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An unbounded integrand on a bounded interval: pull back by $\varepsilon$ and take a one-sided limit, with the $p$-test the other way up.
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 find every point of a closed interval where an integrand is unbounded, including points strictly inside it, split an integral so that each piece has exactly one bad end, pull back from a singularity by $\varepsilon$ and take the one-sided limit, apply the $p$-test at a singularity in the correct direction and say why it points the opposite way from the test at infinity, refuse to cancel a $+\infty$ against a $-\infty$ across an interior singularity, and tell a genuine singularity from a removable discontinuity.
The definite integral needed a bounded interval and a bounded function. The last lesson dropped the first. This one drops the second.
The definition is the same manoeuvre: pull back from the bad point by $\varepsilon$, integrate an ordinary integral, and let $\varepsilon \to 0$. What is genuinely new is that the $p$-test comes out the other way up, and that the bad point can hide in the middle of the interval where nothing in the notation points at it.
A singularity of the integrand is a point where it is undefined or unbounded. It may be an endpoint of the interval or strictly inside it.
The limit is one-sided: $\varepsilon \to 0^{+}$, approaching the bad point from the side the interval lies on. A singularity inside the interval needs two one-sided limits, one from each side, and both must exist.
An integral may be improper in both ways at once — $\int_0^\infty x^{-1/2}(1+x)^{-1}dx$ is unbounded at $0$ and unbounded in extent. Split it into one piece per trouble spot and treat each separately.
Definition. If $f$ is integrable on $[a + \varepsilon, b]$ for every small $\varepsilon > 0$ but unbounded near $a$, $$\int_a^b f := \lim_{\varepsilon \to 0^{+}}\int_{a + \varepsilon}^{b}f,$$ and similarly for a singularity at $b$. The integral converges when that limit is finite.
A singularity inside. If $f$ blows up at $c$ with $a < c < b$, then $$\int_a^b f := \int_a^c f + \int_c^b f,$$ and both pieces must converge. Nothing cancels across the bad point, however symmetric the picture looks.
The $p$-test at zero. $$\int_0^1\frac{dx}{x^q}\text{ converges} \iff q < 1, \text{ and then equals } \frac{1}{1-q}.$$ Compare the test at infinity, which needs $p > 1$. Same borderline, opposite side.
Why they point opposite ways. Out at infinity the danger is a tail that decays too slowly, so you need a large exponent. At zero the danger is a spike that is too tall, so you need a small one. $x^{-1/2}$ is integrable near $0$ and not near $\infty$; $x^{-2}$ is the reverse. Only $x^{-1}$ fails at both, which is what makes it the dividing line.
Removable trouble is not trouble. $\dfrac{\sin x}{x}$ is undefined at $0$ but bounded near it, so $\int_0^1\frac{\sin x}{x}dx$ is a perfectly ordinary integral. What matters is unboundedness, not a missing value.
A test is its hypotheses. Every convergence test in this course is a theorem with conditions, and a test quoted where its conditions fail has not been applied — it has been guessed with. Naming the hypothesis you checked is part of the answer, not a flourish on top of it.
Another way: picture
Two spikes at the origin: $x^{-1/2}$ and $x^{-2}$, both running off the top of the page. The first encloses a finite area — the region is tall but narrows fast — and the second does not. As with the tails, the picture cannot tell them apart, and the exponent can.
Another way: steps
| Integral | Converges when | Value |
|---|---|---|
| $\displaystyle\int_1^\infty x^{-p}dx$ | $p > 1$ | $\dfrac{1}{p-1}$ |
| $\displaystyle\int_0^1 x^{-q}dx$ | $q < 1$ | $\dfrac{1}{1-q}$ |
The pair is worth holding together, because each is the other's mirror. Under the substitution $x \mapsto 1/x$ the interval $[1, \infty)$ becomes $(0, 1]$ and the exponent $p$ becomes $2 - p$ — which is exactly the map that carries one test to the other, and explains why the borderline is fixed at $1$: it is the exponent that maps to itself.
The practical consequence is that $\displaystyle\int_0^\infty x^{-p}dx$ diverges for every $p$. There is no exponent that is small enough at zero and large enough at infinity, because the conditions are $p < 1$ and $p > 1$. Any integral of a pure power over the whole positive axis diverges, and the only way to have a convergent integral over $(0,\infty)$ is for the function to behave differently at the two ends.
Missing a singularity inside the interval. The commonest and the worst, because the fundamental theorem applied across it produces a number, and often a plausible one. $\int_{-1}^{1}x^{-2}dx$ "equals" $-2$ by that route, for a positive integrand.
Using the test from infinity. $q < 1$ at a singularity, $p > 1$ at infinity. Writing down the wrong one gets the answer exactly backwards.
Cancelling across the bad point. $\int_{-1}^{1}\frac{dx}{x}$ is not $0$; both halves diverge, and the definition requires each separately.
Treating a removable discontinuity as a singularity. $\frac{\sin x}{x}$ near $0$ is bounded. There is nothing improper about it.
Splitting at the wrong point. Each piece must have exactly one bad end, so split at the singularity, not near it.
Reporting convergence when only one piece converges. $\int_0^\infty x^{-1/2}dx$ has a convergent piece near zero and a divergent tail. It diverges.
Having learned that $\int_1^\infty x^{-p}$ converges for $p > 1$, learners reliably carry "big exponent good" to the singular case, where it is exactly wrong. The mistake is not carelessness; it comes from remembering a rule instead of what the rule is about.
The two situations have opposite dangers. At infinity the region is infinitely long, so the risk is that the height does not fall fast enough — you want the integrand small, which means a large exponent. At a singularity the region is infinitely tall, so the risk is that the spike is too severe — you want the integrand mild, which means a small exponent.
Hold that and neither rule has to be memorised, because both can be rederived in one line: write the antiderivative $\dfrac{x^{1-p}}{1-p}$ and ask what happens to $x^{1-p}$ at the end in question. At infinity you need the exponent negative; at zero you need it positive. Same antiderivative, same question, opposite answers.
It also explains the fact that surprises everybody: $\int_0^\infty x^{-p}dx$ diverges for every $p$, because no exponent satisfies both conditions at once. A convergent integral over the whole positive axis must be a function that behaves differently at its two ends, and no pure power does.
$\displaystyle\int_0^{\infty}\frac{dx}{\sqrt x\,(1 + x)}$: unbounded at $0$, unbounded interval. Split at $1$.
One trouble spot per piece.
Near $0$: the integrand behaves like $x^{-1/2}$, and $q = \tfrac12 < 1$, so that piece converges. Near $\infty$: it behaves like $x^{-3/2}$, and $p = \tfrac32 > 1$, so the tail converges too.
Each end, its own test.
Both converge, so the whole does. In fact the substitution $u = \sqrt x$ evaluates it: $2\displaystyle\int_0^\infty\frac{du}{1 + u^2} = \pi$. The function is small enough at zero and decays fast enough at infinity, which no single power can do.
$\displaystyle\int_{-2}^{2}\frac{dx}{x}$. An antiderivative is $\ln|x|$, and $\ln 2 - \ln 2 = 0$, which looks tidy and symmetric.
The tempting calculation.
But $1/x$ is undefined at $0$, inside the interval, so the fundamental theorem does not apply and the number means nothing.
Check the interval before evaluating.
Split: $\displaystyle\int_{\varepsilon}^{2}\frac{dx}{x} = \ln 2 - \ln\varepsilon \to +\infty$. The right half diverges, so the integral does; the left half diverges to $-\infty$, and the definition does not permit those to cancel. The symmetric limit $\lim_{\varepsilon\to 0}\left(\int_{-2}^{-\varepsilon} + \int_{\varepsilon}^{2}\right)$ is indeed $0$, and that object has its own name — the principal value — precisely because it is not the integral.
$\ln x \to -\infty$ as $x \to 0^{+}$, so the left end is singular; pull back to $\varepsilon$.
Unbounded, though only logarithmically.
Parts with $u = \ln x$, $dv = dx$: $\displaystyle\int_{\varepsilon}^{1}\ln x\,dx = \left[x\ln x - x\right]_{\varepsilon}^{1} = -1 - \varepsilon\ln\varepsilon + \varepsilon$.
An ordinary integration inside the limit.
As $\varepsilon \to 0^{+}$, $\varepsilon\ln\varepsilon \to 0$ — the linear factor beats the logarithm — so the integral converges to $-1$. Negative, correctly, since $\ln x < 0$ on $(0,1)$.
Evaluate $\displaystyle\int_0^{1}x^{-1/5}\,dx$.
Answer:
Where is $\displaystyle\int_0^{6}\frac{dx}{\sqrt{x}\,(x - 3)}$ improper?
Match each integral to what makes it improper.
| An infinite interval | An unbounded integrand at an endpoint | Both an unbounded integrand and an infinite interval | An unbounded integrand strictly inside the interval | |
|---|---|---|---|---|
| $\displaystyle\int_{7}^{\infty}\frac{dx}{x^{2}}$ | ||||
| $\displaystyle\int_0^{7}\frac{dx}{\sqrt{x}}$ | ||||
| $\displaystyle\int_0^{\infty}\frac{dx}{\sqrt{x}\,(1 + x)}$ | ||||
| $\displaystyle\int_{-7}^{7}\frac{dx}{x^{2}}$ |
For which $q$ does $\displaystyle\int_0^{9}\frac{dx}{x^{q}}$ converge? Give the set of $q$ as an interval.
This task has no paper form; do it on a device.
Evaluate $\displaystyle\int_0^{25}\frac{dx}{\sqrt{x}}$ in stages, with the bad end replaced by $\varepsilon$.
| Value | |
|---|---|
| The antiderivative at the good end | |
| Its limit at the bad end | 0 |
| The value of the integral |
A student computes $\displaystyle\int_{-5}^{5}\frac{dx}{x^{2}} = \left[-\frac{1}{x}\right]_{-5}^{5}$. What number does that calculation produce?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Evaluate $\displaystyle\int_0^{1}x^{-1/4}\,dx$.
Answer:
You can classify an integral improper at either kind of bad point, and find the bad points that are not at the ends. Say in your own words why $\int_0^\infty x^{-p}dx$ diverges for every $p$. Next: what to do when the integral cannot be evaluated at all.
10. Your turn: $\displaystyle\int_0^{1}\ln x\,dx$, step 3