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An integral over an unbounded interval is defined as a limit, and for a plain power the $p$-test settles it: $\int_1^{\infty}x^{-p}dx$ converges exactly when $p > 1$, to $\dfrac{1}{p-1}$.
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 can evaluate an improper integral over an unbounded interval as the limit it is defined to be, writing the limit before you integrate and taking it afterwards. You can apply the $p$-test, including remembering which side of it the boundary case falls on, give the value of a convergent one, say why an exponential tail always converges, and use a comparison in the one direction that settles anything.
You can evaluate a definite integral between two numbers, and you have met limits. An improper integral is those two ideas put together, in the only order that makes sense: integrate between two numbers first, then let one of them run away.
Improper integral: one whose interval is unbounded, or whose integrand is unbounded on it.
Converges: the limit exists and is finite — and the integral is that limit.
Diverges: the limit is infinite or does not exist.
The $p$-test: $\int_1^{\infty}x^{-p}dx$ converges exactly when $p > 1$, and then equals $\dfrac{1}{p-1}$.
$\int_1^{\infty}f(x)\,dx$ is defined as
$$\lim_{b\to\infty}\int_1^{b}f(x)\,dx,$$
and that definition is the whole of the method. Infinity is not a number, so it cannot be substituted into an antiderivative; writing the limit down is not ceremony, it is the only thing that makes the symbol mean anything.
The $p$-test settles almost every example. For $\int_1^{\infty}x^{-p}dx$ the antiderivative is $\dfrac{x^{1-p}}{1-p}$. If $p > 1$ the exponent $1-p$ is negative, so $b^{1-p}\to 0$ and the integral converges to $\dfrac{1}{p-1}$. If $p < 1$ the exponent is positive and $b^{1-p}$ runs away. And if $p = 1$ the antiderivative is $\ln b$, which also runs away — slowly, but without limit.
So the boundary is exactly one, and it belongs to the divergent side. That is the fact to carry out of this lesson. The integrand $\tfrac1x$ goes to zero and its integral is still infinite, which is the first and cheapest warning that an integrand shrinking to nothing guarantees nothing at all.
Exponentials are in a different class. $\int_0^{\infty}ke^{-kx}dx = 1$ for every $k > 0$. A negative exponential eventually falls below every power of $x$, so an exponential tail always converges — which makes it the standard thing to compare an awkward integrand against when no antiderivative can be written.
Another way: picture
Picture painting under the curve with a roller that never lifts. Under $1/x^{2}$ you use a finite amount of paint however far you go, because each new stretch needs so much less than the last. Under $1/x$ you never stop needing more — each doubling of the distance costs the same amount again, for ever.
Another way: steps
| The exponent $p$ | What $b^{1-p}$ does | The integral |
|---|---|---|
| below $1$ | grows without limit | diverges |
| exactly $1$ | the antiderivative is $\ln b$ | diverges |
| above $1$ | falls to zero | converges, to $\dfrac{1}{p-1}$ |
The middle row is the one that has to be learned separately, because the antiderivative changes shape there — a logarithm rather than a power — and no amount of staring at $\dfrac{x^{1-p}}{1-p}$ will tell you what happens when its denominator is zero.
Plenty of integrands have no elementary antiderivative, and for those the definition cannot be applied directly. The move is comparison: if $0 \le f(x) \le g(x)$ on the whole interval and $\int g$ converges, then $\int f$ converges too; if $\int f$ diverges then so does $\int g$.
The direction matters and is easy to get backwards. Being smaller than something divergent tells you nothing, and being larger than something convergent tells you nothing either. The two useful statements are: smaller than convergent means convergent, larger than divergent means divergent.
Substituting infinity. Writing $\left[-\tfrac1x\right]_1^{\infty}$ and putting $\infty$ into the bracket happens to give the right number here and is meaningless. The limit has to be written, and on a paper it is marked.
Putting the boundary on the wrong side. $p = 1$ diverges. Remembering the test as "converges when $p \ge 1$" gets the harmonic case exactly backwards, and that case reappears in every series lesson to come.
Thinking a shrinking integrand is enough. $\tfrac1x$ goes to zero and $\int_1^{\infty}\tfrac{dx}{x}$ is infinite. How fast it shrinks is the only thing that matters.
$\int_1^{\infty}\dfrac{dx}{x^{3}} = \lim_{b\to\infty}\int_1^{b}x^{-3}dx$.
Write the limit first.
$= \lim_{b\to\infty}\left[-\dfrac{1}{2x^{2}}\right]_1^{b} = \lim_{b\to\infty}\left(\dfrac12 - \dfrac{1}{2b^{2}}\right)$.
Now it is ordinary.
$= \tfrac12$, which is $\dfrac{1}{p-1}$ with $p = 3$, as the test promises.
$\int_1^{\infty}\dfrac{dx}{x} = \lim_{b\to\infty}\left[\ln x\right]_1^{b}$.
A logarithm, not a power.
$= \lim_{b\to\infty}\ln b$, which grows without limit.
Slowly, and for ever.
So it diverges. $\ln b$ passes $10$ only around $b = 22000$, which is why divergence here is so easy to disbelieve.
Read the exponent: $p = \tfrac43$, which is above one.
The only question the test asks.
So it converges, and the value is $\dfrac{1}{p-1} = \dfrac{1}{1/3} = 3$.
The test gives the value too.
Notice how close $\tfrac43$ is to the divergent boundary, and how ordinary the answer $3$ is. Nothing dramatic happens on either side of $p = 1$; the behaviour simply changes.
Put the steps of evaluating $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{6}}$ into the order you do them.
Number the steps in order (write the number in the box):
Each integral runs from $1$ to infinity. Say whether it converges or diverges.
| Verdict | |
|---|---|
| $\displaystyle\int_1^{\infty} x^{-1/2}\,dx$ | |
| $\displaystyle\int_1^{\infty} \dfrac{dx}{x}$ | |
| $\displaystyle\int_1^{\infty} x^{-2}\,dx$ | |
| $\displaystyle\int_1^{\infty} x^{-5}\,dx$ |
Match each integral, taken from $1$ to infinity, to the reason it settles as it does.
| Converges: the exponent is above one | Diverges: the exponent is exactly one | Diverges: the exponent is below one | Converges: an exponential falls faster than any power | |
|---|---|---|---|---|
| $\int_1^{\infty} 2x^{-2}\,dx$ | ||||
| $\int_1^{\infty} \dfrac{2}{x}\,dx$ | ||||
| $\int_1^{\infty} \dfrac{2}{\sqrt{x}}\,dx$ | ||||
| $\int_1^{\infty} 2e^{-x}\,dx$ |
Does $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{3}}$ converge or diverge?
$\displaystyle\int_1^{\infty} \dfrac{dx}{x^{3}}$ converges. What is its value?
Answer:
Select every integral below that converges.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of evaluating $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{5}}$ into the order you do them.
Number the steps in order (write the number in the box):
You can evaluate and test an integral that runs to infinity. Without looking: what replaces the infinity before you integrate, for which exponents does the $p$-test converge, and what happens at exactly one?
10. Your turn: does $\int_1^{\infty}\dfrac{dx}{x^{4/3}}$ converge?, step 3