Improper integrals with an infinite integrand
A water clock drains through a hole in its floor, and the last of the water trickles out ever more slowly. The time to empty it is an integral whose integrand runs off to infinity at the bottom. Can such an integral still be a finite number?
Where you are 16 in Integrals
Why this exists
The fundamental theorem computes \int_a^b f(x)\,dx only when f is continuous on the closed interval [a, b], and in particular bounded there. Many natural integrands are not. A water clock is a vessel that drains through a small hole in its floor; the time it takes to empty turns out to be \displaystyle \int_0^H \frac{c}{\sqrt{h}}\,dh, and \displaystyle \frac{c}{\sqrt{h}} runs off to infinity at the bottom, where h = 0. The integral \int_0^1 \ln x\,dx has the same trouble at 0. Each has a vertical asymptote at a finite point, and near it the Riemann sums have no limit, because one rectangle can be made as tall as you like.
The cure is the one that Improper integrals used for an infinite interval: stay a little way back from the trouble, where the integral is ordinary, and then take a limit as you creep up to it.
Definition
Let f be continuous on (a, b] and unbounded near a. The improper integral of f from a to b is
If f is continuous on [a, b) and unbounded near b, then \int_a^b f(x)\,dx = \lim_{t \to b^-} \int_a^t f(x)\,dx. As before, the integral converges if the limit exists as a finite number and diverges otherwise.
If f blows up at a point c strictly inside (a, b), split there:
The integral converges only when both pieces converge, each as its own one-sided limit, and then its value is the sum. If several bad points occur, or a bad point and an infinite limit of integration, split so that each piece has exactly one source of trouble at one end.
When f happens to be continuous on all of [a, b], the limit gives the ordinary integral back (Exercise 20), so the definition extends the old one without changing it.
Intuition
Near 0 the graph of \displaystyle \frac{1}{x^p} is a spike, infinitely tall and infinitely thin. Whether it encloses finite area depends on a race between the two: how fast the height grows as x \to 0^+ against how fast the width shrinks. Figure 1 shows the two curves through (1, 1) that the previous lesson compared. On [1, \infty), \displaystyle \frac{1}{x^2} had finite area and \displaystyle \frac{1}{x} did not. Near 0 the roles turn over: the steeper curve now encloses more, and finite area needs the curve to climb slowly, like \displaystyle \frac{1}{\sqrt{x}}, whose spike from 0 to 1 has area 2.
The water clock gives the same picture in time. When the vessel is nearly empty, the water barely presses on the hole and the flow slows toward nothing, so each last thin layer of depth takes longer than the one above it. The integrand, time per unit of depth, is unbounded. Whether the clock still empties in finite time is exactly the question of whether the improper integral converges.
Results
The p-integrals near 0.
The same split holds for \displaystyle \int_a^{b} \frac{dx}{(x - a)^p} and \displaystyle \int_a^{b} \frac{dx}{(b - x)^p} with b > a: a blow-up like a power of the distance to the bad point converges exactly when that power is less than 1. Compare \displaystyle \int_1^{\infty} \frac{dx}{x^p}, which converges exactly when p > 1.
Comparison at a finite point. The direct and limit comparison tests hold word for word with the bad point in place of infinity. If 0 \le f(x) \le g(x) on (a, b], then \int_a^b g converging makes \int_a^b f converge, and \int_a^b f diverging makes \int_a^b g diverge. If f, g > 0 on (a, b] and \displaystyle \lim_{x \to a^+} \frac{f(x)}{g(x)} = L with 0 < L < \infty, the two integrals converge or diverge together.
Why it is true
The p-integrals. For p \ne 1 and 0 < t < 1,
If p < 1, the exponent 1 - p is positive, so t^{1 - p} \to 0 as t \to 0^+ and the integral tends to \displaystyle \frac{1}{1 - p}. If p > 1, then \displaystyle t^{1 - p} = \frac{1}{t^{p - 1}} \to \infty, and the integral diverges. If p = 1, \displaystyle \int_t^1 \frac{dx}{x} = -\ln t \to \infty. The shifted forms follow by the substitution u = x - a or u = b - x.
The two families are one fact seen twice. The substitution \displaystyle x = \frac{1}{u}, \displaystyle dx = -\frac{du}{u^2}, carries the interval [t, 1] to \displaystyle [1, \frac{1}{t}] and gives
As t \to 0^+ this is the tail p-integral with exponent 2 - p, which converges exactly when 2 - p > 1, that is, when p < 1.
Comparison. The proof in the previous lesson goes through with t \to a^+ in place of b \to \infty. Because f \ge 0, the function t \mapsto \int_t^b f grows as t moves toward a, and f \le g keeps it below \int_a^b g. A function that is monotone and bounded has a finite one-sided limit (the completeness of the real numbers, taken here without proof). Limit comparison then follows as before, from \displaystyle \frac{L}{2}\,g < f < 2L\,g near a.
Worked examples
Two blow-ups at a left endpoint
Evaluate \displaystyle \int_0^8 \dfrac{dx}{\sqrt[3]{x}} and \displaystyle \int_1^5 \dfrac{dx}{\sqrt{x - 1}}.
SolutionThe first integrand is x^{-1/3}, continuous on (0, 8] with x^{-1/3} \to \infty as x \to 0^+. On [t, 8] the integral is ordinary, with antiderivative \displaystyle \frac{3}{2}x^{2/3}:
It converges to 6, as the result for \displaystyle p = \frac{1}{3} < 1 promised. The second blows up at x = 1, where x - 1 = 0. An antiderivative of (x - 1)^{-1/2} is \displaystyle 2\sqrt{x - 1}, so
as t \to 1^+. It converges to 4: the shifted p-integral with \displaystyle p = \frac{1}{2}.
Natia’s water clock
Natia’s water clock is an upright cylinder of cross-sectional area A = 80 square centimeters with a small hole in its floor (Figure 2). By Torricelli’s law, water leaves the hole at a speed proportional to \displaystyle \sqrt{h}, where h is the depth in centimeters, so the volume drains at \displaystyle 4\sqrt{h} cubic centimeters per second for her hole. Filled to 36 centimeters, how long does it take to empty, and how long does the bottom half of the water take?
When the depth falls by a thin layer dh, a volume A\,dh = 80\,dh leaves, at the rate \displaystyle 4\sqrt{h}, so that layer takes about \displaystyle \frac{80\,dh}{4\sqrt{h}} = \frac{20}{\sqrt{h}}\,dh seconds. Adding the layers from the bottom to the top,
The integrand is unbounded as h \to 0^+, which is the slowing trickle. Stop a depth t short of empty:
The integral converges: the clock empties in 240 seconds, 4 minutes. The time to fall from 36 to 18 centimeters is \displaystyle 40(6 - \sqrt{18}) = 240 - 120\sqrt{2} \approx 70.3 seconds, and the bottom 18 centimeters take \displaystyle 120\sqrt{2} \approx 169.7 seconds, well over twice as long. Check by solving the differential equation \displaystyle 80\frac{dh}{dt} = -4\sqrt{h} directly: separating gives \displaystyle 2\sqrt{h} = 12 - \frac{t}{20}, which reaches 0 at t = 240.
Tural integrates \ln x down to 0
Tural wants the signed area between y = \ln x and the x-axis from 0 to 1. Find \int_0^1 \ln x\,dx.
Solution\ln x \to -\infty as x \to 0^+, so the integral is improper at 0. Integration by parts gives \int \ln x\,dx = x\ln x - x + C, so
The term t\ln t is a product of something tending to 0 and something tending to -\infty. Put t = e^{-u}, so u \to \infty as t \to 0^+: then t\ln t = -ue^{-u} \to 0, because the exponential outgrows any power (Comparing growth rates at infinity). So the limit is -1 - 0 + 0, and
The value is negative because \ln x < 0 on (0, 1): the region between the curve and the axis lies below the axis and has area 1. It matches the picture of \ln x as the reflection of e^x in the line y = x: the region under e^{x} for x \le 0 is exactly that region reflected, and \int_{-\infty}^0 e^x\,dx = 1.
A blow-up inside the interval
Evaluate \int_0^3 \dfrac{dx}{(x - 1)^{2/3}}.
SolutionThe integrand is positive and blows up at x = 1, inside [0, 3] (Figure 3). Split there and take each piece as its own limit. An antiderivative is \displaystyle 3\sqrt[3]{x - 1}, the real cube root, which is defined for negative inputs:
as t \to 1^-, and
as s \to 1^+. Both pieces converge, each a shifted p-integral with \displaystyle p = \frac{2}{3}, so
Here plugging 0 and 3 straight into \displaystyle 3\sqrt[3]{x - 1} happens to give the right number, because the antiderivative is continuous across x = 1. The next example shows that this cannot be relied on.
An antiderivative that jumps the asymptote
Decide whether \int_0^3 \dfrac{2x}{x^2 - 1}\,dx converges.
SolutionOn [0, 3] the denominator vanishes at x = 1, so split there. Below 1, x^2 - 1 < 0 and an antiderivative is \ln(1 - x^2):
as t \to 1^-. One piece diverges, so the integral diverges. Nothing else needs computing, but the other piece is instructive: \displaystyle \int_s^3 \frac{2x}{x^2 - 1}\,dx = \ln 8 - \ln(s^2 - 1) \to \infty as s \to 1^+.
The formula \ln\lvert x^2 - 1 \rvert is an antiderivative on each side of 1, and evaluating it at the two ends gives \ln 8 - \ln 1 = \ln 8 \approx 2.079, a finite, harmless-looking number. It is not the value of anything. The fundamental theorem needs an antiderivative on the whole closed interval, and this one is undefined at 1. Nor do the -\infty on the left and the +\infty on the right cancel: each piece must converge on its own, and here neither does.
Daulet compares near the bad point
Daulet settles these without any antiderivative. Decide (a) \displaystyle \int_0^{\pi/2} \dfrac{dx}{\sqrt{\sin x}}, with a bound, and (b) \int_0^1 \dfrac{dx}{e^x - 1}.
Solution(a) The integrand blows up as x \to 0^+, since \sin x \to 0. A benchmark needs a lower bound on \sin x. On \displaystyle [0, \frac{\pi}{2}] the sine graph is concave down, so it lies above its chord from (0, 0) to \displaystyle (\frac{\pi}{2}, 1): \displaystyle \sin x \ge \frac{2x}{\pi}. Then
and the benchmark converges (\displaystyle p = \frac{1}{2}):
By direct comparison the integral converges and is at most \pi. Numerically it is about 2.622.
(b) Near 0, e^x - 1 behaves like x (its derivative there is 1), which suggests \displaystyle g(x) = \frac{1}{x}. Both are positive on (0, 1], and
the reciprocal of the derivative of e^x at 0. Since 0 < 1 < \infty and \displaystyle \int_0^1 \frac{dx}{x} diverges (p = 1), the integral diverges by limit comparison.
Improper at both ends of (0, \infty)
Evaluate \displaystyle \int_0^{\infty} \dfrac{dx}{\sqrt{x}\,(1 + x)}, and decide for which p the integral \displaystyle \int_0^{\infty} \frac{dx}{x^p} converges.
SolutionThe first integrand blows up at 0 and the interval is infinite, two kinds of trouble, so split at 1. Near 0 it is below \displaystyle \frac{1}{\sqrt{x}}, with \displaystyle p = \frac{1}{2} < 1; beyond 1 it is below x^{-3/2}, a tail with \displaystyle p = \frac{3}{2} > 1. So both pieces converge, by comparison. For the value, substitute \displaystyle u = \sqrt{x}, x = u^2, dx = 2u\,du:
Then \displaystyle \int_t^1 = 2\cdot\frac{\pi}{4} - 2\arctan\sqrt{t} \to \frac{\pi}{2} as t \to 0^+, and \displaystyle \int_1^b = 2\arctan\sqrt{b} - \frac{\pi}{2} \to \pi - \frac{\pi}{2} = \frac{\pi}{2} as b \to \infty. So
For \displaystyle \int_0^{\infty} \frac{dx}{x^p}, split at 1 the same way. The piece on [0, 1] needs p < 1 and the piece on [1, \infty) needs p > 1. No p does both, so \displaystyle \int_0^{\infty} \frac{dx}{x^p} diverges for every p. The first integral converges because its integrand behaves like one power near 0 and a different power far out.
Boundaries
- Find the bad points first. The fundamental theorem needs f continuous on the closed interval. An antiderivative evaluated across a vertical asymptote can return a finite number for a divergent integral (Example 5), or, as \displaystyle -\frac{1}{x} does across 0 for \displaystyle \int_{-1}^1 \frac{dx}{x^2}, a negative number for a positive integrand. Neither number means anything.
- Each piece is its own limit. +\infty and -\infty never cancel. \displaystyle \int_{-1}^{1} \frac{dx}{x} diverges, although the symmetric limit \displaystyle \lim_{\varepsilon \to 0^+}\left(\int_{-1}^{-\varepsilon} + \int_{\varepsilon}^{1}\right)\frac{dx}{x} is 0; that symmetric limit is a different quantity, not the improper integral.
- The p cutoff flips. Near a finite point, p < 1 converges; at infinity, p > 1 converges; p = 1 diverges in both places.
- Undefined is not unbounded. If f has a finite limit at the bad point, as \displaystyle \frac{\sin x}{x} does at 0, the gap is removable: fill it in and the integral is ordinary. Only a genuine blow-up needs this lesson.
- Comparison needs f \ge 0 near the bad point, and points one way only: below a divergent integral or above a convergent one, nothing follows.
- The water clock is a model. Torricelli’s law ignores friction at the hole and surface tension, which matter most in the last millimeters. The integral answers the model’s question exactly; a real clock finishes somewhat later.
Where this goes next
Improper integrals at a finite point complete the toolkit begun in Improper integrals. Comparing a sum with an improper integral gives the integral test and the The integral test and p series, which lead into Convergence tests for infinite series. The water clock is a separable differential equation, and integrands that blow up at a point appear throughout probability densities, where a density may be infinite at a point yet still have total area 1.
Exercises
Blow-ups at an endpoint
- Evaluate \int_0^{27} x^{-2/3}\,dx.
- Decide whether \int_0^1 \dfrac{dx}{x^{3/2}} converges.
- Evaluate \displaystyle \int_2^3 \dfrac{dx}{\sqrt{3 - x}}.
- Evaluate \displaystyle \int_0^1 \dfrac{dx}{\sqrt{1 - x^2}}.
- Decide whether \int_0^2 \dfrac{dx}{2 - x} converges.
- Decide whether \int_0^{\pi/2} \tan x\,dx converges.
- Evaluate \displaystyle \int_0^1 \dfrac{\ln x}{\sqrt{x}}\,dx by parts.
- Evaluate \displaystyle \int_1^e \dfrac{dx}{x\sqrt{\ln x}}.
Interior blow-ups and the water clock
- Evaluate \displaystyle \int_{-1}^{27} \dfrac{dx}{\sqrt[3]{x}}.
- Decide whether \int_1^4 \dfrac{dx}{(x - 2)^2} converges, and say what evaluating \displaystyle -\frac{1}{x - 2} at the ends would give.
- Evaluate \displaystyle \int_0^{10} \dfrac{dx}{\sqrt{\lvert x - 1 \rvert}}.
- Decide whether \int_{-1}^{1} \dfrac{dx}{x^3} converges.
- Natia builds a second cylinder of cross-sectional area 150 square centimeters whose hole drains \displaystyle 5\sqrt{h} cubic centimeters per second. Filled to 49 centimeters, how long does it take to empty, and how much of that time do the bottom 16 centimeters take?
- A tank’s sides slope outward, so at depth h its cross-sectional area is 80 + 20h square centimeters. Its hole drains \displaystyle 4\sqrt{h} cubic centimeters per second. Find the time to empty it from a depth of 16 centimeters.
Comparison
- Show that \displaystyle \int_0^1 \dfrac{dx}{x + \sqrt{x}} converges and is at most 2. Then find its value with \displaystyle u = \sqrt{x}.
- Decide whether \int_0^1 \dfrac{dx}{\sin x} converges.
- Decide whether \displaystyle \int_0^1 \dfrac{dx}{\sqrt{x} - x/2} converges. Explain why direct comparison with \displaystyle \frac{1}{\sqrt{x}} points the wrong way, and use limit comparison instead; then evaluate it with \displaystyle u = \sqrt{x}.
- Show that \displaystyle \int_0^2 \dfrac{dx}{\sqrt{8 - x^3}} converges and is at most \displaystyle \sqrt{2}. (Factor 8 - x^3.)
Reasoning
- Give a counterexample to the claim: “If f is continuous on (0, 1] and \lim_{x \to 0^+} f(x) = \infty, then \int_0^1 f(x)\,dx diverges.”
- Let f be continuous on the closed interval [a, b]. Show that \lim_{t \to a^+} \int_t^b f(x)\,dx = \int_a^b f(x)\,dx, so the improper definition agrees with the ordinary integral.
- For any upright cylinder draining by Torricelli’s law, show that the time to empty from depth h is proportional to \displaystyle \sqrt{h}, and conclude that the bottom quarter of the starting depth takes half of the total time.
Answers
- 9: \int_t^{27} x^{-2/3}\,dx = 3\cdot 3 - 3t^{1/3} \to 9 (\displaystyle p = \frac{2}{3} < 1).
- Diverges: \displaystyle \int_t^1 x^{-3/2}\,dx = \frac{2}{\sqrt{t}} - 2 \to \infty (\displaystyle p = \frac{3}{2} \ge 1).
- 2: an antiderivative is \displaystyle -2\sqrt{3 - x}, so \displaystyle \int_2^t = 2 - 2\sqrt{3 - t} \to 2 as t \to 3^-.
- \displaystyle \frac{\pi}{2}: \displaystyle \int_0^t = \arcsin t \to \arcsin 1 = \frac{\pi}{2} as t \to 1^- (Derivatives of inverse trigonometric functions).
- Diverges: \displaystyle \int_0^t \frac{dx}{2 - x} = \ln 2 - \ln(2 - t) \to \infty as t \to 2^- (p = 1).
- Diverges: \int_0^t \tan x\,dx = -\ln(\cos t) \to \infty as \displaystyle t \to \frac{\pi}{2}^-, since \cos t \to 0^+.
- -4: with u = \ln x, dv = x^{-1/2}\,dx, an antiderivative is \displaystyle 2\sqrt{x}\ln x - 4\sqrt{x}. At 1 it is -4; as t \to 0^+, \displaystyle 2\sqrt{t}\ln t \to 0 (power beats logarithm) and \displaystyle 4\sqrt{t} \to 0.
- 2: with u = \ln x it becomes \int_0^1 u^{-1/2}\,du; directly, \displaystyle 2\sqrt{\ln x} is an antiderivative, \displaystyle 2\sqrt{\ln e} - 2\sqrt{\ln t} \to 2 as t \to 1^+.
- 12: split at 0. With antiderivative \displaystyle \frac{3}{2}x^{2/3}, the left piece is \displaystyle 0 - \frac{3}{2} = -\frac{3}{2} and the right piece is \displaystyle \frac{3}{2}\cdot 9 = \frac{27}{2}; both converge, and the sum is 12.
- Diverges: \displaystyle \int_s^4 \frac{dx}{(x - 2)^2} = \frac{1}{s - 2} - \frac{1}{2} \to \infty as s \to 2^+. Evaluating at the ends gives \displaystyle -\frac{1}{2} - 1 = -\frac{3}{2}, negative for a positive integrand and meaningless.
- 8: split at 1. \int_0^1 (1 - x)^{-1/2}\,dx = 2 and \displaystyle \int_1^{10} (x - 1)^{-1/2}\,dx = 2\sqrt{9} = 6.
- Diverges: \displaystyle \int_t^1 x^{-3}\,dx = \frac{1}{2t^2} - \frac{1}{2} \to \infty as t \to 0^+. One piece diverging is enough; the symmetry of \displaystyle \frac{1}{x^3} does not rescue it.
- 420 seconds, of which the bottom 16 centimeters take 240: the integrand is \displaystyle \frac{150}{5\sqrt{h}} = \frac{30}{\sqrt{h}}, and \displaystyle \int_0^{49} \frac{30}{\sqrt{h}}\,dh = 60\cdot 7 = 420, while \displaystyle \int_0^{16} \frac{30}{\sqrt{h}}\,dh = 60\cdot 4 = 240.
- \displaystyle \frac{1120}{3} \approx 373.3 seconds: the integrand is \displaystyle \frac{80 + 20h}{4\sqrt{h}} = \frac{20}{\sqrt{h}} + 5\sqrt{h}, and \displaystyle \int_0^{16} = 160 + \frac{640}{3}.
- 2\ln 2 \approx 1.386: on (0, 1], \displaystyle 0 < \frac{1}{x + \sqrt{x}} < \frac{1}{\sqrt{x}}, and \displaystyle \int_0^1 \frac{dx}{\sqrt{x}} = 2. With \displaystyle u = \sqrt{x}, the integral is \displaystyle \int_0^1 \frac{2u\,du}{u^2 + u} = 2\ln(1 + u)\Big|_0^1.
- Diverges: 0 < \sin x \le x for 0 < x \le 1, so \displaystyle \frac{1}{\sin x} \ge \frac{1}{x}, and \displaystyle \int_0^1 \frac{dx}{x} diverges.
- Converges, to 4\ln 2 \approx 2.773. On (0, 1], \displaystyle \sqrt{x} - \frac{x}{2} < \sqrt{x}, so the integrand is above \displaystyle \frac{1}{\sqrt{x}}, and being above a convergent integral says nothing. The ratio \displaystyle \frac{\sqrt{x}}{\sqrt{x} - x/2} = \frac{1}{1 - \sqrt{x}/2} \to 1 as x \to 0^+, and \displaystyle \int_0^1 \frac{dx}{\sqrt{x}} converges. With \displaystyle u = \sqrt{x} it is \displaystyle \int_0^1 \frac{2\,du}{1 - u/2} = -4\ln\left(1 - \frac{u}{2}\right)\Big|_0^1.
- 8 - x^3 = (2 - x)(4 + 2x + x^2) and 4 + 2x + x^2 \ge 4 on [0, 2], so \displaystyle 0 < \frac{1}{\sqrt{8 - x^3}} \le \frac{1}{2\sqrt{2 - x}}, and \displaystyle \int_0^2 \frac{dx}{2\sqrt{2 - x}} = \sqrt{2}. It converges, to about 0.991.
- \displaystyle f(x) = \frac{1}{\sqrt{x}}: it is continuous on (0, 1] and tends to \infty at 0, yet \displaystyle \int_0^1 \frac{dx}{\sqrt{x}} = 2.
- Let F(t) = \int_t^b f(x)\,dx. Since f is continuous on [a, b], F is differentiable, hence continuous, on [a, b] by the fundamental theorem. So \lim_{t \to a^+} F(t) = F(a) = \int_a^b f(x)\,dx.
- With area A and drain rate \displaystyle k\sqrt{h}, the time from depth h is \displaystyle \int_0^h \frac{A}{k\sqrt{y}}\,dy = \frac{2A}{k}\sqrt{h}. From \displaystyle \frac{H}{4} it is \displaystyle \frac{2A}{k}\cdot\frac{\sqrt{H}}{2}, half of the total \displaystyle \frac{2A}{k}\sqrt{H}.