3.7 Improper Integrals

1. Improper Integrals

Title

Calculus II · Section 3.7

Infinite regions, finite areas, and how to tell which is which

2. What this lesson gives you

Objectives

Every definite integral so far had a finite interval and a function that stayed bounded on it. This lesson drops both requirements, with one idea: integrate to a movable endpoint, then take a limit.

• Evaluate an integral over an infinite interval, and say whether it converges
• Evaluate an integral whose integrand has an infinite discontinuity, at an end or inside the interval
• State the p-integral rules at both ends, and why they point opposite ways
• Use the comparison theorem to decide convergence without evaluating

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 330 — learning objectives 3.7.1 to 3.7.3

Every definite integral you have evaluated so far came with two quiet guarantees: the interval was finite, and the function stayed bounded on it. Riemann sums need both. Yet some of the most useful integrals in science break one or the other: probabilities over all possible waiting times, the escape energy of a rocket pulled by gravity out to infinite distance, areas under curves with vertical asymptotes.

The fix is a single idea you will use in every example today. Do an ordinary integral up to a movable endpoint, then take a limit as that endpoint heads for the trouble. If the limit is a finite number, that number is the integral. If not, the integral diverges.

The last part of the lesson handles integrals you cannot evaluate at all, by comparing them with simpler ones whose behaviour you know. That idea returns in Chapter 5, where it decides whether infinite series converge.

3. Before anything new: the area out to t

Warm-up

Write your answer before revealing it. Nothing here is new: the antiderivative of one over x squared is minus one over x, and you evaluate it between 1 and t in the usual way.

The interesting part is the second question. However large you make t, the answer 1 minus one over t stays below 1, and it gets as close to 1 as you like. So the region under the curve all the way out to infinity, a region that never ends, behaves as if it has area exactly 1.

Keep this computation in mind. The whole of Part 1 is a careful version of what you just did: integrate to a finite t, then let t run.

4. Integrating over an infinite interval

Section

Part 1

5. Two ways an integral can be improper

Concept

The definite integral was built from Riemann sums on a closed, bounded interval, with a function that stays bounded there. A region that runs off to infinity sideways, or shoots up along an asymptote, breaks one of those two assumptions.

improper integral — An integral over an infinite interval, or of a function with an infinite discontinuity on the interval. It is defined as a limit of ordinary integrals.

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 330 — introduction

Look at the two shaded regions. The left one is infinitely long: the interval runs from 1 out to infinity. The right one is infinitely tall: the interval is only from 0 to 1, but the function shoots up along the y-axis. Neither can be chopped into finitely many rectangles of finite height and finite width, so the ordinary definite integral simply does not apply.

These are the two kinds of improper integral, and the book treats them in that order. The first kind has an infinite limit of integration. The second has an integrand with an infinite discontinuity, a vertical asymptote, somewhere in the interval.

The surprising fact, which you will prove for both of these regions today, is that both have finite area. Infinite extent does not force infinite area. Whether the area is finite depends on how quickly the curve thins out.

6. Area out to t, then let t run

Concept

You cannot integrate to infinity directly, but you can integrate to any finite t. Do that, then ask what the answers do as t grows.

\[ \int_1^{\infty} \frac{dx}{x^2} = \lim_{t\to\infty}\int_1^{t}\frac{dx}{x^2} = \lim_{t\to\infty}\left(1 - \frac1t\right) = 1 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 330 — Figure 3.17

The figure shows the region under one over x squared cut off at three different places. Out to 2 the area is one half; out to 4 it is three quarters; out to 8 it is seven eighths. Each time t doubles, the extra area is half of what the previous doubling added.

That pattern is exactly what 1 minus one over t describes, and it never passes 1. So the limit exists and equals 1, and that is what the integral to infinity means. There is no step where you substitute infinity for anything. You compute with finite numbers throughout and let the limit do the rest.

This is Figure 3.17 of the book, made concrete. Every improper integral over an infinite interval is handled this way.

7. Most of the area sits near the start

Intuition

An infinitely long region can have finite area only if its far end is almost empty. Measure how much area lies beyond t:

\[ \int_t^{\infty}\frac{dx}{x^2} = \lim_{s\to\infty}\left(\frac1t - \frac1s\right) = \frac{1}{t} \]

\[ t = 10: \; 0.1 \text{ of the total}, \qquad t = 100: \; 0.01 \text{ of the total} \]

Ninety-nine percent of the area lies between 1 and 100. The rest of the infinite strip holds one hundredth.

Here is a way to believe the finite answer. Compute the area that lies beyond t, the part of the strip you have not yet counted. It is exactly one over t.

So beyond 10, the entire infinite remainder of the strip holds one tenth of the total. Beyond 100 it holds one hundredth. Almost all the area is packed into the first few units, and the endless tail is so thin that together it adds up to very little.

That is the general picture of a convergent improper integral. It converges because its tail becomes negligible. When the tail does not shrink fast enough, as you will see on the next slide with one over x, the leftover area beyond any t is still infinite, and the integral diverges.

8. Reading definitions 3.16 and 3.17

Notation

\[ \int_{-\infty}^{b} f(x)\,dx = \lim_{t\to-\infty}\int_t^{b} f(x)\,dx \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 330 — definition, equations 3.16 and 3.17

Step through the annotations one at a time. The definition has two moves in a fixed order: first an ordinary definite integral with t held fixed, then a limit as t grows. Everything inside the limit is Calculus I.

The words converge and diverge are borrowed from sequences, and they mean the same thing here. If the limit is a finite number, the integral converges to that number. If the limit is infinite, or does not exist at all, the integral diverges.

The second line is equation 3.17, the mirror image. When the infinite end is on the left, you integrate from t to b and let t go to minus infinity. The idea is identical; only the direction changes. Notice that f must be continuous on the whole interval, so the only trouble is the infinite end.

9. Example 3.47: the area under one over x

Worked example

Is the area between y = 1/x and the x-axis over the interval from 1 to infinity finite or infinite?

\[ A = \int_1^{\infty} \frac{1}{x}\,dx \]

\[ A = \lim_{t\to\infty}\int_1^{t}\frac{1}{x}\,dx \]

\[ \int_1^{t}\frac{1}{x}\,dx = \ln|x|\Big|_1^{t} \]

\[ \ln|x|\Big|_1^{t} = \ln t - \ln 1 = \ln t \]

\[ \lim_{t\to\infty} \ln t = +\infty \]

\[ \ln 10 \approx 2.303, \quad \ln 100 \approx 4.605, \quad \ln 1000 \approx 6.908 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, pp. 331-332 — Example 3.47

This is the question the section opened with. The function one over x shrinks to zero, just like one over x squared, so you might expect a finite area. Follow the steps: write the limit, find the antiderivative, evaluate it between 1 and t, and take the limit.

The antiderivative is the natural logarithm, and the logarithm has no ceiling. So the area is infinite. The picture compares the two accumulated areas: the one under one over x squared stalls below 1, the one under one over x keeps climbing.

The numbers in the check are worth noticing. Every time t is multiplied by ten, the area grows by the same amount, about 2.3. That is slow growth, but it never stops. The lesson is that shrinking to zero is not enough; the curve must shrink fast enough.

10. How far out before the area reaches 10?

Estimation

\[ \int_1^{t}\frac{dx}{x} = \ln t \]

\[ \ln t = 10 \iff t = e^{10} \approx 22{,}026 \]

Pick an option before revealing. The integral diverges, so the area passes every number eventually. The question is how long eventually is.

The area out to t is the natural logarithm of t, so it reaches 10 when t is e to the tenth power, about twenty-two thousand. To reach 20 you would need t near five hundred million.

This is why you cannot decide convergence by computing a few numbers. The areas under one over x out to a thousand look as if they might be levelling off near 7. They are not. Only the limit, taken honestly, tells you the truth.

11. Example 3.48: Gabriel's horn

Worked example

Revolve the same infinite region, under y = 1/x from 1 onward, about the x-axis. Find the volume.

\[ V = \pi\int_1^{\infty}\left(\frac1x\right)^2 dx = \pi\int_1^{\infty}\frac{dx}{x^2} \]

\[ V = \pi\lim_{t\to\infty}\int_1^{t}\frac{dx}{x^2} \]

\[ \int_1^{t} x^{-2}\,dx = -\frac{1}{x}\Big|_1^{t} \]

\[ -\frac{1}{x}\Big|_1^{t} = -\frac1t + 1 \]

\[ V = \pi\lim_{t\to\infty}\left(1 - \frac1t\right) = \pi \]

\[ \pi\left(1 - \frac{1}{100}\right) \approx 3.110, \qquad \pi \approx 3.142 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, pp. 332-333 — Example 3.48 and Figure 3.19

Now spin the same infinite region around the x-axis. Each vertical slice sweeps out a disk whose radius is the height of the curve, one over x, so its area is pi over x squared. The volume is pi times the integral of one over x squared from 1 to infinity.

You already know that integral: it is 1. So the volume is exactly pi, a finite number, even though the solid is infinitely long and was built from a region of infinite area.

The check cuts the horn off at x equal to 100. That finite horn already holds 99 percent of pi. The part beyond 100, infinitely long, holds only pi over 100. The figure draws the horn with a few of its circular rims so you can see how quickly the cross-sections shrink.

12. A horn you can fill but cannot paint

Intuition

The region under one over x has infinite area, yet the solid it sweeps out has finite volume. The squaring is what does it: the radius shrinks like one over x, but the disk areas shrink like one over x squared.

\[ \text{area: } \int_1^{\infty}\frac{dx}{x} = \infty \qquad \text{volume: } \pi\int_1^{\infty}\frac{dx}{x^2} = \pi \]

The horn's surface is worse still: its area is at least 2 pi times the area under one over x, so it is infinite. A finite amount of paint fills the inside; no amount covers the outside.

\[ S = 2\pi\int_1^{\infty}\frac1x\sqrt{1 + \frac{1}{x^4}}\,dx \;\ge\; 2\pi\int_1^{\infty}\frac{dx}{x} = \infty \]

This is the famous paradox of Gabriel's horn. The flat region has infinite area, yet the solid made from it has finite volume. The difference is the squaring: the volume adds up disk areas, which shrink like one over x squared, while the flat area adds up heights, which shrink only like one over x.

The surface of the horn is worse still. The surface-area formula from Chapter 2 gives an integrand bigger than 2 pi over x, and the integral of that is infinite. So the horn has infinite surface area.

Put that together and you get the paradox: you could fill the horn with a finite amount of paint, but you could never paint its outside. The resolution is that mathematical paint has no thickness, while any real coat of paint would be too thick to fit down the narrowing horn. It is a vivid reminder that infinite processes do not always behave the way intuition expects.

13. Trap: plugging in infinity

Trap

14. Exercise 363: the area under sine

Prediction

\[ \int_0^{\infty}\sin x\,dx \]

\[ \int_0^{t}\sin x\,dx = 1 - \cos t \quad \text{oscillates between 0 and 2} \]

Commit to an option first. It is tempting to say the positive and negative humps cancel, so the answer must be 0.

Compute honestly. The area from 0 to t is 1 minus cos t. When t is pi it is 2; when t is 2 pi it is back to 0; and it keeps swinging between 0 and 2 for ever. A quantity that never settles has no limit, so the integral diverges.

Notice that this divergence is not to infinity. The areas stay bounded; they just refuse to settle. That is the second way an integral can diverge, and it is the subject of the next slide.

15. Two ways to diverge

Intuition

Convergence means the areas to t settle at one finite number. They can fail by growing without bound, like the logarithm, or by never settling, like 1 minus cos t.

converge / diverge — The improper integral converges if the defining limit exists as a finite number; otherwise it diverges, whether to infinity or by oscillation.

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 330 — definition, last paragraph

The figure puts the two failures side by side. The dashed red curve is the kind of divergence you met in Example 3.47: the areas grow without bound. The orange curve, 1 minus cos t, stays between 0 and 2 but never settles on a single value.

Both are divergence. The definition asks for one thing only: that the limit exists as a finite number. Unbounded growth fails that test, and so does endless oscillation.

When you write up a divergent integral, say which kind it is. Diverges to infinity is more informative, and it matters in the comparison theorem later, which is about areas that grow, never about areas that swing.

16. Example 3.50: from minus infinity to zero

Worked example

Evaluate the integral and state whether it converges or diverges.

\[ \int_{-\infty}^{0}\frac{dx}{x^2+4} \]

\[ \int_{-\infty}^{0}\frac{dx}{x^2+4} = \lim_{t\to-\infty}\int_t^{0}\frac{dx}{x^2+4} \]

\[ \int\frac{dx}{x^2+4} = \frac12\tan^{-1}\frac{x}{2} \]

\[ \frac12\tan^{-1}\frac{x}{2}\Big|_t^{0} = 0 - \frac12\tan^{-1}\frac{t}{2} \]

\[ \lim_{t\to-\infty}\left(-\frac12\tan^{-1}\frac t2\right) = -\frac12\left(-\frac{\pi}{2}\right) = \frac{\pi}{4} \]

\[ \int_{-2000}^{0}\frac{dx}{x^2+4} = \tfrac12\tan^{-1}1000 \approx 0.78490, \quad \frac{\pi}{4} \approx 0.78540 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 334 — Example 3.50

The infinite end is now on the left, so by equation 3.17 the lower limit becomes t and t heads to minus infinity. The rest is familiar.

The antiderivative of one over x squared plus 4 is one half the arctangent of x over 2. The one half matters: if you differentiate the arctangent of x over 2, the chain rule produces a factor of one half, which must be undone. The book's printed solution drops this factor and reports pi over 2; the correct value is pi over 4.

As t goes to minus infinity, the arctangent of t over 2 approaches minus pi over 2, so the limit is one half times pi over 2, which is pi over 4. The numerical check from minus 2000 to 0 lands within a thousandth of pi over 4, and nowhere near pi over 2. That is what a genuine check is for.

17. Checkpoint 3.27: e to the minus x from minus three

Worked example

\[ \int_{-3}^{+\infty} e^{-x}\,dx \]

\[ \int_{-3}^{\infty} e^{-x}\,dx = \lim_{t\to\infty}\int_{-3}^{t} e^{-x}\,dx \]

\[ \int e^{-x}\,dx = -e^{-x} \]

\[ -e^{-x}\Big|_{-3}^{t} = -e^{-t} + e^{3} \]

\[ \lim_{t\to\infty}\left(e^{3} - e^{-t}\right) = e^{3} \approx 20.09 \]

\[ \int_{-3}^{0}e^{-x}\,dx + \int_0^{\infty}e^{-x}\,dx = (e^3 - 1) + 1 = e^3 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 335 — Checkpoint 3.27

Only the right end is infinite; the left end, minus 3, is an ordinary number. So there is one limit, as t goes to infinity.

The antiderivative of e to the minus x is minus e to the minus x. At the lower limit that gives minus e cubed, and subtracting it turns it into plus e cubed. The term e to the minus t vanishes as t grows, leaving e cubed, about 20.09.

The check splits the region at 0 to show where the area lives. The finite piece from minus 3 to 0, where the curve is tall, contributes e cubed minus 1. The infinite tail from 0 onward contributes exactly 1. The picture makes the same point: the infinitely long part of the region is the smallest part.

18. Example 3.49: eight months without an accident

Worked example

Accidents at an intersection used to happen once every three months on average. After the lights were changed, eight months passed with none. How likely is that gap by chance alone? For an average gap of 3 months the waiting time has density one third times e to the minus x over 3.

\[ P(X \ge 8) = \int_8^{\infty}\frac13 e^{-x/3}\,dx \]

\[ P(X \ge 8) = \lim_{t\to\infty}\int_8^{t}\frac13 e^{-x/3}\,dx \]

\[ \int\frac13 e^{-x/3}\,dx = -e^{-x/3} \]

\[ -e^{-x/3}\Big|_8^{t} = e^{-8/3} - e^{-t/3} \]

\[ P(X \ge 8) = e^{-8/3} \approx 0.0695 \]

\[ \int_0^{\infty}\frac13 e^{-x/3}\,dx = 1, \qquad \int_0^{\infty} x\cdot\frac13 e^{-x/3}\,dx = 3 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, pp. 333-334 — Example 3.49

This is the chapter opener. If accidents happen on average once every three months, the waiting time between them is modelled by an exponential density, and the probability of waiting at least eight months is the area under that density from 8 to infinity.

For an average gap of three months the density is one third times e to the minus x over 3. The integral works out to e to the minus eight thirds, about 0.07. So under the old conditions an eight-month quiet spell would happen about 7 percent of the time: unusual, but not so rare that chance can be ruled out.

The book prints the density as 3 times e to the minus 3x and gets about 3.8 times ten to the minus 11. That density has an average gap of one third of a month, not three months. The check shows why the version here is the right one: its total area is 1, as a density must be, and its average waiting time, itself an improper integral, is exactly 3.

19. Both ends infinite

Section

Part 2

20. Split the whole line, and take two separate limits

Concept

When both ends are infinite, there are two limits to take, so the integral is split into two one-sided improper integrals.

\[ \int_{-\infty}^{+\infty} f(x)\,dx = \int_{-\infty}^{0} f(x)\,dx + \int_0^{+\infty} f(x)\,dx \]

The whole-line integral converges only if both pieces converge. If either piece diverges, the whole integral diverges. The split point 0 is a convenience: any number a gives the same answer.

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 331 — definition, equation 3.18

When both ends are infinite, one variable cannot chase both of them honestly, so the definition splits the integral at a convenient point, usually 0, into two integrals with one infinite end each.

The rule for the verdict is strict. The whole-line integral converges only if both pieces converge on their own, and then its value is their sum. If either piece diverges, the whole integral diverges, and you can stop computing the moment you find one divergent piece.

The book adds that the split point does not matter. Moving it from 0 to some other number a shifts a finite piece of area from one side to the other, which cannot change whether either side is finite. You will see on the trap slide why the split itself matters so much.

21. Example 3.51: x times e to the x over the whole line

Worked example

\[ \int_{-\infty}^{+\infty} x e^{x}\,dx \]

\[ \int_{-\infty}^{0} x e^{x}\,dx + \int_0^{\infty} x e^{x}\,dx \]

\[ \int x e^{x}\,dx = x e^{x} - e^{x} \]

\[ (xe^x - e^x)\Big|_t^{0} = -1 - te^{t} + e^{t} \]

\[ \lim_{t\to-\infty} te^{t} = \lim_{t\to-\infty}\frac{t}{e^{-t}} = \lim_{t\to-\infty}\frac{1}{-e^{-t}} = 0 \]

\[ \int_{-\infty}^{0} x e^{x}\,dx = -1 \]

\[ (xe^x - e^x)\Big|_0^{t} = (t - 1)e^{t} + 1 \to +\infty \]

\[ (5 - 1)e^{5} + 1 \approx 594.7 \;\Longrightarrow\; \int_{-\infty}^{\infty} xe^x\,dx \text{ diverges} \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 335 — Example 3.51

Split at 0 first, then deal with the pieces one at a time. Both need the same antiderivative, found by parts with u equal to x: x e to the x minus e to the x.

The left piece is the delicate one. Evaluating from t to 0 leaves a product t e to the t, which as t goes to minus infinity is a zero-times-infinity form. Rewrite it as t over e to the minus t and apply L'Hôpital's rule: it goes to 0. So the left piece converges to minus 1. The picture shows why the value is negative: that region lies below the axis.

The right piece has no such subtlety. The expression t minus 1 times e to the t explodes; at t equal to 5 it is already about 595. One divergent piece is enough, so the whole-line integral diverges, even though half of it was perfectly well behaved.

22. Exercise 359: the whole line under a bell

Worked example

\[ \int_{-\infty}^{\infty}\frac{dx}{x^2+1} \]

\[ \int_{-\infty}^{0}\frac{dx}{x^2+1} + \int_0^{\infty}\frac{dx}{x^2+1} \]

\[ \lim_{t\to\infty}\tan^{-1}x\Big|_0^{t} = \lim_{t\to\infty}\tan^{-1}t = \frac{\pi}{2} \]

\[ \lim_{s\to-\infty}\tan^{-1}x\Big|_s^{0} = 0 - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2} \]

\[ \int_{-\infty}^{\infty}\frac{dx}{x^2+1} = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

\[ 2\tan^{-1}1000 \approx 3.13959, \qquad \pi \approx 3.14159 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 343 — Exercise 359

This time both pieces converge. The integrand is even, so the two pieces are mirror images, each equal to pi over 2, and the total is pi.

Notice that the right piece and the left piece each use their own limit variable, t on the right and s on the left. That is not fussiness; it is what the definition requires, and the next two slides show what goes wrong without it.

The check uses a wide but finite window, from minus 1000 to 1000. The integral there is twice the arctangent of 1000, which is within two thousandths of pi. The missing area, in the two infinite tails beyond 1000, is about two thousandths. Everything agrees.

23. Why one t for both ends is a lie

Picture it

The curve x over x squared plus 1 is odd, so from minus t to t its areas cancel exactly. But each half alone is infinite, and infinity minus infinity has no value.

\[ \int_{-t}^{t}\frac{x\,dx}{x^2+1} = 0, \qquad \int_{-t}^{2t}\frac{x\,dx}{x^2+1} = \frac12\ln\frac{4t^2+1}{t^2+1} \to \ln 2 \]

The curve x over x squared plus 1 is odd: its graph on the left is the upside-down mirror image of its graph on the right. So from minus t to t, the areas above and below the axis cancel exactly, for every t.

But look at each half on its own. The area from 0 to t is one half the logarithm of t squared plus 1, which grows without bound. The left half goes to minus infinity. Cancelling them amounts to claiming that infinity minus infinity is 0.

The second formula shows why that claim is empty. Pair minus t with 2t instead, and the answer settles at ln 2. Pair them differently again and you get another number. An answer that depends on how you choose to pair the ends is no answer at all, which is why the definition insists on two independent limits.

24. Trap: one limit for two infinite ends

Trap

25. An integrand that blows up

Section

Part 3

26. Stop short of the asymptote, then close in

Concept

Same idea, turned on its side. The function is fine on every interval that stays away from the bad point, so integrate there and let the endpoint approach it.

\[ \int_0^{1}\frac{dx}{\sqrt x} = \lim_{t\to0^{+}}\int_t^{1}x^{-1/2}\,dx = \lim_{t\to0^{+}}\left(2 - 2\sqrt t\right) = 2 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 336 — Figure 3.21

Now the second kind of improper integral. The function one over root x is perfectly continuous on any interval from t to 1 with t positive, so integrate there. The answer is 2 minus 2 root t.

The figure shows three cut-off points. Moving the left end from one quarter to four hundredths adds a thin tall sliver, and moving it on to one hundredth adds a thinner one. The area goes from 1 to 1.6 to 1.8, heading for 2, and the limit as t approaches 0 from the right is exactly 2.

So the infinitely tall spike has finite area, for the same reason the long tail under one over x squared did: the extra area near the trouble shrinks fast enough. This is Figure 3.21 of the book, turned into numbers.

27. Reading definitions 3.19 to 3.21

Notation

\[ \int_a^{b} f(x)\,dx = \int_a^{c} f(x)\,dx + \int_c^{b} f(x)\,dx \quad \text{(bad point } c \text{ inside)} \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 336 — definition, equations 3.19 to 3.21

These three definitions are the infinite-discontinuity versions of what you did in Part 1. The trouble is now at a finite point, and t approaches that point from inside the interval.

Pay attention to the one-sided limits. If the bad point is the right end b, t approaches b from the left, written with a small minus sign. If it is the left end a, t approaches a from the right, with a small plus. Approaching from the other side would take you outside the interval, where the integral was never defined.

The third definition is the one that catches people out. If the bad point c is inside the interval, you must split there, and both halves must converge. That rule is the same as the one for the whole line in Part 2, and for the same reason.

28. Example 3.52: a spike at the right end

Worked example

\[ \int_0^{4}\frac{dx}{\sqrt{4-x}} \]

\[ f(x) = (4 - x)^{-1/2} \text{ is continuous on } [0, 4) \]

\[ \int_0^{4}\frac{dx}{\sqrt{4-x}} = \lim_{t\to4^{-}}\int_0^{t}(4-x)^{-1/2}\,dx \]

\[ \int(4-x)^{-1/2}\,dx = -2\sqrt{4-x} \]

\[ -2\sqrt{4-x}\Big|_0^{t} = -2\sqrt{4-t} + 4 \]

\[ \lim_{t\to4^{-}}\left(4 - 2\sqrt{4-t}\right) = 4 \]

\[ \frac{d}{dx}\left(-2(4-x)^{1/2}\right) = -2\cdot\tfrac12(4-x)^{-1/2}\cdot(-1) = (4-x)^{-1/2} \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 337 — Example 3.52

The first job is always to find the bad point. The denominator root of 4 minus x is zero at 4, the right endpoint, so this is equation 3.19 with t approaching 4 from the left.

The antiderivative comes from the substitution u equal to 4 minus x. Because du is minus dx, a minus sign appears in front: minus 2 times root of 4 minus x. Evaluating from 0 to t gives 4 minus 2 root of 4 minus t, and as t approaches 4 the root shrinks to 0, leaving 4.

The check differentiates the antiderivative back. The chain rule produces two minus signs that cancel, and you recover the integrand exactly. The figure shows a spike like the one in the concept slide, but at the right end, with finite area 4.

29. Example 3.53: x ln x down to zero

Worked example

\[ \int_0^{2} x\ln x\,dx \]

\[ f(x) = x\ln x \text{ is continuous on } (0, 2] \]

\[ \int_0^{2} x\ln x\,dx = \lim_{t\to0^{+}}\int_t^{2} x\ln x\,dx \]

\[ \int x\ln x\,dx = \frac12x^2\ln x - \frac14x^2 \]

\[ \left(2\ln2 - 1\right) - \left(\frac12t^2\ln t - \frac14t^2\right) \]

\[ \lim_{t\to0^{+}}\frac{\ln t}{t^{-2}} = \lim_{t\to0^{+}}\frac{1/t}{-2t^{-3}} = \lim_{t\to0^{+}}\left(-\frac{t^2}{2}\right) = 0 \]

\[ \int_0^{2} x\ln x\,dx = 2\ln 2 - 1 \approx 0.386 \]

\[ \frac{d}{dx}\left(\tfrac12x^2\ln x - \tfrac14x^2\right) = x\ln x + \tfrac12x - \tfrac12x = x\ln x \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 337 — Example 3.53

The logarithm has no value at 0, so the integral is improper at its left end, even though, as the figure shows, the curve itself heads calmly for the origin. The hollow dot marks the missing point.

Integration by parts with u equal to ln x gives one half x squared ln x minus one quarter x squared. Evaluating at 2 gives 2 ln 2 minus 1. At the lower end, the term t squared ln t is a zero-times-infinity form. Rewrite it as ln t over t to the minus 2 and apply L'Hôpital: the result simplifies to minus t squared over 2, which goes to 0.

So the integral converges to 2 ln 2 minus 1, about 0.386. The figure shows where that comes from: a negative piece of minus one quarter between 0 and 1, and a positive piece of about 0.636 between 1 and 2. The check differentiates the antiderivative back and confirms the value numerically.

30. Write definition 3.20

Fill the middle

The trouble is at the left end a. Fill in where t sits in the integral, and from which side it approaches a.

\[ \int_a^{b} f(x)\,dx = \lim_{t\to a^{?}}\int_{?}^{b} f(x)\,dx \]

Fill in both blanks before checking. The bad point is the left endpoint a, so it is a that gets replaced by the moving variable t.

That means t sits in the lower limit of the integral, and it approaches a from the right, written a with a small plus sign. From the right means from inside the interval, where the function is continuous and the integral from t to b is an ordinary one.

If you wrote left, picture the number line: approaching a from the left means values smaller than a, which lie outside the interval from a to b. The integral from t to b would then include the bad point itself, which is exactly what the definition is designed to avoid.

31. A bad point can hide inside the interval

Concept

The limits of integration can look perfectly innocent while the integrand blows up somewhere between them. Then the integral must be split at that point, and each half becomes an improper integral with its bad point at an end.

\[ \int_a^{b} f = \lim_{t\to c^{-}}\int_a^{t} f + \lim_{s\to c^{+}}\int_s^{b} f \]

Two separate limit variables, t and s, one for each side. The integral converges only if both one-sided limits exist; they are never tied together.

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 336 — definition, equation 3.21

This is the situation most likely to cost you marks, because nothing in the limits of integration warns you. The integral from minus 1 to 1 of one over x cubed looks like an ordinary definite integral until you notice that the integrand is undefined at 0.

The rule is the same as for the whole line: split at the bad point, and give each half its own limit variable. The left half approaches c from the left, the right half from the right. If either one diverges, the whole integral diverges.

So before you reach for an antiderivative, scan the interval for zeros of denominators, logarithms of zero, and tangents at odd multiples of pi over 2. That scan is step zero of every definite integral from now on.

32. Example 3.54: a bad point inside the interval

Worked example

\[ \int_{-1}^{1}\frac{dx}{x^3} \]

\[ x = 0 \in (-1, 1) \]

\[ \int_{-1}^{1}\frac{dx}{x^3} = \int_{-1}^{0}\frac{dx}{x^3} + \int_0^{1}\frac{dx}{x^3} \]

\[ \int_{-1}^{0}\frac{dx}{x^3} = \lim_{t\to0^{-}}\int_{-1}^{t}x^{-3}\,dx \]

\[ -\frac{1}{2x^2}\Big|_{-1}^{t} = -\frac{1}{2t^2} + \frac12 \]

\[ \lim_{t\to0^{-}}\left(\frac12 - \frac{1}{2t^2}\right) = -\infty \]

\[ t = -0.1: -49.5, \quad t = -0.01: -4999.5 \;\Longrightarrow\; \text{diverges} \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 338 — Example 3.54

The scan finds the bad point at 0, inside the interval from minus 1 to 1, so the integral splits into two improper integrals.

Start with the left piece. The antiderivative is minus one over 2x squared. Evaluated from minus 1 to t it gives one half minus one over 2t squared, and as t approaches 0 from the left, one over t squared blows up, so the piece goes to minus infinity. The book's printed line says plus infinity; the sign is a slip, and the conclusion is unchanged.

The picture shows the tempting error. The two pieces look like mirror images, one minus infinity and one plus infinity, and it is tempting to cancel them to 0. But as on the trap slide in Part 2, infinity minus infinity is not a number. Once one piece diverges, the integral diverges, and you never need to look at the other piece.

33. Checkpoint 3.28: one over x from zero to two

Prediction

\[ \int_0^{2}\frac{dx}{x} \]

\[ \int_t^{2}\frac{dx}{x} = \ln 2 - \ln t \to +\infty \quad\text{as } t \to 0^{+} \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 338 — Checkpoint 3.28

Choose an option before revealing. The bad point is 0, the left end, so the integral is the limit as t approaches 0 from the right of the integral from t to 2.

That integral is ln 2 minus ln t. As t shrinks toward 0, ln t heads to minus infinity, so subtracting it sends the area to plus infinity. The integral diverges.

Compare this with Example 3.47. One over x diverged at infinity because its tail was too fat, and now it diverges near 0 because its spike is too fat. One over x is the borderline case at both ends, and it falls on the divergent side both times. You will see that pattern in general in Part 4.

34. Find the error: integrating straight through an asymptote

Error analysis

Look at the calculation before revealing the notes, and try to find the error yourself. Every algebraic step is correct; the mistake is in using the Fundamental Theorem at all.

The fastest alarm is the sign. One over x squared is positive wherever it is defined, so any area under it must be positive. An answer of minus 2 is impossible. That is a sanity check worth running on every integral: compare the sign of the answer with the sign of the integrand.

The Fundamental Theorem requires a continuous integrand on the whole closed interval, and this one blows up at 0. Split there, and the right piece alone is the limit of one over t minus 1, which is infinite. The careless answer is worse than an obvious error because it looks like a perfectly good number.

35. Before you find an antiderivative

Step zero

\[ \int_{-2}^{2}\frac{dx}{(1+x)^2} \]

Write your answer before revealing. The habit this slide trains is to scan for bad points before doing any calculus at all.

Here the denominator one plus x squared is zero at x equal to minus 1, which sits inside the interval from minus 2 to 2. So the integral must be split at minus 1. The piece from minus 1 to 2 is the limit of one over one plus t, minus one third, as t approaches minus 1 from the right, and one over one plus t blows up. That piece diverges, so the integral diverges.

If you had integrated straight through, you would have got minus four thirds. That is a negative number for a positive integrand, the same warning sign as on the previous slide.

36. Where is each integral improper?

Sorting

For each integral, ask two questions: is either limit infinite, and does the integrand blow up anywhere inside the interval, including its ends? You are sorting by the kind of trouble, not evaluating anything.

Two of the items are traps in the other direction. One over x minus 3 does blow up, but at 3, which is outside the interval from 0 to 2, so that integral is perfectly ordinary. Tangent is continuous from 0 to pi over 4, so that one is ordinary too; only when the interval reaches pi over 2 does it become improper.

The natural logarithm plunges to minus infinity at 0, so an integral of ln x starting at 0 is improper, even though its value turns out to be finite: it is minus 1. Blowing down counts as blowing up.

37. The p-integrals

Section

Part 4

38. The p-integral at infinity

Worked example

Decide, for every real p, whether the integral converges.

\[ \int_1^{\infty}\frac{dx}{x^{p}} \]

\[ \int_1^{t}\frac{dx}{x} = \ln t \to \infty \]

\[ \int_1^{t}x^{-p}\,dx = \frac{t^{1-p} - 1}{1-p} \]

\[ t^{1-p} \to 0 \;\Longrightarrow\; \int_1^{\infty}\frac{dx}{x^{p}} = \frac{1}{p-1} \]

\[ t^{1-p} \to \infty \;\Longrightarrow\; \text{diverges} \]

\[ p = 2: \quad \frac{1}{2 - 1} = 1 = \int_1^{\infty}\frac{dx}{x^2} \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 340 — Example 3.56 and Exercise 370

This slide settles a whole family at once. The case p equal to 1 is Example 3.47 and gives a logarithm, which diverges. For every other p, the power rule gives t to the power one minus p, minus 1, all over one minus p.

Everything depends on the sign of the exponent one minus p. If p is bigger than 1, the exponent is negative, so t to that power shrinks to 0 as t grows, and the integral converges to one over p minus 1. If p is less than 1, the exponent is positive, the power grows without bound, and the integral diverges.

The check puts p equal to 2 into the formula and gets 1, the same integral that gave the horn its volume. The book reaches the p less than 1 half of this result by comparison in Example 3.56, which you will see in Part 5.

39. The p-integral near zero

Worked example

Now the bad point is x equal to 0, the left end. Same function, other end.

\[ \int_0^{1}\frac{dx}{x^{p}}, \quad p > 0 \]

\[ \int_t^{1}\frac{dx}{x} = -\ln t \to \infty \]

\[ \int_t^{1}x^{-p}\,dx = \frac{1 - t^{1-p}}{1-p} \]

\[ t^{1-p} \to 0 \;\Longrightarrow\; \int_0^{1}\frac{dx}{x^{p}} = \frac{1}{1-p} \]

\[ t^{1-p} \to \infty \;\Longrightarrow\; \text{diverges} \]

\[ p = \tfrac12: \quad \frac{1}{1 - 1/2} = 2 = \int_0^{1}\frac{dx}{\sqrt x} \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 343 — Exercises 365 and 375

Same function, same power rule, but now the bad point is at 0 and t approaches it from the right. The antiderivative evaluated from t to 1 is 1 minus t to the power one minus p, all over one minus p.

The sign of one minus p still decides everything, but t is now shrinking instead of growing, so the conclusions swap. If p is less than 1, a positive power of a small number is small, t to that power goes to 0, and the integral converges to one over one minus p. If p is bigger than 1, a negative power of a small number is huge, and the integral diverges.

The case p equal to 1 is Checkpoint 3.28 again: a logarithm, divergent. The check confirms the square-root spike from the start of Part 3, whose area 2 matches the formula at p equal to one half.

40. Same curves, opposite verdicts

Picture it

\[ \int_1^{\infty}\frac{dx}{x^{p}} \text{ converges} \iff p > 1, \qquad \int_0^{1}\frac{dx}{x^{p}} \text{ converges} \iff p < 1 \]

The two panels show the same two curves in two different places. On the left, near 0, one over root x has finite area under it while one over x squared shoots up so steeply that its area is infinite. On the right, far out, the roles reverse: one over x squared thins out fast enough to have area 1, while one over root x lingers and has infinite area.

Read the two rules underneath together. At infinity you need p bigger than 1; near 0 you need p less than 1. The boundary case p equal to 1 fails on both sides.

When you meet a p-integral, the very first thing to identify is which end is bad. The rule you apply depends on it completely.

41. Slide p and push the left end toward zero

Tweak it

The curve plots the area from x to 1 against the left end x, so as you read the graph from right to left you are pushing the left end toward the asymptote. Start with p at one half: the curve rises gently and levels off at 2 as x approaches 0.

Now move p up toward 1. The level the curve heads for, one over one minus p, climbs higher and higher, and at p equal to 1 the curve becomes minus ln x, which rises without limit. Push p past 1 and the curve shoots up even faster.

Compare this with the tail rule two slides back, where large p was the convergent side. Here small p is. There is no gradual transition: every p below 1 gives a finite area, and p equal to 1 does not.

42. Why the two rules point opposite ways

Intuition

At infinity the question is how fast the curve dies away: a bigger p makes the tail thinner, so large p converges.

Near zero the question is how fast the curve blows up: a bigger p makes the spike taller, so small p converges.

\[ x > 1: \; x^{-2} < x^{-1/2}, \qquad 0 < x < 1: \; x^{-2} > x^{-1/2} \]

The one power that fails at both ends is p equal to 1, and so the integral of one over x to the p from 0 to infinity diverges for every p.

Here is the idea in words, without the formulas. At infinity the curve is small and getting smaller, and the question is whether it gets small fast enough. A bigger p makes it smaller faster, so large p is good.

Near 0 the curve is large and getting larger, and the question is whether it stays small enough that the spike has finite area. A bigger p makes it larger faster, so large p is bad. The inequality on the slide says exactly this: beyond 1, x to the minus 2 is the smaller power; between 0 and 1, it is the larger.

One consequence is worth remembering. The integral from 0 to infinity of one over x to the p must be split at some point like 1, and one of the two pieces always diverges, whatever p is. No power of x has finite area over the whole half-line.

43. Trap: using the wrong p-rule

Trap

44. The p-integrals as a measuring stick

Concept

Knowing exactly where the p-integrals switch from convergent to divergent turns them into reference integrals. A messy integrand is judged by the power of x it behaves like at its bad end.

\[ \frac{1}{x^3 + 1} \approx \frac{1}{x^3} \text{ for large } x \;\Longrightarrow\; \text{tail behaves like } p = 3 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 343 — Exercise 353

The p-integrals matter less for their own sake than as benchmarks. Once you know exactly where they switch from convergent to divergent at each end, you can judge an unfamiliar integrand by asking which power of x it resembles near its bad point.

For example, one over x cubed plus 1 is almost exactly one over x cubed when x is large, because the plus 1 is negligible. So its tail should behave like a p-integral with p equal to 3, which converges. Exercise 353 asks you to confirm this with the comparison theorem, and Part 5 shows how.

Keep the table on this slide as your reference list. The comparison theorem needs something to compare with, and nine times out of ten that something is a p-integral or an exponential.

45. Complete the p-integral table

Comparison

Fill in each blank before checking. For every row, identify the bad end first, then compare p with 1 using the rule for that end, then compute the value if it converges.

The cube-root row is a reminder that roots are fractional powers: one over the cube root of x is x to the minus one third, so p is one third, below 1, and near 0 that converges, to one over one minus one third, which is three halves.

The row with p equal to e converges at infinity because e is about 2.718, comfortably above 1, and its value is one over e minus 1. The last row is the one to remember: over the whole half-line, one of the two pieces always diverges, so no p works.

46. The comparison theorem

Section

Part 5

47. Why a finite roof gives a finite area

Concept

Suppose f is never negative and never above g. Then the area to t under f is squeezed below the area to t under g, for every t.

\[ 0 \le f(x) \le g(x) \;\Longrightarrow\; 0 \le \int_a^{t} f(x)\,dx \le \int_a^{t} g(x)\,dx \]

The area under f only grows as t grows, because f is not negative. A growing quantity trapped below a finite ceiling must settle, so it has a limit.

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, pp. 338-339 — Figure 3.22 and the argument before Theorem 3.7

Many integrals cannot be evaluated at all, because their antiderivatives are not elementary. The comparison theorem decides convergence without evaluating anything.

The left panel shows the idea. The function g, the orange roof, has finite area. The function f sits between g and the axis, so its area to any t is less than g's area to that t, and therefore less than g's total. That area also only increases as t grows, since f is not negative. A quantity that only increases, and never passes a fixed ceiling, has to level off. So f's integral converges.

The right panel is the other useful direction. If the lower function already has infinite area, anything above it has at least as much, so it is infinite too. Those are the only two conclusions the theorem draws.

48. Reading Theorem 3.7

Notation

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 339 — Theorem 3.7

Step through the annotations. The hypothesis that f is never negative is easy to forget and essential: without it, the area to t need not increase, and a bounded quantity can oscillate forever without converging, as the integral of sine did.

Part (ii) says convergence passes down from a larger function to a smaller one. Part (i) says divergence passes up from a smaller function to a larger one. The table lists all four possibilities, and two of them say nothing at all.

Those two dead ends are where most comparison mistakes happen. Knowing that f is smaller than something infinite tells you nothing, and knowing that f is bigger than something finite tells you nothing. Before you write a comparison, check that yours is one of the two useful rows.

49. Example 3.55: a comparison that settles it

Worked example

Use a comparison to show that the integral converges.

\[ \int_1^{\infty}\frac{dx}{xe^{x}} \]

\[ x \ge 1: \quad 0 \le \frac{1}{xe^{x}} \le \frac{1}{e^{x}} = e^{-x} \]

\[ \int_1^{\infty}e^{-x}\,dx = \lim_{t\to\infty}\left(-e^{-x}\right)\Big|_1^{t} \]

\[ \lim_{t\to\infty}\left(e^{-1} - e^{-t}\right) = e^{-1} \approx 0.368 \]

\[ \int_1^{\infty}\frac{dx}{xe^{x}} \text{ converges, value } \le e^{-1} \]

\[ \int_1^{60}\frac{dx}{xe^{x}} \approx 0.2194 < 0.3679 \approx e^{-1} \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 340 — Example 3.55

The integrand one over x e to the x has no elementary antiderivative, so evaluating is off the table. But for x at least 1, dividing by x can only shrink e to the minus x, so the integrand sits under e to the minus x.

The roof e to the minus x has a finite integral from 1 to infinity, and it is e to the minus 1, about 0.368. The printed solution writes e to the plus 1 at this step; that is a typo, since e to the minus x is below 1 on the whole interval and cannot enclose area 2.7 there. The conclusion does not depend on it.

By part (ii) the original integral converges, to a value no bigger than 0.368. The check computes it numerically, 0.219, comfortably inside the bound. The theorem gives you convergence and an upper bound; it never gives you the value.

50. Example 3.56: every p below one diverges

Worked example

Use the comparison theorem to show the integral diverges for all p less than 1.

\[ \int_1^{\infty}\frac{dx}{x^{p}}, \quad p < 1 \]

\[ p < 1,\; x \ge 1 \;\Longrightarrow\; x^{p} \le x \]

\[ 0 \le \frac{1}{x} \le \frac{1}{x^{p}} \]

\[ \int_1^{\infty}\frac{dx}{x} = +\infty \]

\[ \int_1^{\infty}\frac{dx}{x^{p}} = +\infty \quad \text{for all } p < 1 \]

\[ \int_1^{100}x^{-1/2}\,dx = 2\sqrt{100} - 2 = 18 > 4.61 \approx \ln 100 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 340 — Example 3.56

This example proves half of the p rule at infinity with the comparison theorem instead of the power rule. The key inequality is that for x at least 1 and p less than 1, x to the p is at most x. Taking reciprocals of positive numbers flips the inequality, so one over x is at most one over x to the p.

Now part (i) applies in its useful direction. The smaller function, one over x, already has infinite area by Example 3.47, so the larger one, one over x to the p, must too.

The figure shows why: every curve with p below 1 runs above one over x, which already encloses infinite area. The check takes p equal to one half and compares directly: out to 100 its area is 18, far above the logarithm's 4.6.

51. Checkpoint 3.29: ln x over x

Worked example

Use a comparison to show the integral diverges.

\[ \int_e^{\infty}\frac{\ln x}{x}\,dx \]

\[ x \ge e: \quad \ln x \ge 1 \;\Longrightarrow\; \frac{\ln x}{x} \ge \frac{1}{x} \ge 0 \]

\[ \int_e^{\infty}\frac{dx}{x} = \lim_{t\to\infty}(\ln t - 1) = \infty \]

\[ \int_e^{\infty}\frac{\ln x}{x}\,dx = \infty \]

\[ \int_e^{t}\frac{\ln x}{x}\,dx = \frac{(\ln t)^2 - 1}{2} \to \infty, \quad t = 100: \; 10.10 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 340 — Checkpoint 3.29

Start at e for a reason: from e onward the natural logarithm is at least 1, so ln x over x is at least one over x. The lower function diverges, so by part (i) the upper one does too.

Notice which direction was needed. You expected divergence, because ln x over x is bigger than one over x for large x, so you needed a smaller divergent function to push it up. If you had tried to find a larger function, no conclusion could have followed.

This integral can also be done directly with u equal to ln x, which makes a nice check. The area from e to t is half of ln t squared minus one half, which grows without bound. At t equal to 100 it is already about 10.1.

52. An integral with no antiderivative: e to the minus x squared

Worked example

The bell curve has no elementary antiderivative, so evaluating is off the table. Decide convergence anyway.

\[ \int_1^{\infty} e^{-x^2}\,dx \]

\[ x \ge 1 \;\Longrightarrow\; x^2 \ge x \;\Longrightarrow\; -x^2 \le -x \]

\[ 0 \le e^{-x^2} \le e^{-x} \]

\[ \int_1^{\infty}e^{-x}\,dx = e^{-1} \]

\[ \int_1^{\infty}e^{-x^2}\,dx \le e^{-1} \approx 0.368 \]

\[ \int_1^{12}e^{-x^2}\,dx \approx 0.1394 < 0.368 \]

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 344 — exercise set, comparison problems 372-373

The bell curve e to the minus x squared is the most important function in statistics, and it has no elementary antiderivative, so no amount of technique from this chapter will evaluate this integral. Comparison still decides it.

For x at least 1, x squared is at least x, so minus x squared is at most minus x, and since the exponential function is increasing, e to the minus x squared is at most e to the minus x. The figure shows why the comparison starts at 1: between 0 and 1 the bell curve is actually on top.

The roof is Example 3.55's, with finite area one over e, so the bell curve's tail converges and is at most 0.368. The numerical check gives 0.139. In statistics this tail area is exactly what a normal-distribution table records, and it is computed numerically for this very reason.

53. Choosing the function to compare with

Concept

Decide first which verdict you expect, by keeping only the dominant part of the integrand at its bad end. Then pick the comparison function in the one direction that proves it.

\[ \frac{1}{\sqrt x + 1} \approx \frac{1}{\sqrt x} \quad\Longrightarrow\quad \text{expect divergence: find a SMALLER divergent } g \]

\[ x \ge 1: \; \sqrt x + 1 \le 2\sqrt x \;\Longrightarrow\; \frac{1}{\sqrt x + 1} \ge \frac{1}{2\sqrt x} \]

The integral of one over 2 root x from 1 on diverges, and the given integrand sits above it, so the given integral diverges too.

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 343 — Exercise 373

The hard part of a comparison is choosing what to compare with, and the trick is to guess the answer first. Near its bad end, keep only the dominant part of the integrand. For one over root x plus 1, the root dominates for large x, so the integrand behaves like one over root x, whose tail diverges.

Now you know you need a smaller divergent function, because divergence only passes upward. One over root x itself is larger, so it is useless. But for x at least 1, root x plus 1 is at most twice root x, so one over root x plus 1 is at least one over 2 root x, which is a constant multiple of a divergent p-integral.

That is Exercise 373: guess with the dominant term, then adjust by a constant so the inequality points in the direction you need.

54. Does this comparison settle it?

Sorting

For each line, decide whether the comparison is in one of the two useful directions: smaller than something convergent, or bigger than something divergent.

Two of the lines are in dead-end directions. One over x squared plus 4x is smaller than one over x, but one over x diverges, so nothing follows. And e to the minus x is bigger than e to the minus x over x squared, which converges, so again nothing follows. Both of these integrals do converge, but you would need a different comparison to prove it, for example comparing the first with one over x squared, as Exercise 372 suggests.

The line with ln x over x squared shows that the inequality only has to hold eventually, for large x. The logarithm grows more slowly than any positive power of x, so the comparison with one over x to the 1.5 holds from some point on, and that is enough.

55. Break the useless direction

Counterexample

Write down a function before revealing the answer. You want something that sits below one over x on the interval from 1 to infinity but has a convergent integral.

One over x squared is the natural choice. It is smaller than one over x for every x above 1, and its integral from 1 to infinity is 1. So being smaller than a divergent integral cannot force divergence.

The deeper point is that a comparison in a useless direction is not a weak argument; it is no argument. Both outcomes are possible, and examples of each exist: one over 2x is also smaller than one over x, and it diverges. That is why Theorem 3.7 has exactly two parts.

56. Match each integral to its value

Matching

Match each integral before checking, and for each one, name the bad point and the tool before you commit to a value.

Two are p-integrals on opposite ends: one over x squared at infinity gives 1, and one over root x near 0 gives 2. The whole-line arctangent integral splits at 0 into two halves of pi over 2 each. The exponential e to the minus 2x has antiderivative minus one half e to the minus 2x, so from 0 to infinity it gives one half.

The last one is the harmonic integral from Example 3.47, which diverges. If you tried to give it a value, look back at the logarithm's graph: it has no ceiling.

57. Probability densities are improper integrals

Real world

OpenStax Calculus Volume 2, §3.7 Improper Integrals §3.7, p. 345 — Exercises 406 and 407

A probability density describes how likely each value of a quantity is, and probabilities are areas under it. Because the total probability must be 1, every density must enclose area exactly 1, which for a quantity like a waiting time is an integral out to infinity.

For the density 7 e to the minus 7x, the total area from 0 to t is 1 minus e to the minus 7t, and the limit is 1. So it is a genuine density. The probability that the wait is between 0 and 0.3 is an ordinary integral, 1 minus e to the minus 2.1, about 0.8775.

The picture shows where the probability sits: most of it in the first third of a unit, with a thin tail of about 12 percent beyond. The same shape, with a different rate, is the accident model of Example 3.49.

58. Putting it together

Section

Part 6

59. Pattern: evaluating an improper integral

Pattern

1. Find every bad point: an infinite end, or a point in the interval where the integrand blows up.
2. Split so that each piece has exactly one bad point, at one of its ends.
3. Replace each bad point by t, integrate as usual, and take the one-sided limit.
4. Every piece finite: add them. Any piece infinite or oscillating: the whole integral diverges.
5. No antiderivative? Compare with a p-integral or an exponential, in a direction that decides it.

This is the procedure for every improper integral in the section, and the pulsing box marks the step people skip. Before any calculus, find all the bad points: infinite ends, and points in the interval where the integrand blows up.

Then split so that each piece has exactly one bad point at one end. Each piece gets its own variable and its own one-sided limit. If every piece converges, the integral is their sum. If any piece diverges, the whole integral diverges, and you can stop.

If you cannot find an antiderivative, or only need a verdict, compare with a p-integral or an exponential. Guess the verdict from the dominant term first, then choose the direction that proves it.

60. Put the steps in order

Ranking

Drag the steps into order before checking. This integral has two bad points, the square root is zero at x equal to 0, and the upper limit is infinite, so it must be split.

Splitting at 1 is convenient but any positive number would do. Each piece then gets its own one-sided limit, near 0 from the right and at infinity. Only when both are finite do you add them. The exit ticket at the end of the lesson finishes this very integral.

The most common wrong order puts evaluating before splitting, with a single variable running from near 0 all the way out to infinity. That is the one-limit-for-two-ends mistake from Part 2 in disguise, and it can hide a divergent piece.

61. Check: which one converges?

Check

For each option, name the bad end first, then apply the p rule for that end. That order is the whole skill being tested.

One over the cube root of x has p equal to one third, and its bad end is 0, where p less than 1 converges. The integral is three halves. One over x squared on the interval from 0 to 1 fails because near 0 it is too steep, and one over root x from 1 to infinity fails because at infinity it is too gentle. The last option has two bad ends, and one of them always fails.

62. Check: what a comparison allows

Check

The key word is smaller. Knowing that f is smaller than something whose integral diverges gives you a ceiling at infinity, which is no ceiling at all.

One over x squared and one over 2x are both smaller than one over x on this interval. The first converges and the second diverges, so the information given cannot decide the question. The two useful directions are smaller than a convergent integral, or larger than a divergent one.

If you chose the last option, notice where the number 1 came from: it is the value of a different integral, the one for one over x squared. A comparison bound only exists when the larger function has a finite integral, and here it does not.

63. Check: an asymptote inside the interval

Check

The careless calculation applies the Fundamental Theorem across x equal to 0, where one over x squared blows up. The sign of the answer is the giveaway: a positive integrand cannot have a negative integral.

Split at 0. The piece from 0 to 2 is the limit of one over t minus one half, and one over t blows up as t approaches 0 from the right. One divergent piece means the integral diverges. Fixing the sign, or looking for cancellation, cannot rescue it.

This is the same mistake as the find-the-error slide in Part 3, with a lopsided interval. The lesson carries over to every definite integral you do from now on: before applying the Fundamental Theorem, check that the integrand is continuous on the whole closed interval.

64. Explain the opposite boundaries

Explain it to yourself

Write your explanation before revealing. The goal is to explain the mechanism, not to restate the two rules.

A good answer names the two different jobs the exponent does. At infinity the curve is already small, and the question is whether it becomes small fast enough to make the infinitely long tail thin. Near 0 the curve is already huge, and the question is whether it stays small enough that the infinitely tall spike is thin. A larger p helps the first and hurts the second, because the same power that shrinks small numbers faster also inflates big numbers faster.

If you can say that clearly, you will never need to memorise which way each rule points.

65. Exit ticket

Exit ticket

\[ \int_0^{\infty}\frac{dx}{\sqrt{x}\,(1+x)} \]

This one integral uses almost everything in the lesson. It has two bad points, so split at 1. Near 0 the factor one plus x is about 1, so the integrand is at most one over root x, a p-integral with p one half, which converges near 0. At infinity the integrand is at most one over x to the three halves, with p three halves, which converges at infinity.

So both pieces converge. To find the value, substitute u equal to root x, so dx is 2u du and the root in the denominator cancels one u. The integral becomes 2 times the integral of one over one plus u squared from 0 to infinity, which is 2 times pi over 2, or pi.

A numerical check with Simpson's rule gives 3.14157, matching pi to four decimal places.

66. Recap

Recap

Chapter 5 uses exactly these integrals: the integral test decides whether a series converges by deciding whether an improper integral does.

Stewart, Calculus: Early Transcendentals 8e, §7.8 Improper Integrals §7.8, pp. 527-537 — the same material in Stewart

Every improper integral is an ordinary integral followed by a limit. Find the bad points, split so each piece has one, replace each bad point by a variable, integrate, and take each limit on its own. The integral converges only if every piece does.

The p-integrals are the benchmarks, and their rules point opposite ways: at infinity p must exceed 1, near 0 p must be less than 1, and p equal to 1 fails at both ends. The comparison theorem transfers a verdict from a benchmark to a harder integrand, but only in two directions: convergence downward from a finite roof, divergence upward from an infinite floor.

These ideas return in Chapter 5, where the integral test decides whether an infinite series converges by deciding whether an improper integral converges.