Explain the "you should realize" comment in a concrete context by stating a reason why \(\displaystyle \int_2^\infty e^{-3 \ln (\ln x)} \, dx\) converges if and only if \(\displaystyle \int_6^\infty e^{-3 \ln (ln x)} \, dx\) converges.
Unit 12.2 Convergence
The central question of this section is: how do we tell whether a limit such as \(\int_b^\infty f(x) \, dx\) exists? If the limit exists, we would like to evaluate it if possible, or estimate it otherwise. When discussing convergence you should realize that \(\int_a^\infty f(x) \, dx\) either diverges for all values of \(a\) or converges for all values of \(a\) as long as \(f\) is defined and continuous on \([a,\infty)\text{.}\) For this reason, we use the notation \(\int^\infty f(x) \, dx\) or, to be really blunt, \(\displaystyle \int_{\rm who cares}^\infty f(x) \, dx\text{.}\)
Checkpoint 173.
Case 1: you know how to compute the definite integral.
Suppose \(\int_b^M f(x) \, dx\) is something for which you know how to compute an explicit formula. The formula will have \(M\) in it. You have to evaluate the limit as \(M \to \infty\text{.}\) How do you do that? There is no one way, but that’s why we studied limits before. Apply what you know. What about \(b\text{,}\) do you have to take a limit in \(b\) as well? I hope you already knew the answer to that. In this definition, \(b\) is any fixed number. You don’t take a limit.
These special cases will become theorems once you have worked them out.
| Name of test | Type of integral | Condition for convergence |
| \(\displaystyle \int_b^\infty e^{kx} \, dx\) | ||
| power test | \(\displaystyle \int_b^\infty x^p \, dx\) | |
| \(\displaystyle \int_b^\infty \frac{(\ln x)^q}{x} \, dx\) |
You will work out these cases in class: write each as a limit, evaluate the limit, state whether it converges, which will depend on the value of the parameter, \(k, p\) or \(q\text{.}\) Go ahead and pencil them in once you’ve done this. The second of these especially, is worth remembering because it is not obvious until you do the computation where the break should be between convergence and not.
Checkpoint 174.
Case 2: you don’t know how to compute the integral.
In this case you can’t even get to the point of having a difficult limit to evaluate. So probably you can’t evaluate the improper integral. But you can and should still try to answer whether the integral has a finite value versus being undefined. This is where comparison tests come in. You build up a library of cases where you do know the answer and then, for the rest of functions, you try to compare them to functions in your library.
Sometimes a comparison is informative, sometimes it isn’t. Suppose that \(f\) and \(g\) are positive functions and \(f (x) \leq g(x)\) for all \(x\text{.}\) Consider several pieces of information you might have about these functions.
| what you know | conclusion | |
| (a) | \(\int_b^\infty f(x) \, dx\) converges to a finite value \(L\text{.}\) | |
| (b) | \(\int_b^\infty f(x) \, dx\) does not converge. | |
| (c) | \(\int_b^\infty g(x) \, dx\) converges to a finite value \(L\text{.}\) | |
| (d) | \(\int_b^\infty g(x) \, dx\) does not converge. |
In which cases can you conclude something about the other function? We are doing this in class. Once you have the answer, either by working it out yourself or from the class discussion, please pencil it in here so you’ll have it for later reference.
Checkpoint 175.
Suppose you want to show that \(\displaystyle \int_1^\infty \frac{3 + \sin(x)}{x^2} \, dx\) converges. Which pair of facts allows you to do this?
-
\(\frac{3 + \sin x}{x^2} \leq \frac{4}{x^2}\) and \(\int^\infty \frac{4}{x^2} \, dx\) does not converge
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\(\frac{3 + \sin x}{x^2} \leq \frac{4}{x^2}\) and \(\int^\infty \frac{2}{x^2} \, dx\) does not converge
Asymptotic comparison tests.
Here are two key ideas that help your comparison tests work more of the time, based on the fact that the question "convergent or not?" is not sensitive to certain differences between integrands.
Multiplying by a constant does not change whether an integral converges. That’s because if \(\displaystyle \lim_{M \to \infty} \int_b^M f(x) \, dx\) converges to the finite constant \(L\) then \(\displaystyle \lim_{M \to \infty}
\int_b^M K f(x) \, dx\) converges to the finite constant \(KL\text{.}\)
Checkpoint 176.
Does \(\displaystyle \int_1^\infty \frac{10}{x} \, dx\) converge or not?
-
yes, this integral converges.
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no, this integral diverges.
In either case, give a reason why.
If the integral does not converge, is the value \(\infty\) or \(-\infty\) or is it truly undefined?
-
the integral converges.
-
the integral diverges to \(-\infty\text{.}\)
-
the integral diverges to \(\infty\text{.}\)
-
the integral is ’truly undefined’.
It doesn’t matter if \(f(x) \leq g(x)\) for every single \(x\) as long as the inequality is true for sufficiently large \(x\text{.}\) If \(f(x) \leq g(x)\) once \(x \geq 100\text{,}\) then you can apply the comparison test to compare \(\int_b^\infty f(x) \, dx\) to \(\int_b^\infty g(x) \, dx\) as long as \(b \geq 100\text{.}\) But even if not, once you compare \(\int_{100}^\infty f(x) \, dx\) to \(\int_{100}^\infty
g(x) \, dx\text{,}\) then adding the finite quantity \(\int_b^{100} f(x) \, dx\) or \(\int_b^{100} g(s) \, dx\) will not change whether either of these converges.
Putting these two ideas together leads to the conclusion that if \(f(x) \leq K g(x)\) from some point onward and \(\int_b^\infty g(x) \, dx\) converges, then so does \(\int_b^\infty f(x) \, dx\text{.}\) The theorem we just proved is:
Theorem 12.10. asymptotic comparison.
If \(f\) and \(g\) are positive functions on some interval \((b,\infty)\) and if there are some constants \(M\) and \(K\) such that
\begin{equation}
f(x) \leq K g(x) \mbox{ for all } x \geq M\tag{12.4}
\end{equation}
then convergence of the integral \(\int_b^\infty g(x) \, dx\) implies convergence of the integral \(\int_b^\infty f(x) \, dx\text{.}\)
Checkpoint 177.
Let \(f(x) := 3x^3 / (x - 17)\) and \(g(x) := x^2\text{.}\) Is it true that \(f(x) \leq K g(x)\) from some point onward?
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Yes, if we pick \(K\) large enough, then at some point \(f(x)\) starts being less than \(Kg(x)\) and stays that way.
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No. No matter how large we pick \(K\text{,}\) we’ll be able to find large \(x\) so that \(f(x)\gt Kg(x)\text{.}\)
Explain your answer.
Example 12.11.
Does \(\displaystyle \int_1^\infty x^8 e^{-x} \, dx\) converge? One way to do this is by computing the integral exactly. This takes eight integrations by parts, and is probably too messy unless you figured out how to do "tabular" integration (optional when you learned integration by parts). In any case, there’s an easier way if you only want to know whether it converges, but not to what.
We claim that \(x^8 e^{-x} \ll e^{- (1/2) x}\) (you could use \(e^{-\beta x}\) in this argument for any \(\beta \in (0,1)\)). It follows from the asymptotic comparison test that convergence of \(\int_1^\infty e^{-(1/2) x}\) implies convergence of \(\int x^8 e^{-x} \, dx\text{.}\) We check the claim by evaluating
\begin{equation*}
\displaystyle\lim_{x \to \infty} \frac{x^8 e^{-x}}{e^{-(1/2) x}} =
\lim_{x \to \infty} \frac{x^8}{e^{(1/2) x}} = 0
\end{equation*}
because we know the power \(x^8\) is much less than the exponential \(e^{(1/2) x }\text{.}\)
Checkpoint 178.
A particular case of Theorem 12.10 is when \(f(x) \sim g(x)\text{.}\) When two functions are asymptotically equivalent, then each can be upper bounded by a constant multiple of the other, hence we have the following proposition.
Proposition 12.12.
If \(f\) and \(g\) are positive functions and \(f \sim g\) then \(\int^\infty f(x) \, dx\) converges if and only if \(\int^\infty g(x) \, dx\) converges.
Example 12.13.
- Does \(\displaystyle \int_1^\infty \frac{dx}{x^2 + 3x}\) converge?
- We can use comparison test (c) here: \(\displaystyle \frac{1}{x_2 + 3x} \leq \frac{1}{x^2}\) and we know \(\displaystyle \int_1^\infty \frac{dx}{x^2}\) converges, hence so does \(\displaystyle \int_1^\infty \frac{dx}{x^2 + 3x}\text{.}\)
- Does \(\displaystyle \int_4^\infty \frac{dx}{x^2 - 3x}\) converge?
- Now the inequatlity goes the other way, so we are in case (c) of the comparison test and we cannot conclude anything from direct comparison. However, we also know \(\displaystyle \frac{1}{x^2 - 3x} \sim \frac{1}{x^2}\) as \(x \to \infty\text{,}\) therefore we can conclude convergence again by Proposition 12.12.
Did you wonder about the lower limit of 4 in the second part? That wasn’t just randomly added so you’d be more flexible about the lower limits of integrals to infinity. It was put there to ensure that \(f\) was continuous; note the discontinuity at \(x=3\text{.}\)
