A definite integral becomes improper when the interval is unbounded or the integrand becomes unbounded somewhere on the interval. In either case, the integral is defined through a limit, and the geometric region may extend infinitely far or rise without bound and still have finite area.

An infinite interval

Consider \(\textstyle\int_1^\infty \tfrac{1}{x^2}\,dx\). The infinity symbol is not an endpoint at which an antiderivative can be evaluated, so instead write

\[\begin{aligned} \int_1^\infty \frac{1}{x^2}\,dx &= \lim_{T\to\infty} \int_1^T \frac{1}{x^2}\,dx\\ &= \lim_{T\to\infty} \left[-\frac1x\right]_1^T = \lim_{T\to\infty} \left(1-\frac1T\right) = 1 \end{aligned}\]

The interval has infinite length, but the area is finite, and the integral converges.

An unbounded integrand

Now consider \(\textstyle\int_0^1\tfrac{1}{\sqrt{x}}\,dx\). The interval is finite, but the integrand grows without bound as \(x\to0^+\), so write

\[\begin{aligned} \int_0^1\frac{1}{\sqrt{x}}\,dx &= \lim_{s\to0^+} \int_s^1 x^{-1/2}\,dx\\ &= \lim_{s\to0^+} \left[2\sqrt{x}\right]_s^1 = \lim_{s\to0^+} \left(2-2\sqrt{s}\right) = 2 \end{aligned}\]

Again, the integral converges, and the vertical asymptote does not force the area to be infinite.

The \(p\)-integral thresholds

Top: y = x⁻ᵖ drawn on logarithmic axes, where every power function is a straight line of slope −p. The dashed diagonal is p = 1, the dividing case. Bottom: the two partial integrals plotted against how far the cutoff has been pushed, in decades, outward toward infinity, and inward toward zero. Both start at zero and climb. A curve that flattens is converging to the value the panel names; a curve that keeps rising is diverging, and the panel says so rather than printing a number. Slide p through 1 and watch the two verdicts trade places.

Two standard families are worth knowing. For the tail integral \(\textstyle\int_1^\infty \tfrac{1}{x^p}\,dx\), the integral converges exactly when \(p>1\), and when it converges, \(\textstyle\int_1^\infty x^{-p}\,dx = \tfrac{1}{p-1}\). For the integral near zero, \(\textstyle\int_0^1\tfrac{1}{x^p}\,dx\), the condition reverses, so it converges exactly when \(p<1\), and when it converges, \(\textstyle\int_0^1 x^{-p}\,dx = \tfrac{1}{1-p}\). The boundary case \(p=1\) diverges in both settings, and the same expression, \(\tfrac1x\), therefore sits at the convergence threshold both near zero and at infinity.

A discontinuity inside the interval

Suppose \(\textstyle\int_{-1}^{1}\tfrac{1}{x^2}\,dx\). The integrand is unbounded at \(x=0\), so the integral must be split there:

\[\int_{-1}^{1}\frac{1}{x^2}\,dx = \int_{-1}^{0}\frac{1}{x^2}\,dx + \int_{0}^{1}\frac{1}{x^2}\,dx\]

Each piece is improper and needs its own limit. Both diverge, so the original integral diverges. If we ignore the discontinuity and apply the Fundamental Theorem mechanically, \(\left[-\tfrac1x\right]_{-1}^{1} = -2\), and that answer is impossible as geometric area under a positive function. The problem is not the antiderivative. The Fundamental Theorem was applied across a point where the hypotheses failed.

Two habits that prevent most errors

First, inspect the interval and the integrand before integrating, looking for infinite bounds, denominator zeros, logarithmic singularities, and endpoints where a root or power becomes unbounded. Second, write the limit explicitly, because the limit is part of the definition of the improper integral.

If the limit is finite, the integral converges, and if the limit is infinite or fails to exist, the integral diverges.