3.7 Improper Integrals

Big idea

The definite integral was built for a bounded function on a closed bounded interval. Both conditions are used in the construction: you need finitely many subintervals of finite width, and you need the function to have finite values on each. Drop either condition and the Riemann sum that defines the integral no longer makes sense, yet the questions people ask do not respect that boundary. Total work done by a force that acts forever, total probability under a curve that never touches the axis, the electric field of a point charge: all of them want an integral over an unbounded region.

The extension is done the way every extension in calculus is done, with a limit. Integrate over a region where the ordinary definition applies, then let the offending boundary move toward the trouble. If the resulting limit exists and is finite, you call the improper integral convergent and assign it that value. If the limit is infinite or fails to exist, the integral diverges and has no value.

The result is genuinely surprising the first time. A region of infinite extent can have finite area. The curve $y = 1/x^2$ stretches to the right forever, yet the area under it from $1$ onward is exactly $1$. The curve $y = 1/x$ looks almost identical and its area is infinite. What separates them is how fast the function decays, and the p-test makes that comparison precise.

Not every improper integral needs an exact value. Often you only need to know whether it is finite, and comparison against a known integral answers that without computing anything. If your integrand is squeezed below one that converges, yours converges too. That kind of reasoning is the backbone of the convergence tests for series in the chapters ahead.

Decoder

An improper integral is defined as the limit of proper integrals taken over regions that approach the problem point, and it converges exactly when that limit exists and is finite.

The load-bearing word is limit. The symbol $\int_1^{\infty}$ is not an integral with a strange upper limit; it is notation for $\lim_{t\to\infty}\int_1^t$. You always evaluate the ordinary integral first, obtaining a function of $t$, and only then take the limit. Reversing that order, by substituting infinity into an antiderivative, is how sign errors and nonsense values appear.

The second phrase to take seriously is the problem point, singular in each piece. If an integral is improper in two ways at once, or at an interior point, you must split it so that each piece has exactly one source of trouble, and every piece must converge on its own. One convergent piece does not rescue a divergent one, and the pieces are not allowed to cancel.

The classic mistake is applying the fundamental theorem across a point where the integrand blows up. The theorem requires continuity on the closed interval. Ignore that and you can produce a negative number for the area under a positive curve, which is the signal that something illegal happened.

Definitions and results

Infinite upper limit. If $f$ is continuous on $[a,\infty)$,

$$ \int_a^{\infty} f(x)\,dx = \lim_{t\to\infty}\int_a^{t} f(x)\,dx $$

and the integral converges when the limit exists and is finite. The analogous definition with $t \to -\infty$ handles an infinite lower limit.

Both limits infinite. Choose any convenient real number $c$ and split:

$$ \int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^{c} f(x)\,dx + \int_{c}^{\infty} f(x)\,dx $$

Both pieces must converge separately. The value does not depend on the choice of $c$. For example $\int_{-\infty}^{\infty}\frac{dx}{1+x^2} = \pi$, since each half contributes $\frac{\pi}{2}$.

Unbounded at the right endpoint. If $f$ is continuous on $[a,b)$ and unbounded as $x$ approaches $b$,

$$ \int_a^b f(x)\,dx = \lim_{t\to b^-}\int_a^{t} f(x)\,dx $$

with the mirror-image definition when the trouble is at $a$.

Unbounded at an interior point. If $f$ is unbounded near $c$ with $a < c < b$, split at $c$ and require both one-sided improper integrals to converge:

$$ \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx $$

If either piece diverges, the whole integral diverges.

The p-test on an infinite interval. For $a > 0$,

$$ \int_a^{\infty}\frac{dx}{x^p} \ \text{ converges exactly when } p > 1, \ \text{ with value } \ \frac{a^{1-p}}{p-1} $$

The p-test near zero.

$$ \int_0^{a}\frac{dx}{x^p} \ \text{ converges exactly when } p < 1, \ \text{ with value } \ \frac{a^{1-p}}{1-p} $$

The two conditions point opposite ways, which is worth pausing on: far out you need fast decay, and near a singularity you need a mild one. The borderline case $p = 1$ diverges in both settings.

Comparison. Suppose $0 \le f(x) \le g(x)$ on $[a,\infty)$. If $\int_a^{\infty} g$ converges then so does $\int_a^{\infty} f$; if $\int_a^{\infty} f$ diverges then so does $\int_a^{\infty} g$. This decides convergence without producing a value, and the p-test supplies the comparison functions.

Worked examples

Decay fast enough, and not fast enough

Evaluate $\int_1^{\infty}\frac{dx}{x^3}$, then test $\int_1^{\infty}\frac{dx}{x}$.

Replace the infinite limit by $t$ and integrate:

$$ \int_1^{t} x^{-3}\,dx = \left[-\frac{1}{2x^2}\right]_1^{t} = \frac{1}{2} - \frac{1}{2t^2} $$

As $t\to\infty$ the subtracted term vanishes, so the integral converges to $\frac{1}{2}$. This agrees with the p-test: $p = 3 > 1$ and $\frac{a^{1-p}}{p-1} = \frac{1}{2}$.

The second integral gives $\int_1^t \frac{dx}{x} = \ln t$, which grows without bound, so it diverges. Here $p = 1$, the borderline case. The two curves look similar for large $x$, but one encloses area $\frac12$ and the other encloses infinite area.

An unbounded integrand at an endpoint

Evaluate $\int_0^1 \frac{dx}{x^{2/3}}$.

The integrand blows up at the left endpoint, so cut the interval short there and take a one-sided limit:

$$ \int_t^1 x^{-2/3}\,dx = \Big[3x^{1/3}\Big]_t^1 = 3 - 3t^{1/3} $$

As $t \to 0^+$ the second term goes to $0$, so the integral converges to $3$. The p-test near zero confirms it: $p = \frac{2}{3} < 1$, and $\frac{a^{1-p}}{1-p} = \frac{1}{1/3} = 3$.

The region here is unbounded upward yet has finite area, which is the vertical counterpart of the first example.

An infinite limit handled with parts

Evaluate $\int_0^{\infty} x e^{-x}\,dx$.

Work on $[0,t]$ first. Integration by parts with $u = x$ and $dv = e^{-x}dx$ gives $v = -e^{-x}$, so an antiderivative is $-xe^{-x} - e^{-x} = -(x+1)e^{-x}$. Then

$$ \int_0^{t} x e^{-x}\,dx = \Big[-(x+1)e^{-x}\Big]_0^{t} = 1 - (t+1)e^{-t} $$

The exponential decays faster than $t+1$ grows, so $(t+1)e^{-t} \to 0$ and the integral converges to $1$. Check the antiderivative by differentiating: $-(x+1)e^{-x}$ has derivative $-e^{-x} + (x+1)e^{-x} = xe^{-x}$.

A discontinuity hiding inside the interval

Test $\int_{-1}^{2}\frac{dx}{x^2}$.

Applying the fundamental theorem without thinking gives $\left[-\frac{1}{x}\right]_{-1}^{2} = -\frac{1}{2} - 1 = -\frac{3}{2}$. That answer is impossible: the integrand is positive everywhere it is defined, so no correct value can be negative. The error is that $\frac{1}{x^2}$ is not continuous on $[-1,2]$; it blows up at $x = 0$, which sits inside the interval.

Split at the bad point and test each side. On the right,

$$ \int_t^{2} x^{-2}\,dx = \left[-\frac{1}{x}\right]_t^{2} = \frac{1}{t} - \frac{1}{2} $$

which grows without bound as $t \to 0^+$. That piece diverges, so the original integral diverges and has no value. The p-test near zero gives the same verdict at a glance, since $p = 2 \ge 1$.

Practice

These problems ask you to identify what makes each integral improper, write the correct limit or splitting, and decide convergence. Some want an exact value and some only want a verdict, so read the question before computing.

Quiz

Five items on infinite limits of integration, unbounded integrands at endpoints and interior points, convergence as a limit, and the p-test.