Can a region that stretches forever, or shoots up to infinity, have a finite area? Define the integral as a limit and find out.
Every definite integral we have met so far had two features: a finite interval $[a,b]$ and a bounded, continuous integrand. Real problems break both rules. How much work does it take to launch a spacecraft so that it never falls back to Earth? What is the total exposure of a patient to a drug whose concentration decays but never quite reaches zero? What is the probability that a light bulb lasts longer than three years, when in principle it could last for any length of time? Each question is an integral over an infinite interval.
Surprisingly, an infinitely long region can have a finite area — the region under $y=1/x^2$ to the right of $x=1$ has area exactly $1$ — while the region under $y=1/x$, which looks almost the same, has infinite area. Similarly a graph can shoot up to infinity at a point and still enclose a finite area. These improper integrals are defined as limits of ordinary integrals, and the whole art is deciding when the limit exists.
Integrals over Infinite Intervals
The idea is simple: integrate up to a finite point $b$, then let $b\to\infty$ and see whether the areas settle down.
📘 Definition 5.6 — Improper integrals over infinite intervals
If $f$ is continuous on $[a,\infty)$, then $\displaystyle\int_a^{\infty}f(x)\,dx=\lim_{b\to\infty}\int_a^bf(x)\,dx$.
If $f$ is continuous on $(-\infty,b]$, then $\displaystyle\int_{-\infty}^{b}f(x)\,dx=\lim_{a\to-\infty}\int_a^bf(x)\,dx$.
The improper integral converges if the limit exists as a finite number, and diverges otherwise.
If $f$ is continuous on $(-\infty,\infty)$, then $\displaystyle\int_{-\infty}^{\infty}f(x)\,dx=\int_{-\infty}^{c}f(x)\,dx+\int_c^{\infty}f(x)\,dx$ for any real number $c$. It converges only if both integrals on the right converge (and its value does not depend on $c$).
✏️ Example 1 — Two famous tails
Determine whether each integral converges, and find its value if it does: (a) $\displaystyle\int_1^{\infty}\frac{dx}{x^2}$ $\quad$ (b) $\displaystyle\int_1^{\infty}\frac{dx}{x}$.
💡 Show solution▶
(a) $1/x^2$ is continuous on $[1,\infty)$. $$\int_1^b\frac{dx}{x^2}=\Big[-\frac1x\Big]_1^b=1-\frac1b\ \xrightarrow{\ b\to\infty\ }\ 1.$$ The integral converges and $\displaystyle\int_1^\infty\frac{dx}{x^2}=1$.
(b) $$\int_1^b\frac{dx}{x}=\Big[\ln x\Big]_1^b=\ln b\ \xrightarrow{\ b\to\infty\ }\ \infty.$$ The integral diverges.
Interpret Both curves tend to $0$, but $1/x^2$ does so fast enough that the extra area beyond $x=b$, namely $1/b$, shrinks to nothing; for $1/x$ the area beyond any point is still infinite. Tending to zero is not enough.
Split The integrand is continuous everywhere. Split at $c=0$ and treat each half separately.
Right half $\displaystyle\int_0^{\infty}\frac{dx}{1+x^2}=\lim_{b\to\infty}\Big[\tan^{-1}x\Big]_0^b=\lim_{b\to\infty}\tan^{-1}b=\frac\pi2.$
Left half $\displaystyle\int_{-\infty}^{0}\frac{dx}{1+x^2}=\lim_{a\to-\infty}\left(0-\tan^{-1}a\right)=-\left(-\frac\pi2\right)=\frac\pi2.$
Combine Both halves converge, so $\displaystyle\int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\frac\pi2+\frac\pi2=\pi$. An infinitely wide region of area $\pi$!
⚠️ A tempting shortcut that is wrong
It is not correct to compute $\displaystyle\int_{-\infty}^{\infty}f(x)\,dx$ as $\displaystyle\lim_{b\to\infty}\int_{-b}^{b}f(x)\,dx$. For $f(x)=x$ that symmetric limit is $\lim_{b\to\infty}0=0$, yet $\displaystyle\int_0^\infty x\,dx$ diverges, so by Definition 5.6 the whole integral diverges. The two tails must converge independently. (Exercise I.8 is another example.)
📐 Theorem 5.10 — The $p$-integrals
Let $p$ be a real constant.
$$\int_1^{\infty}\frac{dx}{x^p}\ \text{ converges if } p\gt1\ \left(\text{to }\frac{1}{p-1}\right)\ \text{ and diverges if } p\le1;$$
$$\int_0^{1}\frac{dx}{x^p}\ \text{ converges if } p\lt1\ \left(\text{to }\frac{1}{1-p}\right)\ \text{ and diverges if } p\ge1.$$
Proof sketch. For $p\ne1$, $\displaystyle\int_1^b x^{-p}dx=\frac{b^{1-p}-1}{1-p}$, and $b^{1-p}\to0$ when $p\gt1$ but $\to\infty$ when $p\lt1$; the case $p=1$ gives $\ln b\to\infty$. The second statement follows in the same way with $t\to0^+$ in place of $b\to\infty$ (Definition 5.7 below).
Notice the switch: on $[1,\infty)$ we need the curve to fall fast ($p\gt1$); on $(0,1]$ we need it to blow up slowly ($p\lt1$). No value of $p$ works on both, so $\int_0^\infty x^{-p}\,dx$ always diverges. Interactive 1 lets you watch both limits.
Interactive 1 — Does the Area Settle Down?
The shaded region is $\int_1^b x^{-p}\,dx$ (or $\int_t^1x^{-p}\,dx$). The purple curve shows the running total as the endpoint moves; when the integral converges it levels off at the dashed limit. Play sends $b$ from $1$ to $10\,000$ (or $t$ from $1$ down to $0.0001$) — the readouts keep counting after the endpoint leaves the screen. Try $p=1$, then $p=1.1$ and $p=0.9$.
2
endpoint—
area so far—
limit—
still to come—
Integrands with Infinite Discontinuities
The second kind of improper integral has a finite interval but an integrand that becomes unbounded — typically at a vertical asymptote. We stay away from the bad point, integrate, and then let the endpoint approach it.
📘 Definition 5.7 — Improper integrals with infinite discontinuities
If $f$ is continuous on $[a,b)$ and unbounded near $b$, then $\displaystyle\int_a^bf(x)\,dx=\lim_{t\to b^-}\int_a^tf(x)\,dx$.
If $f$ is continuous on $(a,b]$ and unbounded near $a$, then $\displaystyle\int_a^bf(x)\,dx=\lim_{t\to a^+}\int_t^bf(x)\,dx$.
If $f$ is continuous on $[a,b]$ except at an interior point $c$, where it is unbounded, then $\displaystyle\int_a^bf(x)\,dx=\int_a^cf(x)\,dx+\int_c^bf(x)\,dx$, and it converges only if both pieces converge.
As before, the integral converges if the limit(s) exist as finite numbers and diverges otherwise.
Spot it The integrand is continuous on $[0,4)$ and $\to\infty$ as $x\to4^-$: improper at the right endpoint.
Limit With $u=4-x$, an antiderivative is $-2\sqrt{4-x}$: $$\int_0^t\frac{dx}{\sqrt{4-x}}=\Big[-2\sqrt{4-x}\Big]_0^t=-2\sqrt{4-t}+4\ \xrightarrow{\ t\to4^-\ }\ 4.$$
Answer The integral converges, with value $4$. (This is Theorem 5.10 in disguise: $\frac{1}{(4-x)^{1/2}}$ has $p=\tfrac12\lt1$.)
✏️ Example 4 — A discontinuity inside the interval
Split $x^{-1/3}=\dfrac{1}{\sqrt[3]{x}}$ is unbounded at $x=0$, which lies inside $[-1,8]$. So $\displaystyle\int_{-1}^{8}=\int_{-1}^{0}+\int_0^8$, and both must converge. An antiderivative is $\tfrac32x^{2/3}$.
Left piece $\displaystyle\lim_{t\to0^-}\int_{-1}^{t}x^{-1/3}dx=\lim_{t\to0^-}\tfrac32\left(t^{2/3}-(-1)^{2/3}\right)=\tfrac32(0-1)=-\tfrac32.$
Right piece $\displaystyle\lim_{t\to0^+}\int_t^{8}x^{-1/3}dx=\lim_{t\to0^+}\tfrac32\left(8^{2/3}-t^{2/3}\right)=\tfrac32\cdot4=6.$
Answer Both converge, so $\displaystyle\int_{-1}^8x^{-1/3}dx=-\tfrac32+6=\tfrac92$. The negative piece is a (signed) area below the axis.
⚠️ Always look for discontinuities first
Applying the Fundamental Theorem blindly to $\displaystyle\int_{-1}^{1}\frac{dx}{x^2}$ gives $\Big[-\frac1x\Big]_{-1}^{1}=-1-1=-2$ — a negative answer for a positive integrand, which is absurd. The FTC needs $f$ continuous on the whole interval, and $1/x^2$ is unbounded at $0$. Correctly: $\displaystyle\int_0^1\frac{dx}{x^2}$ is a $p$-integral with $p=2\ge1$, so it diverges, and therefore $\displaystyle\int_{-1}^1\frac{dx}{x^2}$ diverges.
The Comparison Test
Often we cannot find an antiderivative — $e^{-x^2}$ is the famous example — but we only need to know whether the integral converges (perhaps to approximate it afterwards with the methods of Section 5.4). The idea: if a positive function lies below one with finite area, its area is finite too; if it lies above one with infinite area, its area is infinite too.
📐 Theorem 5.11 — Comparison test for improper integrals
Suppose $f$ and $g$ are continuous on $[a,\infty)$ and $0\le f(x)\le g(x)$ for all $x\ge a$.
If $\displaystyle\int_a^\infty g(x)\,dx$ converges, then $\displaystyle\int_a^\infty f(x)\,dx$ converges, and $\displaystyle\int_a^\infty f\le\int_a^\infty g$.
If $\displaystyle\int_a^\infty f(x)\,dx$ diverges, then $\displaystyle\int_a^\infty g(x)\,dx$ diverges.
The same statements hold for integrals with an infinite discontinuity (for example on $(a,b]$ with both functions continuous there).
🧭 Using the comparison test
Guess the answer by looking at how the integrand behaves for large $x$ (or near the bad point): keep only the dominant terms. E.g. $\dfrac{1}{\sqrt{x^3+1}}\approx\dfrac{1}{x^{3/2}}$, a convergent $p$-integral.
To prove convergence, find a bigger function whose integral converges. To prove divergence, find a smaller (non-negative) function whose integral diverges.
Check the inequality really holds on the whole interval.
Beware: being smaller than a divergent integral, or bigger than a convergent one, tells you nothing.
✏️ Example 5 — Comparison in both directions
Determine whether (a) $\displaystyle\int_1^{\infty}e^{-x^2}dx$ and (b) $\displaystyle\int_1^{\infty}\frac{2+\cos x}{x}\,dx$ converge.
💡 Show solution▶
(a) For $x\ge1$ we have $x^2\ge x$, so $0\lt e^{-x^2}\le e^{-x}$. And $\displaystyle\int_1^\infty e^{-x}dx=\lim_{b\to\infty}\left(e^{-1}-e^{-b}\right)=\frac1e$ converges. By Theorem 5.11, $\displaystyle\int_1^\infty e^{-x^2}dx$ converges, with value less than $\frac1e\approx0.368$. (Numerically it is about $0.1394$.)
(b) Since $\cos x\ge-1$, we have $\dfrac{2+\cos x}{x}\ge\dfrac1x\gt0$ for $x\ge1$. As $\displaystyle\int_1^\infty\frac{dx}{x}$ diverges (Example 1), the bigger integral diverges too.
Bonus Because $\displaystyle\int_0^1e^{-x^2}dx$ is an ordinary integral, (a) shows that $\displaystyle\int_0^\infty e^{-x^2}dx$ converges. Its exact value, $\tfrac{\sqrt\pi}{2}$, needs multivariable calculus — but convergence needed only a comparison.
Interactive 2 — Comparison Test Overlay
The red function $f$ lies below the blue function $g$ for $x\ge1$. As $b$ grows (Play sends it out to about $300$), compare the two running areas $\int_1^bf$ and $\int_1^bg$. A finite area on top traps the bottom one; an infinite area underneath pushes the top one to infinity. The last two pairs show when the test says nothing.
$b$—
$\int_1^bf(x)\,dx$—
$\int_1^bg(x)\,dx$—
conclusion—
Applications
✏️ Example 6 — Escape velocity
The gravitational force on a mass $m$ at distance $r$ from the centre of the Earth is $F(r)=\dfrac{GMm}{r^2}$, where $GM\approx3.986\times10^{14}$ m³/s². The Earth's radius is $R\approx6.371\times10^6$ m. Find the work needed to move a $1$ kg mass from the surface "to infinity", and the launch speed needed to supply that energy.
💡 Show solution▶
Set up Work is force integrated over distance. For $m=1$ kg: $$W=\int_R^\infty\frac{GM}{r^2}\,dr=\lim_{b\to\infty}\Big[-\frac{GM}{r}\Big]_R^b=\lim_{b\to\infty}\left(\frac{GM}{R}-\frac{GM}{b}\right)=\frac{GM}{R}.$$ The improper integral converges ($p=2$): the total work to escape is finite, even though gravity never switches off.
Evaluate $W=\dfrac{3.986\times10^{14}}{6.371\times10^{6}}\approx6.26\times10^{7}$ J per kilogram.
Speed Setting the kinetic energy $\tfrac12v^2$ equal to $W$: $v=\sqrt{\dfrac{2GM}{R}}\approx\sqrt{1.251\times10^{8}}\approx1.12\times10^{4}$ m/s, i.e. about $11.2$ km/s (ignoring air resistance and the Earth's rotation).
Interpret If the force fell off only like $1/r$, the integral would diverge and nothing could ever escape. The $1/r^2$ law is what makes space travel possible. Exercise IV.4 repeats the calculation for the Moon.
Improper integrals also appear throughout probability. A probability density function is a function $f\ge0$ with $\displaystyle\int_{-\infty}^{\infty}f(x)\,dx=1$; probabilities are areas under it, such as $P(X\gt3)=\displaystyle\int_3^\infty f(x)\,dx$, and the mean is $\displaystyle\int_{-\infty}^{\infty}xf(x)\,dx$. Exercises IV.2–3 work with the exponential density used for lifetimes of components, and the Research–Teaching Nexus uses the same idea in pharmacology.
Exercises
I
Infinite intervals. Determine whether each integral converges or diverges; evaluate those that converge.
5 By parts, $\int xe^{-x}dx=-(x+1)e^{-x}$. $\displaystyle\lim_{b\to\infty}\left(1-(b+1)e^{-b}\right)=1$, since $\displaystyle\lim_{b\to\infty}\frac{b+1}{e^b}=\lim_{b\to\infty}\frac{1}{e^b}=0$ by L'Hôpital's rule. Converges to $1$.
7 $u=\ln x$: $\displaystyle\int\frac{dx}{x(\ln x)^2}=-\frac{1}{\ln x}$. $\displaystyle\lim_{b\to\infty}\left(\frac1{\ln2}-\frac1{\ln b}\right)=\frac{1}{\ln2}\approx1.4427$. Converges (also slowly — see the check-lab).
8 $\displaystyle\int_0^b\frac{x\,dx}{1+x^2}=\tfrac12\ln\left(1+b^2\right)\to\infty$, so the right half diverges, and therefore the whole integral diverges — even though $\int_{-b}^{b}=0$ for every $b$ (see the warning after Example 2).
10 $\dfrac{1}{x(x+1)}=\dfrac1x-\dfrac1{x+1}$, so $\displaystyle\int_1^b=\Big[\ln\frac{x}{x+1}\Big]_1^b=\ln\frac{b}{b+1}-\ln\frac12\ \to\ 0+\ln2$. Converges to $\ln2\approx0.6931$. (Each of $\int\frac{dx}{x}$ and $\int\frac{dx}{x+1}$ diverges on its own; never split a convergent improper integral into two divergent ones.)
II
Infinite discontinuities. Determine whether each integral converges or diverges; evaluate those that converge.
2 $p=1$: diverges. ($\int_t^1\frac{dx}{x}=-\ln t\to\infty$ as $t\to0^+$.)
3 $\displaystyle\int_t^1\ln x\,dx=\Big[x\ln x-x\Big]_t^1=-1-t\ln t+t$. Since $\displaystyle\lim_{t\to0^+}t\ln t=\lim_{t\to0^+}\frac{\ln t}{1/t}=\lim_{t\to0^+}\frac{1/t}{-1/t^2}=0$ (L'Hôpital), the integral converges to $-1$.
4 Unbounded at $x=2$: $\displaystyle\lim_{t\to2^+}\Big[\tfrac32(x-2)^{2/3}\Big]_t^3=\tfrac32$. Converges.
5 Unbounded at $x=1$: $\displaystyle\lim_{t\to1^-}\sin^{-1}t=\frac\pi2$. Converges.
6 Interior discontinuity at $x=1$. $\displaystyle\int_1^2\frac{dx}{(x-1)^2}$ is a $p$-integral with $p=2\ge1$ (shifted), so it diverges; hence the whole integral diverges. (The careless answer $\Big[-\frac{1}{x-1}\Big]_0^2=-2$ is meaningless.)
7 Interior point $0$; antiderivative $\tfrac32x^{2/3}$. $\displaystyle\int_{-8}^{0}=\tfrac32\left(0-(-8)^{2/3}\right)=\tfrac32(-4)=-6$ and $\displaystyle\int_0^1=\tfrac32$. Both converge: total $-\tfrac92$.
8 $\displaystyle\int_0^t\tan x\,dx=-\ln(\cos t)\to\infty$ as $t\to\frac{\pi}{2}^-$ (because $\cos t\to0^+$). Diverges.
III
Comparison test. Use Theorem 5.11 to decide whether each integral converges or diverges. State the function you compare with.
4 For $x\ge2$, $0\lt\sqrt x-1\lt\sqrt x$, so $\dfrac{1}{\sqrt x-1}\gt\dfrac{1}{\sqrt x}$, and $\int_2^\infty x^{-1/2}dx$ diverges ($p=\tfrac12$). Diverges.
5 For $x\ge1$, $0\lt\dfrac{e^{-x}}{x}\le e^{-x}$, and $\int_1^\infty e^{-x}dx=\frac1e$ converges. Converges.
6 Improper at $0$. On $(0,1]$, $0\lt\dfrac{e^x}{\sqrt x}\le\dfrac{e}{\sqrt x}$, and $\int_0^1\frac{e\,dx}{\sqrt x}=2e$ converges ($p=\tfrac12\lt1$). Converges (to less than $2e$).
7 $\sqrt{x^6+1}\gt\sqrt{x^6}=x^3$, so $0\lt\dfrac{x}{\sqrt{x^6+1}}\lt\dfrac{x}{x^3}=\dfrac1{x^2}$. Converges.
8 Improper at $0$. $x^2+\sqrt x\gt\sqrt x$ on $(0,1]$, so $0\lt\dfrac{1}{x^2+\sqrt x}\lt\dfrac{1}{\sqrt x}$, and $\int_0^1x^{-1/2}dx=2$. Converges.
IV
Applications and theory.
1Pharmacology. After an injection the plasma concentration of a drug is $C(t)=8e^{-0.25t}$ mg/L ($t$ in hours). The total exposure is $\text{AUC}=\displaystyle\int_0^\infty C(t)\,dt$. Find it.
2Reliability. The lifetime $X$ (years) of a sensor has density $f(x)=0.5e^{-0.5x}$ for $x\ge0$ (and $0$ for $x\lt0$). Show that $\displaystyle\int_{-\infty}^{\infty}f(x)\,dx=1$, and find $P(X\gt3)=\displaystyle\int_3^\infty f(x)\,dx$.
3 Find the mean lifetime $\displaystyle\int_0^\infty x\,f(x)\,dx$ for the sensor in part 2.
4Space. For the Moon, $GM\approx4.905\times10^{12}$ m³/s² and $R\approx1.737\times10^6$ m. Using Example 6, find the escape speed from the Moon's surface.
5Finance. A perpetual endowment pays $5000$ OMR per year forever, continuously. With continuous discount rate $4\%$, its present value is $\displaystyle\int_0^\infty5000e^{-0.04t}\,dt$. Find it.
6Gabriel's horn. The curve $y=1/x$, $x\ge1$, is rotated about the $x$-axis. Show that the volume $\pi\displaystyle\int_1^\infty\frac{dx}{x^2}$ is finite, but that the surface area $2\pi\displaystyle\int_1^\infty\frac1x\sqrt{1+\frac1{x^4}}\,dx$ is infinite.
7Theory. Show that $\displaystyle\int_0^\infty\frac{dx}{x^p}$ diverges for every real $p$.
💡 Show solutions▶
1 $\displaystyle\lim_{b\to\infty}\Big[-32e^{-0.25t}\Big]_0^b=32$ mg·h/L. (In general $\int_0^\infty C_0e^{-kt}dt=C_0/k$.)
2 $\displaystyle\int_{-\infty}^0 0\,dx+\lim_{b\to\infty}\Big[-e^{-0.5x}\Big]_0^b=0+1=1$. ✓ And $P(X\gt3)=\displaystyle\lim_{b\to\infty}\left(e^{-1.5}-e^{-0.5b}\right)=e^{-1.5}\approx0.2231$: about a $22\%$ chance of lasting more than three years.
3 By parts with $u=x$, $dv=0.5e^{-0.5x}dx$: $\displaystyle\int_0^b0.5xe^{-0.5x}dx=\Big[-xe^{-0.5x}-2e^{-0.5x}\Big]_0^b=2-(b+2)e^{-0.5b}\to2$. Mean lifetime $2$ years $=1/\lambda$.
4 $v=\sqrt{\dfrac{2GM}{R}}=\sqrt{\dfrac{2(4.905\times10^{12})}{1.737\times10^6}}\approx\sqrt{5.648\times10^6}\approx2.38\times10^3$ m/s, about $2.4$ km/s — under a quarter of the Earth's value.
5 $\displaystyle\lim_{b\to\infty}\Big[-125\,000e^{-0.04t}\Big]_0^b=125\,000$ OMR. A finite price for an infinite stream of payments.
6 Volume $=\pi\cdot1=\pi$ (Example 1). For the surface, $\sqrt{1+x^{-4}}\ge1$, so the integrand is $\ge\dfrac{2\pi}{x}$, whose integral on $[1,\infty)$ diverges. By comparison the area is infinite: a horn you could fill with paint but never paint.
7 $\displaystyle\int_0^\infty=\int_0^1+\int_1^\infty$ needs both to converge. The first needs $p\lt1$, the second needs $p\gt1$ (Theorem 5.10). No $p$ satisfies both, so the integral always diverges.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. A drug trial measures blood concentration for only 12 hours, but the drug stays in the body much longer. How do pharmacologists estimate the total exposure $\text{AUC}_{0\to\infty}$ — an improper integral — from finite data?
Why it matters. AUC is used worldwide by medicine regulators to compare generic and brand-name drugs ("bioequivalence") and to set safe doses, including for children and patients with kidney disease.
What to do.
Sample data (hours, mg/L): $(0,0),\ (1,6.0),\ (2,7.2),\ (4,5.6),\ (6,3.9),\ (8,2.7),\ (12,1.3)$. Note the uneven spacing.
Estimate $\text{AUC}_{0\to12}$ with the trapezoidal rule, one trapezoid per interval (Section 5.4).
Assume that after 12 h the concentration decays exponentially: $C(t)=C_{12}e^{-k(t-12)}$. Estimate $k$ from the last two measurements, then show $\displaystyle\int_{12}^\infty C(t)\,dt=\frac{C_{12}}{k}$.
Add the two parts. What percentage of the total exposure lies beyond the last measurement?
Tools: Calculator or spreadsheet.
Hand in: A half-page note with the two AUC parts, the total, the "extrapolated percentage", and one assumption that could make the estimate wrong.
Going further. Regulators usually reject a study if the extrapolated part exceeds about 20% of the total. Why might they insist on that? Would your sample study pass?
Instructor note: $\text{AUC}_{0\to12}=3+6.6+12.8+9.5+6.6+8.0=46.5$ mg·h/L; $k=\tfrac14\ln\tfrac{2.7}{1.3}\approx0.183$ h⁻¹; tail $\approx1.3/0.183\approx7.1$; total $\approx53.6$ mg·h/L, of which about 13% is extrapolated — acceptable.