What is an improper integral?

An integral in which one of the two assumptions behind a definite integral breaks: either the interval of integration is unbounded, or the integrand is unbounded at one of the endpoints. In both cases it cannot be evaluated directly, so it is defined as the limit of an ordinary definite integral whose interval is only then stretched out.

When is an improper integral convergent?

Exactly when the limit in its definition exists and is finite. If the limit is plus or minus infinity, or does not exist at all, the integral is called divergent and no value is assigned to it. The mere fact that a region is unbounded settles nothing — an unbounded region may well have a finite area.

How can an infinitely long region have a finite area?

By shrinking fast enough. The region under y = 1/x² from 1 rightwards has area 1: each further stretch adds less than the one before, and it all sums to a number. Under y = 1/x, a curve that looks all but identical, the area already grows without bound. The difference is invisible on the graph — only the limit decides it.

A comparison of the exponent with 1, with the inequality pointing the opposite way in the two situations. Far from the origin ∫ dx/x^p from 1 to infinity converges for p > 1, because the function has to shrink quickly. At a pole at zero ∫ dx/x^p from 0 to 1 converges for p < 1, because the function has to grow slowly.

Why are the conditions reversed at zero and at infinity?

Because a different thing causes trouble in each place. At infinity the problem is the length of the interval, so the function has to decay faster than 1/x to fade away in time. At zero the problem is the height of the function, so it has to grow more slowly than 1/x to still fit. The exponent p = 1 is the boundary and gives divergence in both cases.

What do I do when the function blows up in the middle of the interval?

Split the integral at that point and study each half separately; the whole converges only if both halves do. Substituting blindly into the Newton–Leibniz formula gives nonsense there: for ∫ dx/x² from −1 to 1 it produces −2, a negative area for a region lying entirely above the axis.

Is every integral of a function decaying to zero convergent?

No. The function 1/x decays to zero, and yet its integral from 1 to infinity diverges — its value grows like a logarithm, slowly but without bound. Tending to zero is a necessary condition for convergence, not a sufficient one; what counts is the rate.

What are improper integrals actually used for?

Anywhere a quantity is accumulated over unbounded time: the present value of an endless stream of payments, the total radiation dose from an isotope, the mean working time of a device before failure, the distance covered while speed decays exponentially. In every one of those the answer is a finite number, even though the integration runs to infinity.

Advanced level

Improper integrals

Q: Why are the conditions reversed at zero and at infinity?

Because a different thing causes trouble in each place. At infinity the problem is the length of the interval, so the function has to decay faster than 1/x to fade away in time. At zero the problem is the height of the function, so it has to grow more slowly than 1/x to still fit. The exponent p = 1 is the boundary and gives divergence in both cases.

Q: What do I do when the function blows up in the middle of the interval?

Split the integral at that point and study each half separately; the whole converges only if both halves do. Substituting blindly into the Newton–Leibniz formula gives nonsense there: for ∫ dx/x² from −1 to 1 it produces −2, a negative area for a region lying entirely above the axis.

Q: Is every integral of a function decaying to zero convergent?

No. The function 1/x decays to zero, and yet its integral from 1 to infinity diverges — its value grows like a logarithm, slowly but without bound. Tending to zero is a necessary condition for convergence, not a sufficient one; what counts is the rate.

Q: What are improper integrals actually used for?

Anywhere a quantity is accumulated over unbounded time: the present value of an endless stream of payments, the total radiation dose from an isotope, the mean working time of a device before failure, the distance covered while speed decays exponentially. In every one of those the answer is a finite number, even though the integration runs to infinity.

A definite integral quietly assumes two things: that the interval is finite and that the function is bounded on it. When either assumption fails, the integral is defined through a limit — and it turns out that an unbounded region can have a finite area. Whether it does is decided by a single number: the exponent p.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Integration by parts and substitutionThe table of integrals stops at polynomials, sine and cosine — and ∫ x·eˣ dx already has no row in it. Two techniques close the gap, because each of them is a differentiation rule read backwards: the product rule gives integration by parts, and the chain rule gives integration by substitution.
Limit of a functionA limit says where the values of a function are heading as the argument closes in on a number — even when the function has no value at that number at all. It is the first idea of analysis and the ground the derivative stands on.

Where this is used

Real situations where you count exactly the way this lesson teaches:

How much further the bike will roll
Once you stop pedalling, the speed decays exponentially: v(t) = 6·e^(−t/12) metres per second. The distance to a standstill is the integral of speed over all remaining time — an improper integral over an infinite interval — and it comes out as a finite 6·12 = 72 metres. In theory the bike never quite stops, and it still travels exactly 72 metres.
Present value of a perpetuity
A fund pays out 12,000 a year forever, and money is discounted at 5 percent a year. The present value is the integral of 12000·e^(−0.05t) from zero to infinity, which is 12000/0.05 = 240,000. An infinite stream of payments has a finite price, and that is exactly what the convergence of this integral means.
Total radiation dose
After a contamination the dose rate decays exponentially: D′(t) = 0.8·e^(−t/6) millisieverts per hour. The dose absorbed over all remaining time is the integral of that rate, namely 0.8·6 = 4.8 mSv — the number a radiation protection officer compares against the annual limit. Had the dose rate fallen off like 1/t, the same integral would diverge and the limit would certainly be exceeded.
Mean time between failures
The reliability of a device is described by R(t) = e^(−t/4000), with time in hours. The mean time to failure is the integral of R over an infinite interval, namely 4000 hours — the MTBF quoted on the datasheet. The whole definition of that quantity rests on this improper integral being convergent.

All formulas

Improper integral, first kind
$\int_a^{\infty} f(x)\,dx = \lim_{T \to \infty} \int_a^{T} f(x)\,dx$
an unbounded interval of integration
Improper integral, second kind
$\int_a^b f(x)\,dx = \lim_{\varepsilon \to 0^+} \int_{a+\varepsilon}^{b} f(x)\,dx$
the function is unbounded at the endpoint a
The p-test at infinity
$\int_1^{\infty} \frac{dx}{x^p} = \frac{1}{p-1} \quad \text{for } p > 1$
divergent for p ≤ 1
The p-test at zero
$\int_0^1 \frac{dx}{x^p} = \frac{1}{1-p} \quad \text{for } p < 1$
divergent for p ≥ 1
An exponential tail
$\int_0^{\infty} e^{-ax}\,dx = \frac{1}{a} \quad (a > 0)$
convergent for every positive a
Finite area, infinite height
$\int_0^1 \frac{dx}{\sqrt{x}} = 2$
the flagship example of the second kind

The definite integral $\int_a^b f(x)\,dx$ quietly assumes two things: that the interval $\langle a;\, b \rangle$ is finite and that the function is bounded on it. When either assumption fails, the integral sign has to be defined afresh — and that is how improper integrals arise.

what fails	example	name
the interval is unbounded	$\int_1^{\infty} \dfrac{dx}{x^2}$	improper integral of the first kind
the function is unbounded at an endpoint	$\int_0^1 \dfrac{dx}{\sqrt{x}}$	improper integral of the second kind

The prescription is the same in both cases: evaluate an ordinary definite integral, then pass to a limit.

First kind: an unbounded interval

Infinity cannot be substituted into the Newton–Leibniz formula, so instead we integrate up to a moving edge $T$ and watch where the result heads:

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

If that limit is finite, the integral is called convergent and is given that value. If the limit is infinite or fails to exist, the integral is divergent and has no value at all.

Decide whether ∫₁^∞ dx/x² converges and find its value.

The shaded region does not end at the right-hand edge of the drawing — the frame does. The region itself carries on, and its area is still exactly 1.

No drawing will ever show infinity; it shows the beginning of it, and the computation carries the rest. That is the most important thing in this lesson: convergence cannot be settled by eye.

Convergent or divergent

Change the exponent from $2$ to $1$ . The curves look all but the same, and the outcome is entirely different.

Decide whether ∫₁^∞ dx/x converges.

Two curves and two regions with the same bounds. The upper one has infinite area, the lower one has area 1. On the drawing that difference is invisible, because both curves decay to zero.

Decaying to zero is not enough by itself. What counts is the rate — and that is precisely what the p-test measures.

The p-test at infinity

\int_1^{\infty} \frac{dx}{x^p} = \begin{cases} \dfrac{1}{p-1}, & p > 1 \quad \text{(convergent)} \\[2mm] \infty, & p \leq 1 \quad \text{(divergent)} \end{cases}

The computation is one line. For $p \neq 1$ the antiderivative is $\tfrac{x^{1-p}}{1-p}$ , so

\int_1^{T} x^{-p}\,dx = \frac{T^{1-p} - 1}{1-p}

Everything hangs on the sign of the exponent $1-p$ : for $p > 1$ it is negative, so $T^{1-p} \to 0$ and the limit is finite; for $p < 1$ it is positive, so $T^{1-p} \to \infty$ . The case $p = 1$ was done separately — it gives a logarithm, and divergence.

$p$	$\int_1^{\infty} \tfrac{dx}{x^p}$
$0.5$	divergent
$1$	divergent
$2$	$1$
$3$	$\tfrac{1}{2}$
$4$	$\tfrac{1}{3}$

Second kind: a function unbounded at an endpoint

The second kind breaks not at the far end of the interval but inside it: the interval is short and the function runs off to infinity within it. So we step away from the awkward endpoint by $\varepsilon$ and squeeze that gap to zero:

\int_a^b f(x)\,dx = \lim_{\varepsilon \to 0^+} \int_{a+\varepsilon}^{b} f(x)\,dx

Decide whether ∫₀¹ dx/√x converges and find its value.

The region between y = 1/√x and the axis, from 0 to 1. There is no ceiling at the left edge — the curve escapes past the top of the drawing — and the area is still 2.

The p-test at zero

The same family of functions, the same exponent, but a different endpoint — and the inequality flips over:

\int_0^1 \frac{dx}{x^p} = \begin{cases} \dfrac{1}{1-p}, & p < 1 \quad \text{(convergent)} \\[2mm] \infty, & p \geq 1 \quad \text{(divergent)} \end{cases}

The two rules are worth keeping side by side, because it is their contrast that actually teaches something:

where	convergent when	what the function must do
$\int_1^{\infty} \tfrac{dx}{x^p}$	$p > 1$	decay fast
$\int_0^1 \tfrac{dx}{x^p}$	$p < 1$	grow slowly

The directions are opposite because the trouble is different in each place. Far from the origin it is the length of the interval, so the function has to vanish quickly enough. At a pole it is the height of the function, so it must not grow too steeply. The exponent $p = 1$ is the boundary neither rule crosses — which is why $\int_0^{\infty} \tfrac{dx}{x}$ diverges at both ends at once.

When the pole is inside the interval

This is where computing without thinking produces nonsense.

What is wrong with the computation ∫ dx/x² from −1 to 1 = [−1/x] from −1 to 1 = −1 − 1 = −2?

Exercises

The set asks two things. The first is the value of a convergent integral — the answer is a number or a fraction, e.g. 1/3. The second is the threshold on the exponent: rather than asking whether an integral converges (which would be answered with a word), it asks for which $p$ it converges, and you type the answer as an inequality, e.g. p>1 or p<1.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

∫₁^∞ 4/x² dx =

The bare verdict of convergent-or-divergent for one particular integral stays in the lesson text: it is a conclusion drawn from the number you are practising here, not a separate computation.

Common mistakes

Substituting infinity into the Newton–Leibniz formula — the symbol $\infty$ is not a number; the route runs through a moving edge $T$ and a limit.
Assuming that a function decaying to zero makes the integral convergent — $\tfrac{1}{x}$ decays to zero and its integral diverges. The rate is what counts, not the fact.
Flipping the p-test the wrong way — at infinity convergence needs $p > 1$ , at zero it needs $p < 1$ ; the two cases look identical and differ in everything.
Missing a pole inside the interval — integrating straight through a point where the function has no value can hand you a negative area for a region lying above the axis.
Assigning a value to a divergent integral — if the limit fails to exist or is infinite, the integral has no value, and $\infty$ must not be written as though it were one.
Forgetting which endpoint moves — for $\int_a^b$ with the pole at the upper endpoint it is $b$ that is stepped away from, not $a$ ; the formula has two mirror versions.

Formula card

Topic: Improper integrals

Improper integral, first kind
$\int_a^{\infty} f(x)\,dx = \lim_{T \to \infty} \int_a^{T} f(x)\,dx$
an unbounded interval of integration
Improper integral, second kind
$\int_a^b f(x)\,dx = \lim_{\varepsilon \to 0^+} \int_{a+\varepsilon}^{b} f(x)\,dx$
the function is unbounded at the endpoint a
The p-test at infinity
$\int_1^{\infty} \frac{dx}{x^p} = \frac{1}{p-1} \quad \text{for } p > 1$
divergent for p ≤ 1
The p-test at zero
$\int_0^1 \frac{dx}{x^p} = \frac{1}{1-p} \quad \text{for } p < 1$
divergent for p ≥ 1
An exponential tail
$\int_0^{\infty} e^{-ax}\,dx = \frac{1}{a} \quad (a > 0)$
convergent for every positive a
Finite area, infinite height
$\int_0^1 \frac{dx}{\sqrt{x}} = 2$
the flagship example of the second kind

The region under y = 1/x² from x = 1 rightwards. It runs on without end, yet its area is exactly 1 — the drawing shows only its beginning, because the frame is finite and the region is not.

The region under y = 1/√x between 0 and 1. At the left end the function escapes to infinity, so the region has no ceiling — and its area is 2.

Frequently asked questions

A luminous parabola and a straight line crossing above a faint coordinate grid, the vertex flaring like a star, against a cosmic nebula in violet and magenta

Functions and analysis

Why a function's graph can never double back

The whole definition of a function comes down to one word: uniqueness. One argument, one value — and not a single one more. That seemingly minor condition decides why a circle is not the graph of a function, why a line and a parabola describe so much of the world, and why four datasets with identical summary statistics can look nothing alike.

12 min