Improper Integrals

No end to substitute

The region under $y=\frac{1}{x^2}$y = 1/x² from x = 1 runs on to the right without end. The curve gets closer and closer to the x-axis and never meets it, so the region has no right-hand edge.

An ordinary definite integral is worked by putting the two limits into an antiderivative. Here there is no upper limit to put in: infinity is not a number, and it cannot be substituted for x.

The gold curve $y=\frac{1}{x^2}$y = 1/x², shaded from x = 1 to x = 6. The shaded part has area $1-\frac{1}{6}\approx 0.833$1 − 1/6 ≈ 0.833. The region it belongs to carries on past the edge of the frame, beyond any value of x.

Stop at t, then let t grow

Stop the region at a finite x = t, where the integral is an ordinary one: ${\int }_1^t\frac{1}{x^2}dx={\left[-\frac{1}{x}\right]}_1^t=-\frac{1}{t}+1=1-\frac{1}{t}$∫₁ᵗ 1/x² dx = [−1/x]₁ᵗ = −1/t + 1 = 1 − 1/t.

Now let t grow. At t = 2 the area is 0.5; at t = 10 it is 0.9; at t = 100 it is 0.99; at t = 1000 it is 0.999. As t grows without bound, $\frac{1}{t}$1/t shrinks to 0, so the area tends to 1.

That limit is what the integral of $\frac{1}{x^2}$1/x² from 1 to infinity means. It is written ${\int }_1^{\infty }\frac{1}{x^2}dx=1$∫₁^∞ 1/x² dx = 1, and the integral is said to converge. When the limit does not exist, the integral diverges and has no value.

The region under $y=\frac{1}{x^2}$y = 1/x² from 1 to b = 10, with its area as the gold bar: $1-\frac{1}{10}=0.9$1 − 1/10 = 0.9, below the dashed ceiling at 1. Drag b along its scale to 100 and then 1000: the area becomes 0.99 and then 0.999, and the bar never reaches the ceiling. Switch to $\frac{1}{x}$1/x: at b = 10 the area is $\ln 10\approx 2.303$ln 10 ≈ 2.303, and the bar is already through the ceiling.

Endless, but finite

So a region of endless width can have a finite area. The strips far out are thin enough that, all added together, they never push the total past 1. The region from 1 to 2 has area $1-\frac{1}{2}=\frac{1}{2}$1 − 1/2 = 1/2, and the whole of the endless rest, beyond x = 2, has area $\frac{1}{2}$1/2 as well.

The same happens under $y=e^{-x}$y = e^(−x) from x = 0. The area to t is ${\left[-e^{-x}\right]}_0^t=1-e^{-t}$[−e^(−x)]₀ᵗ = 1 − e^(−t), and $e^{-t}$e^(−t) shrinks to 0 as t grows, so the integral converges to 1. At t = 10 the area is already 0.99995.

$\frac{1}{x}$1/x does not settle

Under $y=\frac{1}{x}$y = 1/x the area from 1 to t is ${\left[\ln x\right]}_1^t=\ln t$[ln x]₁ᵗ = ln t. At t = 10 it is 2.303; at t = 1000 it is 6.908; at t = 1000000 it is 13.82. A logarithm grows without bound, slowly, so this area passes every number in the end. The integral of $\frac{1}{x}$1/x from 1 to infinity diverges.

Both $\frac{1}{x}$1/x and $\frac{1}{x^2}$1/x² shrink to 0, so shrinking to 0 does not decide it; how fast they shrink does. At x = 100, $\frac{1}{x}$1/x is 0.01 while $\frac{1}{x^2}$1/x² is 0.0001.

The gold curve $y=\frac{1}{x}$y = 1/x, shaded from x = 1 to x = 8, and the dashed curve $y=\frac{1}{x^2}$y = 1/x² below it. To x = 8 the area under $\frac{1}{x}$1/x is $\ln 8\approx 2.079$ln 8 ≈ 2.079, and under $\frac{1}{x^2}$1/x² it is $1-\frac{1}{8}=0.875$1 − 1/8 = 0.875. Carried on without end, the first grows without bound and the second tends to 1.

Any power of x

For a power p other than 1, the area under $\frac{1}{x^p}$1/xᵖ from 1 to t is $\frac{t^{1-p}-1}{1-p}$(t^(1 − p) − 1)/(1 − p). When p > 1 the power 1 − p is negative, so $t^{1-p}$t^(1 − p) shrinks to 0 and the area tends to $\frac{1}{p-1}$1/(p − 1). When p < 1 the power is positive, $t^{1-p}$t^(1 − p) grows without bound, and the integral diverges. At p = 1 the area is ln t, which also diverges.

So the integral of $\frac{1}{x^p}$1/xᵖ from 1 to infinity converges exactly when p > 1. For p = 1.5 the area to t = 100 is 1.8, to t = 10000 it is 1.98, and the limit is $\frac{1}{0.5}=2$1/0.5 = 2. For p = 0.5 the area is $2\sqrt{t}-2$2√t − 2, which is 18 at t = 100 and 198 at t = 10000.

Infinitely tall

The other kind of improper integral has a finite width but an integrand that blows up at one end. The integral of $\frac{1}{\sqrt{x}}$1/√x from 0 to 4 is one: the width is only 4, but $\frac{1}{\sqrt{x}}$1/√x grows without bound as x approaches 0, and at x = 0 it has no value at all.

The cure is the same, taken at the bad end. Start at a small t > 0 and integrate from there: the antiderivative of $x^{-\frac{1}{2}}$x^(−1/2) is $2\sqrt{x}$2√x, so the area from t to 4 is $2\sqrt{4}-2\sqrt{t}=4-2\sqrt{t}$2√4 − 2√t = 4 − 2√t. Then let t approach 0 from above. At t = 0.25 the area is 3; at t = 0.01 it is 3.8; at t = 0.0001 it is 3.98. The limit is 4, so the integral converges to 4.

Under $\frac{1}{x}$1/x the same move fails. The area from t to 1 is ln 1 − ln t = −ln t, which is 4.605 at t = 0.01 and 9.210 at t = 0.0001, and it grows without bound as t approaches 0. Near 0 the test on powers turns round: the integral of $\frac{1}{x^p}$1/xᵖ from 0 to 1 converges exactly when p < 1.

The gold curve $y=\frac{1}{\sqrt{x}}$y = 1/√x, which runs off the top of the frame as x approaches 0. The shaded region runs from t = 0.25 to 4 and has area $4-2\sqrt{0.25}=3$4 − 2√0.25 = 3. As t moves left toward 0, the shaded area tends to 4.

Both ends, or a break inside

A region can be endless in both directions. Under $y=\frac{1}{1+x^2}$y = 1/(1 + x²) the area from 0 to t is arctan t, which tends to $\frac{\pi }{2}$π/2 as t grows. The curve is symmetric about the y-axis, so the area from −t to 0 tends to $\frac{\pi }{2}$π/2 as well. Split at 0, take each limit on its own, and the whole region has area $\pi$π.

An integrand can also blow up inside the interval. The integral of $\frac{1}{x^2}$1/x² from −1 to 1 looks ordinary, and putting the limits into $-\frac{1}{x}$−1/x gives −1 − 1 = −2. That cannot be right: $\frac{1}{x^2}$1/x² is positive everywhere, so no area under it is negative. The function blows up at x = 0, so split there. The area from t to 1 is $\frac{1}{t}-1$1/t − 1, which is 99 at t = 0.01 and grows without bound, so the integral diverges.

The usual mistakes

Substituting infinity as if it were a number. The integral is a limit of ordinary integrals, and only the limit says whether it has a value at all.

Deciding that an integrand which shrinks to 0 must have a finite integral. $\frac{1}{x}$1/x shrinks to 0, and its area from 1 grows without bound.

Missing the infinitely tall kind. The integral of $\frac{1}{\sqrt{x}}$1/√x from 0 to 4 has finite limits, but $\frac{1}{\sqrt{x}}$1/√x blows up at 0, so that end needs a limit too.

Putting limits straight into an antiderivative across a point where the integrand blows up. The integral of $\frac{1}{x^2}$1/x² from −1 to 1 is not −2; it diverges.

Two thrusters left running

In the application below, two thrusts both fall away to nothing. One falls exponentially and its total impulse converges; the other falls like a reciprocal and its total passes any figure.