Improper integrals
An oil well pumps a little less every year but never quite stops. Can the oil it will ever pump be a finite amount, and how do you add up a rate over a time that never ends?
Where you are 16 in Integrals
Why this exists
A definite integral \int_a^b f(x)\,dx is built on a closed, bounded interval: cut [a, b] into finitely many pieces of finite width, add, take a limit. Many honest questions run past any bound. An oil well whose output declines but never stops: how much will it pump in all? A payment that arrives every year forever: what is it worth today? The area under \displaystyle y = \frac{1}{x^2} to the right of x = 1: is it a number at all?
None of these is a definite integral, because the interval [a, \infty) cannot be cut into finitely many pieces of finite width. What can be done is to integrate out to a finite point b, which is an ordinary integral, and then ask what happens as b moves off to infinity. That is a limit at infinity, and the whole idea of this lesson is to put the two together.
Definition
Day 1: integrals over an unbounded interval
Let f be continuous on [a, \infty). The improper integral of f from a to infinity is
If the limit exists as a finite number, the integral converges and its value is that number. Otherwise it diverges: the limit may be \infty or -\infty (which the notation records but which is not a number), or there may be no limit at all, as with \int_0^b \cos x\,dx = \sin b, which keeps swinging between -1 and 1.
For f continuous on (-\infty, b], the left tail is defined the same way:
For f continuous on the whole line, choose any number c and split:
The integral over the whole line converges only when both tails converge, each as its own limit, and then its value is the sum. If either tail diverges, the whole integral diverges. Which c is used does not matter (see Why it is true).
Day 2: comparing integrands
Often the antiderivative is out of reach, but the question “converges or diverges?” can still be settled by comparing f with a simpler function g whose integral is known. The tool is the comparison test in Results. It applies to integrands that are nonnegative, and it only ever needs the inequality from some point on.
Intuition
Day 1: an infinitely long region with finite area
Take the well of Example 1, pumping r(t) = 1{,}200e^{-0.15t} barrels per year t years from now. The oil pumped in the first b years is the area under the rate curve from 0 to b (Figure 1). That area is a function of b. It grows as b grows, since the curve is above the axis, but the rate falls off so fast that the added area shrinks year by year, and the total levels off: it has a horizontal asymptote, and the height of that asymptote is the total output forever.
So an improper integral is a running total watched from far away. It converges when the running total settles toward a number and diverges when it keeps climbing (or keeps swinging). Knowing that f(x) \to 0 is not enough to decide: Figure 2 shows \displaystyle \frac{1}{x} and \displaystyle \frac{1}{x^2}, both falling to 0, one with finite area to the right of x = 1 and one without. The difference is in how fast they fall.
Day 2: a region inside a region
If 0 \le f(x) \le g(x), the region under f sits inside the region under g (Figure 3). If the bigger region has finite area, so does the smaller. If the smaller one has infinite area, so does the bigger. Neither fact needs an antiderivative of f. What it needs is a known benchmark, and the p-integrals and exponentials of Day 1 supply them.
The setting for Day 2 is money. When money earns interest at 7\% a year compounded continuously, a dollar invested today grows to e^{0.07t} dollars in t years, so a dollar paid t years from now is worth e^{-0.07t} dollars today. A stream of payments at c(t) dollars per year, running forever, has present value
each thin slice of payment c(t)\,dt discounted back to today and the slices added.
Results
Exponential tails. For k > 0 and any a,
In particular \displaystyle \int_0^{\infty} e^{-kx}\,dx = \frac{1}{k}.
The p-integrals.
The same split holds with any lower limit a > 0 in place of 1.
Direct comparison test. Let f and g be continuous on [a, \infty) with 0 \le f(x) \le g(x) for all x \ge a.
- If \int_a^{\infty} g(x)\,dx converges, then \int_a^{\infty} f(x)\,dx converges, and \int_a^{\infty} f(x)\,dx \le \int_a^{\infty} g(x)\,dx.
- If \int_a^{\infty} f(x)\,dx diverges, then \int_a^{\infty} g(x)\,dx diverges.
Limit comparison test. Let f and g be continuous and positive on [a, \infty), and suppose \lim_{x \to \infty} \dfrac{f(x)}{g(x)} = L with 0 < L < \infty. Then \int_a^{\infty} f(x)\,dx and \int_a^{\infty} g(x)\,dx both converge or both diverge.
Why it is true
Exponential tails. \int_a^b e^{-kx}\,dx = \dfrac{e^{-ka} - e^{-kb}}{k}, and e^{-kb} \to 0 as b \to \infty because k > 0.
The p-integrals. For p \ne 1 the power rule gives
If p > 1, the exponent 1 - p is negative, so \displaystyle b^{1-p} = \frac{1}{b^{p-1}} \to 0 and the integral tends to \displaystyle \frac{-1}{1 - p} = \frac{1}{p - 1}. If p < 1, the exponent is positive, b^{1-p} \to \infty, and the integral diverges. If p = 1, \displaystyle \int_1^b \frac{dx}{x} = \ln b \to \infty, so it diverges too. Moving the lower limit to another a > 0 adds or removes the finite amount \int_1^a x^{-p}\,dx, which cannot change whether the limit exists.
The split point does not matter. Suppose the two tails converge when split at c, and let d be another number. By additivity, \int_a^d f = \int_a^c f + \int_c^d f for every a < \min(c, d), so letting a \to -\infty, the left tail at d converges and equals the left tail at c plus the fixed number \int_c^d f. In the same way the right tail at d equals the right tail at c minus \int_c^d f. The two corrections cancel in the sum.
Direct comparison. Let F(b) = \int_a^b f(x)\,dx. Because f \ge 0, F never decreases as b grows. Because f \le g, F(b) \le \int_a^b g(x)\,dx \le \int_a^{\infty} g(x)\,dx, the last step since g \ge 0 too. So F is nondecreasing and bounded above. A nondecreasing function that is bounded above has a finite limit as b \to \infty, namely its least upper bound; this is the completeness of the real numbers, taken here without proof. So \int_a^{\infty} f converges, and the bound passes to the limit. The second statement is the first one read backward: if \int_a^{\infty} g converged, \int_a^{\infty} f would too.
Limit comparison. Since \displaystyle \frac{f(x)}{g(x)} \to L > 0, there is a number X \ge a beyond which \displaystyle \frac{L}{2} < \frac{f(x)}{g(x)} < 2L, that is, \displaystyle \frac{L}{2}\,g(x) < f(x) < 2L\,g(x). If \int g converges, so does \int 2L\,g, and the right inequality with direct comparison makes \int_X^{\infty} f converge. If \int g diverges, so does \displaystyle \int \frac{L}{2}\,g, and the left inequality makes \int_X^{\infty} f diverge. The piece from a to X is an ordinary finite integral and changes nothing.
Worked examples
Day 1: Ximun’s well, pumped forever
Ximun models a well whose output declines as r(t) = 1{,}200e^{-0.15t} barrels per year, t years from now. Find the total oil the well will ever pump, and the time by which it has pumped 90\% of that total.
SolutionThe oil pumped in the first b years is an ordinary integral, by the The fundamental theorem of calculus:
As b \to \infty, e^{-0.15b} \to 0, so
The integral converges: a well pumping forever yields a finite total. The oil still to come after year b is 8000e^{-0.15b}, which is the 10\% left over when e^{-0.15b} = 0.1, that is, \displaystyle b = \frac{\ln 10}{0.15} \approx 15.35 years. Check: the first year alone gives 8000(1 - e^{-0.15}) \approx 1114 barrels, a little less than the starting rate of 1{,}200 barrels per year, as it should be for a falling rate.
Day 1: Ayumu tests three powers
Ayumu decides each integral: (a) \int_1^{\infty} \dfrac{dx}{x^{1.01}}, (b) \int_1^{\infty} \dfrac{dx}{x^{0.99}}, (c) \int_4^{\infty} x^{-3/2}\,dx.
Solution(a) Here p = 1.01 > 1, so the integral converges to \displaystyle \frac{1}{p - 1} = \frac{1}{0.01} = 100.
(b) Here p = 0.99 < 1, so it diverges. The partial integrals are \displaystyle \int_1^b x^{-0.99}\,dx = \frac{b^{0.01} - 1}{0.01}, which grow without bound, but very slowly: at b = 10^{100} the value is only \displaystyle \frac{10 - 1}{0.01} = 900. No table of partial integrals computed on a machine would reveal the divergence; the limit does.
(c) The lower limit is 4, not 1, so compute directly. With \displaystyle p = \frac{3}{2} the antiderivative is -2x^{-1/2}:
It converges to 1. Starting at 4 rather than 1 removed the finite piece \int_1^4 x^{-3/2}\,dx = 1 from the total \displaystyle \frac{1}{p - 1} = 2.
Day 1: both tails of a bell-shaped curve
Evaluate \int_{-\infty}^{\infty} \dfrac{dx}{x^2 + 4}.
SolutionThe integrand is continuous on the whole line, so split at c = 0 and treat each tail as its own limit. With \displaystyle \int \frac{dx}{x^2 + 4} = \frac{1}{2}\arctan\frac{x}{2} + C:
because \displaystyle \arctan u \to \frac{\pi}{2} as u \to \infty. On the left, \displaystyle \int_a^0 \frac{dx}{x^2 + 4} = -\frac{1}{2}\arctan\frac{a}{2} \to \frac{\pi}{4} as a \to -\infty. Both tails converge, so
Splitting at c = 2 instead gives \displaystyle \frac{1}{2}\left(\frac{\pi}{4} + \frac{\pi}{2}\right) = \frac{3\pi}{8} on the left and \displaystyle \frac{1}{2}\left(\frac{\pi}{2} - \frac{\pi}{4}\right) = \frac{\pi}{8} on the right: the same total, as the theory says.
Day 1: two logarithms, one convergent
Decide \int_2^{\infty} \dfrac{dx}{x(\ln x)^2} and \int_2^{\infty} \dfrac{dx}{x\ln x}.
SolutionSubstitute u = \ln x, \displaystyle du = \frac{dx}{x}; as x runs from 2 to b, u runs from \ln 2 to \ln b (u-substitution). The first becomes
since \ln b \to \infty. It converges. The second becomes
It diverges. Both integrands fall to 0 faster than \displaystyle \frac{1}{x}, yet only the first converges: after the substitution they are the p-integrals with p = 2 and p = 1. The divergence of the second is extremely slow; \ln(\ln b) passes 5 only when b > e^{e^5} \approx 10^{64}.
Day 2: Kotone traps an integral she cannot evaluate
Kotone wants \displaystyle \int_1^{\infty} \dfrac{dx}{x^3 + \sqrt{x}}. Show it converges and find bounds for its value.
SolutionAn antiderivative is not practical, so compare. For x \ge 1, \displaystyle \sqrt{x} > 0, so \displaystyle x^3 + \sqrt{x} \ge x^3 and
The benchmark \displaystyle \int_1^{\infty} \frac{dx}{x^3} is a p-integral with p = 3, convergent with value \displaystyle \frac{1}{2}. By direct comparison Kotone’s integral converges, and it is at most \displaystyle \frac{1}{2} (Figure 3).
For a lower bound, note \displaystyle \sqrt{x} \le x^3 when x \ge 1, so \displaystyle x^3 + \sqrt{x} \le 2x^3 and \displaystyle \frac{1}{x^3 + \sqrt{x}} \ge \frac{1}{2x^3}. Integrating, the value is at least \displaystyle \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}. So
and numerical integration gives about 0.361, inside the bounds.
Day 2: an oscillating numerator, and a ratio that settles
Decide (a) \int_1^{\infty} \dfrac{2 + \cos x}{x}\,dx and (b) \int_3^{\infty} \dfrac{dx}{x^2 - 4}.
Solution(a) Since \cos x \ge -1, the numerator is at least 1, so for x \ge 1
The smaller integral \displaystyle \int_1^{\infty} \frac{dx}{x} diverges (p = 1), so the larger one diverges. The upper bound \displaystyle \frac{2 + \cos x}{x} \le \frac{3}{x} is also true but useless here: being below a divergent integral says nothing. A comparison decides the question only when it points the right way.
(b) For large x the integrand behaves like \displaystyle \frac{1}{x^2}, but \displaystyle \frac{1}{x^2 - 4} > \frac{1}{x^2}, so direct comparison with \displaystyle \frac{1}{x^2} points the wrong way. Limit comparison does not care about direction:
which is positive and finite, and \displaystyle \int_3^{\infty} \frac{dx}{x^2} converges, so the integral converges. Here partial fractions also give the value: \displaystyle \frac{1}{x^2 - 4} = \frac{1}{4}\left(\frac{1}{x - 2} - \frac{1}{x + 2}\right), so
since \displaystyle \frac{b - 2}{b + 2} \to 1 and \ln 1 = 0.
Day 2: Ximun prices a payment that grows forever
A contract pays 6000 + 300t dollars per year, continuously, starting now and never ending. With money earning 7\% a year compounded continuously, find its present value.
SolutionThe present value is \int_0^{\infty} (6000 + 300t)e^{-0.07t}\,dt. The payments grow without bound, so convergence is not obvious; the exponential discount must beat the linear growth. Split the integrand. The constant part is an exponential tail:
For the growing part, integrate by parts with u = t, dv = e^{-kt}\,dt, k = 0.07:
As b \to \infty, be^{-kb} \to 0 because the exponential outgrows any power (Comparing growth rates at infinity), so \displaystyle \int_0^{\infty} te^{-kt}\,dt = \frac{1}{k^2}. Then
Both pieces converge, so the sum does: the present value is about 146{,}938.78 dollars. A payment stream that rises forever is worth a finite amount today, because each later dollar is discounted more than the stream grows.
Boundaries
- Continuity on the whole interval. The definition needs f continuous on [a, \infty), so that every \int_a^b f is an ordinary integral. If f blows up at a finite point, as \displaystyle \frac{1}{x^2} does at 0, the integral is improper for a second reason, handled in Improper integrals with an infinite integrand; \displaystyle \int_0^{\infty} \frac{dx}{x^2} must be split at a point and both parts treated.
- Each tail is its own limit. \int_{-\infty}^{\infty} f is not \lim_{R \to \infty} \int_{-R}^R f. For f(x) = x the symmetric integrals are all 0, yet \int_0^{\infty} x\,dx diverges, so \int_{-\infty}^{\infty} x\,dx diverges.
- f(x) \to 0 does not decide convergence. It is not enough (\displaystyle \frac{1}{x} falls to 0 and diverges). If instead f is continuous and f(x) \to L \ne 0, the integral diverges (Exercise 25).
- The p-result is about the tail at infinity. It holds for any lower limit a > 0. With lower limit 0 the behavior near 0 also matters, and for every p > 0 that part is improper too.
- Comparison needs f \ge 0, eventually. The inequality 0 \le f \le g has to hold from some point on. For an integrand that changes sign, such as \displaystyle \frac{\sin x}{x^2}, the usual route is to compare |f| and use the fact, stated here without proof, that if \int_a^{\infty} |f(x)|\,dx converges then so does \int_a^{\infty} f(x)\,dx.
- Comparison in the wrong direction is silent. Below a divergent integral, or above a convergent one, nothing follows.
- Limit comparison needs 0 < L < \infty. With L = 0 or L = \infty the test gives only one direction, and with no limit it gives nothing.
Where this goes next
The next lesson, Improper integrals with an infinite integrand, applies the same limit idea to integrands that blow up at a finite point, where the p-integrals split the other way: \displaystyle \int_0^1 \frac{dx}{x^p} converges exactly when p < 1. The comparison of an infinite sum with an improper integral gives the integral test and the The integral test and p series, and the direct and The direct and limit comparison tests for series are the comparison tests of this lesson with sums in place of integrals; together they lead into Convergence tests. Improper integrals also define the probability density functions of statistics, whose total area over the whole line must be 1.
Exercises
Day 1: one tail
- Evaluate \int_1^{\infty} \dfrac{dx}{x^4}.
- Evaluate \int_0^{\infty} e^{-x/4}\,dx.
- Decide whether \displaystyle \int_9^{\infty} \dfrac{dx}{\sqrt{x}} converges.
- Evaluate \int_{-\infty}^0 e^{2x}\,dx.
- Evaluate \int_1^{\infty} \dfrac{dx}{(2x + 1)^2}.
- Decide whether \int_0^{\infty} \dfrac{x}{x^2 + 1}\,dx converges.
- Decide whether \int_0^{\infty} \cos x\,dx converges.
- Evaluate \int_1^{\infty} \dfrac{\ln x}{x^2}\,dx by parts.
Day 1: both tails, substitution, and the well
- Evaluate \int_{-\infty}^{\infty} \dfrac{dx}{x^2 + 6x + 10}.
- Evaluate \int_{-\infty}^{\infty} xe^{-x^2}\,dx.
- Evaluate \int_e^{\infty} \dfrac{dx}{x(\ln x)^3}.
- For which numbers p does \int_1^{\infty} \dfrac{dx}{x^{2p - 1}} converge, and to what?
- A second well pumps 900e^{-0.1t} barrels per year. Find its total output forever and the time by which it has pumped half of it.
Day 1: reasoning
- Show that \int_{-\infty}^{\infty} x^3\,dx diverges, even though \int_{-R}^{R} x^3\,dx = 0 for every R. What goes wrong with \int_{-R}^{R+1} x^3\,dx?
- Let f be continuous on [1, \infty). Show that \int_1^{\infty} f(x)\,dx and \int_5^{\infty} f(x)\,dx both converge or both diverge, and that when they converge they differ by \int_1^5 f(x)\,dx.
Day 2: comparison
- Show that \int_1^{\infty} \dfrac{dx}{x^4 + x + 1} converges and give an upper bound.
- Decide whether \displaystyle \int_2^{\infty} \dfrac{dx}{\sqrt{x} - 1} converges.
- Show that \int_1^{\infty} \dfrac{\sin^2 x}{x^2}\,dx converges and give an upper bound.
- Show that \int_1^{\infty} e^{-x^2}\,dx converges and is less than \displaystyle \frac{1}{e}.
- Decide whether \int_1^{\infty} \dfrac{1 + e^{-x}}{x}\,dx converges.
- Decide whether \int_2^{\infty} \dfrac{x}{x^3 - 1}\,dx converges, using limit comparison.
Day 2: present value
- Money earns 7\% a year compounded continuously. (a) Find the present value of 4200 dollars per year paid continuously forever. (b) Oil from the well of Example 1 sells for 70 dollars a barrel. Find the present value of all its sales.
- At the same 7\%, a stream pays 6000e^{0.03t} dollars per year, rising 3\% a year forever. Find its present value. What happens if it rises at 9\% a year, 6000e^{0.09t}?
Day 2: reasoning
- Give a counterexample to the claim: “If 0 \le f(x) \le g(x) on [1, \infty) and \int_1^{\infty} g(x)\,dx diverges, then \int_1^{\infty} f(x)\,dx diverges.”
- Let f be continuous on [a, \infty) with \lim_{x \to \infty} f(x) = L > 0. Prove that \int_a^{\infty} f(x)\,dx diverges.
Answers
- \displaystyle \frac{1}{3}: the p-integral with p = 4, value \displaystyle \frac{1}{4 - 1}; directly, \displaystyle \int_1^b x^{-4}\,dx = \frac{1}{3} - \frac{1}{3b^3} \to \frac{1}{3}.
- 4: \int_0^b e^{-x/4}\,dx = 4(1 - e^{-b/4}) \to 4, the exponential tail with \displaystyle k = \frac{1}{4}.
- Diverges: \displaystyle \int_9^b x^{-1/2}\,dx = 2\sqrt{b} - 6 \to \infty (\displaystyle p = \frac{1}{2} \le 1).
- \displaystyle \frac{1}{2}: \displaystyle \int_a^0 e^{2x}\,dx = \frac{1}{2}(1 - e^{2a}) \to \frac{1}{2} as a \to -\infty.
- \displaystyle \frac{1}{6}: an antiderivative is \displaystyle -\frac{1}{2(2x + 1)}, so \displaystyle \int_1^b = \frac{1}{6} - \frac{1}{2(2b + 1)} \to \frac{1}{6}.
- Diverges: \displaystyle \int_0^b \frac{x}{x^2 + 1}\,dx = \frac{1}{2}\ln(b^2 + 1) \to \infty.
- Diverges: \int_0^b \cos x\,dx = \sin b, which has no limit as b \to \infty; it is neither \infty nor a number.
- 1: with u = \ln x, dv = x^{-2}\,dx, an antiderivative is \displaystyle -\frac{\ln x}{x} - \frac{1}{x}, so \displaystyle \int_1^b = 1 - \frac{\ln b}{b} - \frac{1}{b} \to 1, since \displaystyle \frac{\ln b}{b} \to 0.
- \pi: x^2 + 6x + 10 = (x + 3)^2 + 1, an antiderivative is \arctan(x + 3). Split at -3: each tail is \displaystyle \frac{\pi}{2}.
- 0: an antiderivative is \displaystyle -\frac{1}{2}e^{-x^2}. Split at 0: the right tail is \displaystyle \frac{1}{2}, the left tail \displaystyle -\frac{1}{2}; both converge, so the total is 0.
- \displaystyle \frac{1}{2}: with u = \ln x the integral is \displaystyle \int_1^{\ln b} u^{-3}\,du = \frac{1}{2} - \frac{1}{2(\ln b)^2} \to \frac{1}{2}.
- Converges exactly when p > 1, to \displaystyle \frac{1}{2p - 2}: it is the p-integral with exponent 2p - 1, which must exceed 1, and the value is \displaystyle \frac{1}{(2p - 1) - 1}.
- 9000 barrels, half by 10\ln 2 \approx 6.93 years: \int_0^b 900e^{-0.1t}\,dt = 9000(1 - e^{-0.1b}) \to 9000, and \displaystyle 1 - e^{-0.1b} = \frac{1}{2} gives \displaystyle b = \frac{\ln 2}{0.1}.
- Split at 0: \displaystyle \int_0^b x^3\,dx = \frac{b^4}{4} \to \infty, so the right tail diverges, and by definition the whole integral diverges. The symmetric integrals vanish only because the window is balanced: \int_{-R}^{R+1} x^3\,dx equals \displaystyle \frac{(R + 1)^4 - R^4}{4}, which tends to \infty. Different ways of widening the window give different answers, which is why each tail must converge on its own.
- For b > 5, additivity gives \int_1^b f = \int_1^5 f + \int_5^b f. The first term on the right is a fixed number, so the left side has a finite limit as b \to \infty exactly when \int_5^b f does, and the limits differ by \int_1^5 f.
- Converges, and the value is less than \displaystyle \frac{1}{3}: for x \ge 1, x^4 + x + 1 > x^4, so \displaystyle 0 < f(x) < \frac{1}{x^4}, and \displaystyle \int_1^{\infty} \frac{dx}{x^4} = \frac{1}{3}. (Numerically about 0.191.)
- Diverges: for x \ge 2, \displaystyle 0 < \sqrt{x} - 1 < \sqrt{x}, so \displaystyle \frac{1}{\sqrt{x} - 1} > \frac{1}{\sqrt{x}} > 0, and \displaystyle \int_2^{\infty} \frac{dx}{\sqrt{x}} diverges (\displaystyle p = \frac{1}{2}).
- Converges, and the value is at most 1: 0 \le \sin^2 x \le 1, so \displaystyle 0 \le \frac{\sin^2 x}{x^2} \le \frac{1}{x^2}, and \displaystyle \int_1^{\infty} \frac{dx}{x^2} = 1. (Numerically about 0.673.)
- For x \ge 1, x^2 \ge x, so 0 < e^{-x^2} \le e^{-x}, and \displaystyle \int_1^{\infty} e^{-x}\,dx = \frac{1}{e}. The integral converges and is at most \displaystyle \frac{1}{e}, in fact strictly less since e^{-x^2} < e^{-x} for x > 1. (Numerically about 0.139.)
- Diverges: e^{-x} > 0, so \displaystyle \frac{1 + e^{-x}}{x} > \frac{1}{x} > 0 on [1, \infty), and \displaystyle \int_1^{\infty} \frac{dx}{x} diverges.
- Converges: the ratio of the integrand to \displaystyle \frac{1}{x^2} is \displaystyle \frac{x^3}{x^3 - 1}, which tends to 1 as x \to \infty, and \displaystyle \int_2^{\infty} \frac{dx}{x^2} converges. (Numerically about 0.517.)
- (a) 60{,}000 dollars: \displaystyle \int_0^{\infty} 4200e^{-0.07t}\,dt = \frac{4200}{0.07}. (b) About 381{,}818.18 dollars: sales run at 70 \cdot 1{,}200e^{-0.15t} = 84{,}000e^{-0.15t} dollars per year, so the present value is the integral of 84{,}000e^{-0.22t} from 0 to \infty, which is \displaystyle \frac{84{,}000}{0.22}.
- 150{,}000 dollars: the discounted rate is 6000e^{0.03t}e^{-0.07t} = 6000e^{-0.04t}, and \displaystyle \int_0^{\infty} 6000e^{-0.04t}\,dt = \frac{6000}{0.04}. At 9\% the discounted rate is 6000e^{0.02t}, whose integral from 0 to b is 300{,}000(e^{0.02b} - 1), which tends to \infty: the integral diverges, and the stream has no finite present value.
- \displaystyle f(x) = \frac{1}{x^2} and \displaystyle g(x) = \frac{1}{x}: 0 \le f \le g on [1, \infty) and \int_1^{\infty} g diverges, but \int_1^{\infty} f = 1 converges. Being below a divergent integral decides nothing.
- Since f(x) \to L, there is X \ge a with \displaystyle f(x) \ge \frac{L}{2} for all x \ge X. Then \displaystyle \int_X^b f(x)\,dx \ge \frac{L}{2}(b - X) \to \infty, so \int_X^{\infty} f diverges (direct comparison with the constant \displaystyle \frac{L}{2}). By Exercise 15’s reasoning, adding the finite \int_a^X f does not change this, so \int_a^{\infty} f diverges.