Back to the on-screen lesson ·
An integral to infinity is defined as a limit of proper ones, and $\int_1^\infty x^{-p}$ converges exactly when $p > 1$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state the definition of an improper integral over an infinite interval as a limit, replace an infinite endpoint by a letter rather than substituting into an antiderivative, apply the $p$-test at infinity and say why the borderline $p = 1$ diverges, split an integral with two infinite ends and require both pieces to converge rather than cancelling them, say why a divergent integral has no value rather than the value $\infty$, and read the test backwards to find the exponent that produces a given value.
The definite integral $\int_a^b f$ was built for a bounded function on a bounded interval. Both assumptions were used: the Darboux sums need finitely many finite rectangles, and the fundamental theorem needs $f$ continuous on a closed bounded interval.
Drop either and the construction does not apply — not because the answer is hard to find, but because nothing has been defined. $\int_1^\infty x^{-2}dx$ is not a difficult integral; until this lesson it is not an integral at all. What follows is the definition that gives it a meaning, and the definition is a limit.
An integral is improper when an endpoint is infinite (type I) or the integrand is unbounded near an endpoint (type II). This lesson is type I; the next is type II.
It converges when the defining limit exists and is finite, and its value is that limit. It diverges otherwise — including when the limit is $+\infty$, which is a description of how it diverges rather than a value.
The tail of a convergent improper integral is $\int_c^\infty f$ for large $c$. Convergence is exactly the statement that the tails shrink to nothing, which is what makes truncation an approximation with a bound.
Definition. If $f$ is integrable on $[a, b]$ for every $b > a$, $$\int_a^\infty f(x)\,dx := \lim_{b \to \infty}\int_a^b f(x)\,dx,$$ and the integral converges when that limit exists and is finite. Similarly $\int_{-\infty}^{b}$ with $a \to -\infty$.
Two infinite ends. $\int_{-\infty}^{\infty}f$ is defined as $\int_{-\infty}^{c}f + \int_c^{\infty}f$ for any $c$, and it converges only if both pieces do. It is emphatically not $\lim_{R\to\infty}\int_{-R}^{R}f$: that symmetric limit exists for $f(x) = x$, where it is $0$, while the integral itself diverges. The symmetric version is a different object with its own name (the principal value), and one of the pieces being $+\infty$ while the other is $-\infty$ is not a cancellation the definition permits.
The $p$-test at infinity. $$\int_1^\infty \frac{dx}{x^p} \text{ converges} \iff p > 1, \text{ and then equals } \frac{1}{p-1}.$$ At $p = 1$ the antiderivative is $\ln b$, which grows without bound — slowly, but without bound. This is the yardstick everything else is compared against.
Exponentials beat every power. $\int_0^\infty e^{-kx}dx = \tfrac1k$ for every $k > 0$, and $\int_1^\infty x^n e^{-x}dx$ converges for every $n$. Decay of the form $e^{-x}$ is the strongest kind you will meet here.
Finite area, infinite extent. There is nothing paradoxical about a region of infinite length and finite area: it only says the height falls fast enough. The same region rotated about the axis can have finite volume and infinite surface area, which is likewise not a paradox but a statement about two different integrals.
Another way: picture
Two tails on the same axes, $1/x$ and $1/x^2$, both heading to zero and looking much alike. The area under $1/x^2$ out to $b$ approaches $1$ and stops; the area under $1/x$ grows like $\ln b$ and never settles, though it takes $b \approx 22000$ to reach $10$. The picture cannot tell them apart, and that is exactly why a test is needed: the eye is not a convergence test.
Another way: steps
For $p \ne 1$ the antiderivative of $x^{-p}$ is $\dfrac{x^{1-p}}{1-p}$, and everything turns on the sign of $1 - p$. If $p > 1$ that exponent is negative, so $b^{1-p} \to 0$ and the integral settles. If $p < 1$ it is positive, so $b^{1-p} \to \infty$.
At $p = 1$ the power rule has nothing to say — dividing by $1 - p$ would be dividing by zero — and the antiderivative is $\ln x$ instead. That is not a technicality: the shape of the antiderivative genuinely changes at that exponent, and it changes to something unbounded. So the borderline diverges.
How marginal is it? $\int_1^b \frac{dx}{x^{1.01}}$ converges, to $100$. $\int_1^b\frac{dx}{x^{0.99}}$ diverges. And $\int_1^b\frac{dx}{x}= \ln b$ reaches only about $14$ by $b = 10^6$ — it diverges more slowly than any positive power grows. Nothing about the graphs of these three functions distinguishes them by eye; only the exponent does, which is the argument for having a test rather than an intuition.
Substituting $\infty$ into the antiderivative. Writing $\left[-\tfrac1x\right]_1^{\infty} = 0 - (-1)$ gets the right number by an argument that is not one, and it fails silently on $\int_1^\infty \frac{dx}{x}$, where $\left[\ln x\right]_1^\infty$ looks equally writable. Take a limit.
Cancelling infinities across a symmetric interval. $\int_{-\infty}^{\infty}x\,dx$ diverges. That the two halves are equal and opposite is not a reason to call it zero; the definition requires each half separately.
Splitting at the wrong place, or not at all. $\int_0^\infty \frac{dx}{x^2}$ is improper at both ends, and it must be split at some interior point. It diverges, because of the end at zero.
Applying the $p$-test to the wrong end. $p > 1$ is the condition at infinity. At a singular endpoint it is the opposite, as the next lesson shows.
Reporting $\infty$ as a value. A divergent integral has no value. "Diverges to $+\infty$" describes the behaviour; "equals $\infty$" asserts something the definition does not allow.
Assuming $f \to 0$ is enough. $1/x \to 0$ and its integral diverges. The terms going to zero is necessary and not sufficient — which is the same lesson the series tests will teach in four lessons' time.
The notation encourages the error. $\left[-\tfrac1x\right]_1^{\infty}$ looks like every other evaluation, and substituting "$\infty$" to get $0$ produces the right answer. So the habit forms, and it is not a habit — it is a sentence with no meaning, that happens to coincide with one that has.
Where it costs you is at the borderline. $\left[\ln x\right]_1^\infty$ is written just as easily and there is nothing to substitute: $\ln \infty$ is not a number, and the integral diverges. A learner substituting symbols has no way to notice, because the symbols look the same. A learner taking limits sees immediately that one limit exists and the other does not.
The deeper point is what the definition is doing. The integral was constructed for a bounded interval, and no construction so far assigns a meaning to an infinite one. The limit is not a technique for evaluating a pre-existing object; it is the definition of the object. That is why an integral whose limit does not exist has no value at all, rather than a value of $\infty$ — the definition simply fails to name anything.
So write the letter. It costs one symbol, it makes the divergent cases announce themselves, and it keeps the object and its definition attached to each other.
$\displaystyle\int_{-\infty}^{\infty}\frac{dx}{1 + x^2}$: two infinite ends, so split at $0$ and treat each separately.
One trouble spot per limit.
$\displaystyle\int_0^{b}\frac{dx}{1+x^2} = \arctan b \to \frac{\pi}{2}$, and by symmetry the other half is also $\frac\pi2$.
Each half converges on its own.
Both converge, so the whole does, to $\pi$. Had one half diverged the answer would be "diverges", however tidy the other half looked.
$\displaystyle\int_0^\infty x e^{-x}dx = \lim_{b\to\infty}\int_0^b x e^{-x}dx$. Parts with $u = x$, $dv = e^{-x}dx$ gives $\left[-xe^{-x} - e^{-x}\right]_0^b$.
Set the limit up first, then integrate ordinarily inside it.
$= 1 - be^{-b} - e^{-b}$. As $b \to \infty$, $be^{-b} \to 0$ — the exponential beats the linear factor — so the integral is $1$.
The limit is the last step, not the first.
The same argument gives $\int_0^\infty x^n e^{-x}dx = n!$ for every $n$, by the reduction formula of lesson 2. An exponential tail is strong enough to absorb any power.
Replace the infinity: $\displaystyle\int_2^{b}$, and substitute $u = \ln x$, $du = \dfrac{dx}{x}$.
The substitution is ordinary; only the ceiling is a letter.
The integral becomes $\displaystyle\int_{\ln 2}^{\ln b}\frac{du}{u^{2}} = \frac{1}{\ln 2} - \frac{1}{\ln b}$.
Change the limits with the variable.
As $b \to \infty$, $\ln b \to \infty$, so the integral converges to $\dfrac{1}{\ln 2}$. Compare $\int \frac{dx}{x\ln x}$, which gives $\ln\ln b$ and diverges: a second logarithm is enough to change the verdict.
Evaluate $\displaystyle\int_1^{\infty}\frac{dx}{x^{2}}$.
Answer:
Does $\int_1^\infty \dfrac{dx}{x^{3/2}}$ converge?
Put the steps of evaluating $\displaystyle\int_{7}^{\infty}\frac{dx}{x^{3}}$ into the order you do them.
Number the steps in order (write the number in the box):
For which $p$ does $\displaystyle\int_{6}^{\infty}\frac{dx}{x^{p}}$ converge? Give the set of $p$ as an interval.
This task has no paper form; do it on a device.
Evaluate $\displaystyle\int_{4}^{\infty}\frac{8\,dx}{x^{3}}$ in stages.
| Value | |
|---|---|
| The antiderivative at the lower limit | |
| The limit of the antiderivative at the ceiling | 0 |
| The value of the integral |
For which $p$ does $\displaystyle\int_1^{\infty}\frac{dx}{x^{p}} = \frac{1}{5}$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Evaluate $\displaystyle\int_1^{\infty}\frac{dx}{x^{9}}$.
Answer:
You can evaluate or classify an integral over an infinite interval, and say what the limit in the definition is doing. Say in your own words why $\int_{-\infty}^{\infty}x\,dx$ diverges although its symmetric partial integrals are all zero. Next: the other way an integral can be improper — an integrand that blows up.
10. Your turn: $\displaystyle\int_2^{\infty}\frac{dx}{x\,(\ln x)^{2}}$, step 3