (or with an unbounded integrand) as a limit, evaluate it, decide convergence — and recall on sight that \(\int_1^\infty x^{-p}dx\) converges exactly when \(p > 1\).

Predict: for p = 2 the shaded area approaches a finite value as b grows. For p = 1, does it? Slide b out to check — then slide p across 1 and watch the verdict flip.

The top chart shades \(\int_1^b x^{-p}dx\) under the curve \(y = x^{-p}\) (p-integral). The bottom chart plots that running total against b: it either levels off at the dashed limit line (converges) or climbs right off the top of the chart (diverges). Hover the bottom curve or the dot for exact values.

Area from 1 to b = 0.900 · as b → ∞ it converges to 1/(p−1) = 1.000converges

Top: y = x−p with the area from 1 to b shaded · bottom: running total of that area as b slides toward ∞
shaded area / running total limit 1/(p−1) when p > 1 current b

Improper integrals extend integration to infinite intervals or unbounded integrands by taking limits of proper integrals; the integral converges if the limit is finite and diverges otherwise.

Visual

Picture the demo above: the shaded region never stops extending to the right, yet its total area can still be finite — the tail gets thin fast enough that each extra mile of region adds almost nothing. That's convergence: the running total levels off at the limit line. When the tail is too fat (p ≤ 1), every stretch of road keeps adding real area and the total climbs forever. The surprise worth internalizing: an integrand can go to 0 and the integral can still diverge — shrinking isn't enough, it has to shrink fast.

Formal

Type 1 (infinite interval): \(\displaystyle\int_a^{\infty} f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx\), and similarly toward \(-\infty\); if both ends are infinite, split at any point and require both pieces to converge. Type 2 (unbounded integrand): if \(f\) blows up at an endpoint, replace that endpoint with a variable and take a one-sided limit, e.g. \(\displaystyle\int_a^b f(x)\,dx = \lim_{t\to b^-}\int_a^t f(x)\,dx\); an interior blow-up forces a split into two such pieces, both of which must converge. In every case: finite limit ⇒ converges; otherwise ⇒ diverges. The benchmark family is \(\displaystyle\int_1^\infty x^{-p}dx\), which converges to \(\frac{1}{p-1}\) for \(p>1\) and diverges for \(p \le 1\).

Worked example — Type 1, convergent

Evaluate \(\displaystyle\int_1^\infty \frac{1}{x^2}\,dx\). Replace \(\infty\) with \(t\) and take a limit: \(\displaystyle\int_1^\infty x^{-2}dx = \lim_{t\to\infty}\int_1^t x^{-2}dx = \lim_{t\to\infty}\Big[-\frac1x\Big]_1^t = \lim_{t\to\infty}\left(-\frac1t + 1\right) = 0 + 1 = \) 1. The limit is finite, so the integral converges to 1 — exactly where the running-total curve levels off in the demo at p = 2. Contrast \(p = 1\): \(\lim_{t\to\infty}[\ln x]_1^t = \lim_{t\to\infty}\ln t = \infty\), so \(\int_1^\infty \frac{dx}{x}\) diverges.

Your turn

Same recipe, a different power: \(\displaystyle\int_1^\infty \frac{1}{x^4}\,dx\). Set up the limit: \(\displaystyle\lim_{t\to\infty}\int_1^t x^{-4}dx = \lim_{t\to\infty}\Big[-\frac{1}{3x^3}\Big]_1^t = \lim_{t\to\infty}\left( -\frac{1}{3t^3} + \;\_\_\_\_ \right) = \) ____ — does it converge, and to what?

Reveal the answer

The blank is \(\frac13\): evaluating \(-\frac{1}{3x^3}\) at the lower limit 1 gives \(+\frac13\). As \(t\to\infty\), \(-\frac{1}{3t^3}\to 0\), so the integral converges to \(1/3\) — which matches the shortcut \(\frac{1}{p-1} = \frac{1}{4-1}\). Set p = 2.5 (the slider's max) in the demo and watch the running total level off near \(1/(2.5-1) \approx 0.667\) the same way.

More info — why p = 1 is the exact boundary

For \(p \ne 1\) the antiderivative of \(x^{-p}\) is \(\dfrac{x^{1-p}}{1-p}\), so \(\displaystyle\int_1^t x^{-p}dx = \frac{t^{1-p}-1}{1-p}\). Everything hinges on \(t^{1-p}\): if \(p>1\) the exponent \(1-p\) is negative, \(t^{1-p}\to 0\), and the area settles at \(\frac{1}{p-1}\); if \(p<1\) the exponent is positive and \(t^{1-p}\to\infty\). Exactly at \(p=1\) the power-rule antiderivative breaks down and is replaced by \(\ln t\) — which also grows without bound, just very slowly. That slow growth is why, in the demo, p = 1 looks almost flat over a short window yet still climbs off the chart: divergence is about the limit, not the speed. The same \(p\)-benchmark reappears when you study infinite series. For more worked boundary cases, see the Paul's Online Math Notes link under Dive deeper.

More info — when there's no antiderivative at all

Convergent improper integrals like \(\int_1^\infty e^{-x^2}dx\) often have no elementary antiderivative, so the limit-of-antiderivatives recipe stalls. The fix combines this lesson with the previous one: use convergence to truncate — choose a finite \(T\) where the leftover tail \(\int_T^\infty\) is provably tiny (compare it against a benchmark you can integrate) — then approximate the proper integral \(\int_1^T\) with the trapezoidal or Simpson's rule from the numerical integration lesson. Convergence guarantees the truncation error can be made as small as you like.

Check your understanding

Question 1 of 4

Determine whether \(\int_1^\infty \dfrac{1}{x^3}\,dx\) converges, and if so find its value.

Question 2 of 4

A student evaluates \(\int_{-1}^{1} \dfrac{1}{x^2}\,dx\) as \(\left[-\tfrac1x\right]_{-1}^{1} = -1 - 1 = -2\). What actually happens?

Question 3 of 4

For which values of \(p\) does \(\int_1^\infty \dfrac{1}{x^p}\,dx\) converge?

Question 4 of 4

\(\int_1^\infty e^{-x^2}dx\) converges but \(e^{-x^2}\) has no elementary antiderivative. What's a sound way to get a numerical value?

Recap

  • An integral is improper when the interval is infinite (Type 1) or the integrand is unbounded on the interval (Type 2). Both are defined as limits of proper integrals.
  • Finite limit ⇒ the integral converges to that value; infinite or nonexistent limit ⇒ it diverges.
  • \(\int_1^\infty x^{-p}dx\) converges to \(\frac{1}{p-1}\) exactly when \(p>1\) and diverges when \(p\le 1\) — with \(p=1\) (\(\ln t\)) as the boundary case.
  • When two pieces are required (both ends infinite, or an interior blow-up), the whole integral converges only if both pieces converge — never integrate straight across a discontinuity.
  • An integrand tending to 0 does not guarantee convergence; the tail must shrink fast enough.

Dive deeper

Sources

  • Improper Integrals