Chapter 08

Medium

Improper Integrals

00 · Symbol Glossary

$\int_a^\infty f(x)\,dx$improper integral, infinite upper limit

A definite integral (Chapter 06) where the upper limit of integration is $\infty$ rather than a finite number. Defined as a limit of ordinary definite integrals over a growing finite interval.

$\int_a^b f(x)\,dx,\ f \text{ unbounded on } [a,b]$improper integral, infinite discontinuity

A definite integral where the integrand has a vertical asymptote (Chapter 01) somewhere in $[a,b]$ , including possibly at an endpoint — making the region under the curve unbounded in height even though the interval itself is finite in width.

$\text{converges / diverges}$convergence and divergence

An improper integral converges if its defining limit exists and is a finite number; it diverges if the limit is $\pm\infty$ or fails to exist at all.

$t$t — limit-of-integration variable

A temporary variable used to hold the place of the "moving" limit of integration while the limit is taken, so as not to confuse it with $x$ , the integration variable itself.

01 · Infinite Limits of Integration

FTC Part 2 (Chapter 06) evaluates $\int_a^b f(x)\,dx$ using $F(b)-F(a)$ — but that formula assumes $b$ is an actual number. To integrate all the way out to infinity, replace $\infty$ with a finite placeholder $t$ , evaluate normally, then take the limit as $t\to\infty$ .

Definition — Improper Integral, Infinite Interval

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

If the limit exists and is finite, the improper integral converges to that value. If the limit is $\pm\infty$ or does not exist, the improper integral diverges.

Step-by-step — Evaluate $\int_1^\infty \dfrac{1}{x^2}\,dx$

Rewrite as a limit:

\displaystyle\int_1^\infty \frac{1}{x^2}\,dx = \lim_{t\to\infty}\int_1^t x^{-2}\,dx

Find the antiderivative (Chapter 06 power rule, $n=-2$ ):

\int x^{-2}\,dx = \dfrac{x^{-1}}{-1}+C = -\dfrac1x+C

Apply FTC Part 2 on $[1,t]$ :

\left[-\dfrac1x\right]_1^t = -\dfrac1t - \left(-\dfrac11\right) = -\dfrac1t+1

Take the limit as $t\to\infty$ :

\lim_{t\to\infty}\left(1-\dfrac1t\right) = 1-0=1

, since

\dfrac1t\to0

t

grows without bound.

Conclude:

\displaystyle\int_1^\infty \frac{1}{x^2}\,dx = 1

— converges to a finite value, even though the region extends infinitely far to the right.

❌ A very similar-looking integral that diverges

Evaluate $\displaystyle\int_1^\infty \frac{1}{x}\,dx$ the same way. Antiderivative (Chapter 07, Section 02): $\int x^{-1}\,dx = \ln\lvert x\rvert + C$ .

$\displaystyle\int_1^t \frac1x\,dx = \big[\ln\lvert x\rvert\big]_1^t = \ln t - \ln 1 = \ln t - 0 = \ln t$ .

Why it breaks: as $t\to\infty$ , $\ln t \to \infty$ as well — the natural logarithm grows without bound, just slowly.

Consequence: $\displaystyle\int_1^\infty \frac1x\,dx$ diverges, despite $\frac1x$ and $\frac{1}{x^2}$ looking almost identical and both shrinking toward $0$ as $x\to\infty$ . The exponent matters: $\frac1x$ shrinks too slowly to enclose a finite area over an infinite interval, while $\frac{1}{x^2}$ shrinks fast enough. This exact contrast is generalized in Section 03.

02 · Improper Integrals from an Infinite Discontinuity

The second way an integral can be "improper" is when the integrand itself becomes unbounded somewhere in a finite interval — an infinite discontinuity (Chapter 01, Section 05), rather than an infinite interval.

Definition — Improper Integral, Infinite Discontinuity at an Endpoint

If $f$ has an infinite discontinuity at $x=a$ but is continuous on $(a,b]$ :

\int_a^b f(x)\,dx = \lim_{t\to a^+}\int_t^b f(x)\,dx

(An analogous definition with $t\to b^-$ applies when the discontinuity is at the upper endpoint $b$ instead.)

Step-by-step — Evaluate $\int_0^1 \dfrac{1}{\sqrt{x}}\,dx$

Identify the problem point:

f(x)=x^{-1/2}

is undefined at

x=0

(division by

\sqrt0=0

) — an infinite discontinuity at the lower endpoint.

Rewrite as a limit:

\displaystyle\int_0^1 x^{-1/2}\,dx = \lim_{t\to0^+}\int_t^1 x^{-1/2}\,dx

Find the antiderivative (power rule, $n=-\frac12$ ):

\int x^{-1/2}\,dx = \dfrac{x^{1/2}}{1/2}+C = 2\sqrt{x}+C

Apply FTC Part 2 on $[t,1]$ :

\big[2\sqrt x\big]_t^1 = 2\sqrt1 - 2\sqrt t = 2-2\sqrt t

Take the limit as $t\to0^+$ :

\lim_{t\to0^+}(2-2\sqrt t) = 2-0=2

, since

\sqrt t\to0

t\to0^+

Conclude:

\displaystyle\int_0^1 \frac{1}{\sqrt x}\,dx = 2

— converges, even though

f(x)\to\infty

x\to0^+

❌ Applying FTC Part 2 directly, ignoring the discontinuity

A tempting shortcut on the same integral: plug straight into the antiderivative, $\big[2\sqrt x\big]_0^1 = 2\sqrt1-2\sqrt0 = 2-0=2$ — which happens to give the correct numeric answer here, but for the wrong reason.

Why it breaks: FTC Part 2 (Chapter 06) explicitly requires $f$ to be continuous on the entire closed interval $[a,b]$ . Since $f(x)=x^{-1/2}$ is not even defined at $x=0$ , the hypothesis fails, and direct substitution is not justified — it must be verified by the limit process instead.

Consequence: the same shortcut applied to $\int_0^1 \dfrac{1}{x}\,dx$ would give $\big[\ln\lvert x\rvert\big]_0^1 = \ln1 - \ln0 = 0 - (-\infty)$ — nonsensical, since $\ln 0$ is undefined. Only the limit process correctly reveals that this particular improper integral diverges (mirroring Section 01's $\int_1^\infty \frac1x\,dx$ result): $\lim_{t\to0^+}\int_t^1\frac1x\,dx = \lim_{t\to0^+}\big(0-\ln t\big) = 0-(-\infty)=\infty$ .

03 · The p-Integral Test

Sections 01–02 both involved powers of $x$ . Generalizing them gives a reusable convergence rule, without needing to redo the limit computation from scratch every time.

Definition — p-Integral (infinite interval)

\int_1^\infty \frac{1}{x^p}\,dx \ \text{ converges if } p>1, \quad \text{ diverges if } p\leq1

Step-by-step — Prove the p-integral rule for $p\neq1$

Find the antiderivative for $p\neq1$ :

\int x^{-p}\,dx = \dfrac{x^{1-p}}{1-p}+C

(power rule, valid since

-p\neq-1

Evaluate on $[1,t]$ :

\left[\dfrac{x^{1-p}}{1-p}\right]_1^t = \dfrac{t^{1-p}}{1-p} - \dfrac{1}{1-p}

Case $p>1$ : then

1-p<0

, so

t^{1-p} = \dfrac{1}{t^{p-1}} \to 0

t\to\infty

(a positive power of

t

in the denominator grows without bound). The limit becomes

0 - \dfrac{1}{1-p} = \dfrac{1}{p-1}

— finite. Converges.

Case $p<1$ : then

1-p>0

, so

t^{1-p}\to\infty

t\to\infty

(a positive power of a growing quantity). The limit diverges to

\infty

. Diverges.

Case $p=1$ : handled separately in Section 01 — logarithm, diverges.

04 · Comparison Test

Some integrands have no elementary antiderivative, but their convergence can still be determined by comparing them to a simpler function whose behavior is already known — most often a p-integral.

Definition — Comparison Test for Improper Integrals

Suppose $0\leq f(x)\leq g(x)$ for all $x\geq a$ .

If $\displaystyle\int_a^\infty g(x)\,dx$ converges, then $\displaystyle\int_a^\infty f(x)\,dx$ also converges (a smaller non-negative quantity under a finite bound is itself finite).
If $\displaystyle\int_a^\infty f(x)\,dx$ diverges, then $\displaystyle\int_a^\infty g(x)\,dx$ also diverges (a larger quantity can't be finite if the smaller one already isn't).

Example — Using comparison to avoid computing an antiderivative

Determine whether $\displaystyle\int_1^\infty \frac{1}{x^2+3}\,dx$ converges, without finding its antiderivative directly.

For all $x\geq1$ : $x^2+3 > x^2$ , so $\dfrac{1}{x^2+3} < \dfrac{1}{x^2}$ (a larger denominator makes a smaller fraction — both sides positive). By Section 03, $\displaystyle\int_1^\infty \frac{1}{x^2}\,dx$ converges (it's a p-integral with $p=2>1$ ). Since $\frac{1}{x^2+3}$ is squeezed below a convergent integral, $\displaystyle\int_1^\infty \frac{1}{x^2+3}\,dx$ converges too — without ever computing its exact value.

❌ Comparing in the wrong direction

For $\displaystyle\int_1^\infty \frac{1}{\sqrt{x}-0.5}\,dx$ , a student notes $\dfrac{1}{\sqrt x-0.5} > \dfrac{1}{\sqrt x}$ for $x>1$ (smaller denominator gives a bigger fraction), and $\int_1^\infty x^{-1/2}\,dx$ is a p-integral with $p=\frac12\leq1$ , which diverges (Section 03).

Why this direction of the comparison test does apply, but only correctly: here $f(x) = \frac{1}{\sqrt x} \leq g(x) = \frac{1}{\sqrt x - 0.5}$ , and the smaller function's integral already diverges — by the comparison test's second bullet, this correctly forces the larger function's integral to diverge too.

Where the error would occur: if instead a student tried to conclude convergence from $g(x)$ being bounded below by a divergent $f(x)$ , or tried to bound $g(x)$ above by a convergent function without checking the inequality direction algebraically first, the conclusion would not follow. The comparison test only transmits convergence downward (from a larger convergent function to a smaller one) and divergence upward (from a smaller divergent function to a larger one) — never the reverse.

05 · Quant Application — Perpetual Cash Flows

An infinite-interval improper integral is the continuous-time analogue of summing an indefinitely long stream of payments — a perpetuity.

Example — Present value of a perpetual cash flow stream

A cash flow pays at a continuous rate of $r(t) = 100e^{-0.05t}$ dollars per year, discounted for all future time $t\geq0$ (the $e^{-0.05t}$ decay factor represents both discounting and the given payment schedule combined). Total present value:

\int_0^\infty 100e^{-0.05t}\,dt = \lim_{T\to\infty}\int_0^T 100e^{-0.05t}\,dt

Antiderivative: $\int 100e^{-0.05t}\,dt = 100\cdot\dfrac{e^{-0.05t}}{-0.05} = -2000e^{-0.05t}$ (verify: $\frac{d}{dx}[-2000e^{-0.05t}] = -2000\cdot(-0.05)e^{-0.05t}=100e^{-0.05t}$ ✓, using the chain rule from Chapter 03 with inner function $-0.05t$ ).

Evaluate on $[0,T]$ : $\big[-2000e^{-0.05t}\big]_0^T = -2000e^{-0.05T} - (-2000e^0) = -2000e^{-0.05T}+2000$ .

As $T\to\infty$ , $e^{-0.05T}\to0$ (exponential decay), so the limit is $0+2000=2000$ . The perpetual stream has a finite present value of $2000$ dollars, even though payments continue forever — the discounting shrinks distant payments fast enough for the total to converge, exactly analogous to the p-integral test in Section 03.

06 · Exercises

EXERCISE 8.1

Rewrite the infinite upper limit as a variable $t$ , integrate normally, then take the limit as $t\to\infty$ .

$\int_1^\infty x^{-3}\,dx = \lim_{t\to\infty}\left[\dfrac{x^{-2}}{-2}\right]_1^t = \lim_{t\to\infty}\left(-\dfrac{1}{2t^2}+\dfrac12\right) = 0+\dfrac12=\dfrac12$ . Converges to $\frac12$ (consistent with Section 03's rule: $p=3>1$ ).

Evaluate $\int_1^\infty x^{-3}\,dx$ , showing the limit step explicitly.

EXERCISE 8.2

Apply the p-integral rule from Section 03 directly by identifying $p$ .

$\int_1^\infty x^{-0.8}\,dx$ has $p=0.8\leq1$ , so by Section 03 it diverges. Confirm: antiderivative $\dfrac{x^{0.2}}{0.2}$ , and as $t\to\infty$ , $t^{0.2}\to\infty$ , so the limit diverges.

Without fully computing it, determine whether $\int_1^\infty x^{-0.8}\,dx$ converges or diverges, and justify using the p-integral rule.

EXERCISE 8.3

Locate where the integrand is undefined within the interval, then rewrite as a one-sided limit approaching that point.

$\int_0^8 x^{-2/3}\,dx$ : undefined at $x=0$ . $\lim_{t\to0^+}\int_t^8 x^{-2/3}\,dx$ . Antiderivative: $\dfrac{x^{1/3}}{1/3}=3x^{1/3}$ . $\big[3x^{1/3}\big]_t^8 = 3(8)^{1/3} - 3t^{1/3} = 3(2)-3t^{1/3}=6-3t^{1/3}$ . As $t\to0^+$ : $6-0=6$ . Converges to $6$ .

Evaluate $\int_0^8 x^{-2/3}\,dx$ , showing why it is improper and how the limit is set up.

EXERCISE 8.4

Bound the integrand above by a simpler function whose p-integral convergence is already known, keeping both sides non-negative for the comparison to apply.

For $\int_1^\infty \dfrac{1}{x^3+x}\,dx$ : for $x\geq1$ , $x^3+x>x^3$ , so $\dfrac{1}{x^3+x}<\dfrac{1}{x^3}$ . Since $\int_1^\infty x^{-3}\,dx$ converges ( $p=3>1$ , confirmed in Exercise 8.1), and $0\leq\frac{1}{x^3+x}<\frac{1}{x^3}$ , the comparison test gives that $\int_1^\infty \dfrac{1}{x^3+x}\,dx$ converges too.

Use the comparison test to determine whether $\int_1^\infty \dfrac{1}{x^3+x}\,dx$ converges.

EXERCISE 8.5

Check both ends: is the interval infinite, and is the integrand itself unbounded anywhere inside it?

$\int_1^\infty \dfrac{1}{x\sqrt{x-1}}\,dx$ is improper in two ways: the interval extends to $\infty$ , and the integrand is undefined at $x=1$ (division by $\sqrt0=0$ ). It must be split: $\int_1^\infty = \int_1^2 + \int_2^\infty$ (or any interior split point), and each piece must converge separately for the whole integral to converge — a single divergent piece makes the entire integral diverge, regardless of the other piece.

Explain why $\int_1^\infty \dfrac{1}{x\sqrt{x-1}}\,dx$ is improper in two separate ways, and describe (without fully evaluating) how the integral must be split to handle both.

EXERCISE 8.6

Set up the perpetual cash flow as an improper integral, find the antiderivative using the chain rule fact for exponential decay, then take the limit as the upper bound grows.

$r(t) = 300e^{-0.06t}$ , $t\geq0$ . $\int_0^\infty 300e^{-0.06t}\,dt = \lim_{T\to\infty}\left[\dfrac{300}{-0.06}e^{-0.06t}\right]_0^T = \lim_{T\to\infty}\big(-5000e^{-0.06T}+5000\big)$ . As $T\to\infty$ , $e^{-0.06T}\to0$ , so the limit is $5000$ . Present value: $5000$ dollars.

A perpetual cash flow pays at continuous rate $r(t)=300e^{-0.06t}$ dollars/year. Find its total present value using an improper integral.

07 · Chapter Summary

Concept	Formula / Rule
Improper integral, infinite interval	$\int_a^\infty f\,dx = \lim_{t\to\infty}\int_a^t f\,dx$
Improper integral, infinite discontinuity	$\int_a^b f\,dx = \lim_{t\to a^+}\int_t^b f\,dx$ (or symmetric at $b$ )
Convergence / divergence	Finite limit = converges; $\pm\infty$ or no limit = diverges
p-integral test	$\int_1^\infty x^{-p}dx$ converges iff $p>1$
Comparison test	$0\leq f\leq g$ : $g$ converges $\Rightarrow f$ converges; $f$ diverges $\Rightarrow g$ diverges
Perpetual cash flow	$\int_0^\infty (\text{rate})\,dt$ ; finite present value when decay outpaces growth

This closes the eight-chapter core sequence: limits and continuity, derivatives and their rules, the chain rule and implicit differentiation, core and optimization applications of derivatives, and the definition, techniques, and improper cases of integration — the full single-variable calculus toolkit used throughout applied quantitative work.