Definition 5.3.1.
These two types of integral (where there is a vertical asymptote at one of the endpoint, or where the interval of integration is infinite) are called improper integrals.
Section 5.3 Improper Integrals
Subsection 5.3.1 Asymptotes and Infinite Domains
It is sometimes useful to consider definite integrals where the function has an asymptote at one (or both!) of the bounds of integration. A function which is undefined in the middle of the interval cannot be integrated, but a function which has an asymptote at the edge might still have a reasonable area under its curve. Similarly, it is also useful to consider integration over infinite intervals, such as the entire real number line or \([0, \infty)\text{.}\) Both of these goals present problems for the integral as defined so far. To solve them, I need an additional technique. First, though, let me define a term to refer to these integrals.
The approach to improper integrals is quite direct. If there is an asymptote at one edge of the domain of integration, say at \(c_0\text{,}\) I set the bounds of integation to be a variable \(a\) and take the limit \(a \rightarrow c_0\text{.}\) Similarly, if the domain extends to infinity, as in \([0,\infty)\text{,}\) I integrate over \([0, a]\) and take the limit as \(a
\rightarrow \infty\text{.}\) In either case, a limit of the bounds of integration solves the problem.
Since there is a limit involved, an improper integral may or may not converge. The area under the curve might be infinite, or just undefined if the limit does not exist. The usual question for an improper integral is the convergence question: is the described area finite or infinite?
Definition 5.3.2.
If there is a finite area, if the limit exists, the improper integral converges. Otherwise, the improper integral diverges.
There will be some interesting discussion of the implications of divergence and convergence this in Section 8.1. For now, let me move on to some examples.
Example 5.3.3.
Let me try to calculate the area under the curve of \(\frac{1}{\sqrt{x}}\) on the interval \([0,2]\text{.}\) There is an asymptote at the left endpoint, so this is an improper integral. The function is positive, so there is a well-defined positive area under the curve. The question is whether this area is finite or infinite as the function grows along the asymptote.
I calculate the integral by taking a limit approaching the left endpoint. Note that I need a one-sided limit here: I can only approach \(0\) from the positive side.
\begin{equation*}
\int_0^2 \frac{1}{\sqrt{x}} dx = \lim_{a \rightarrow 0^+}
\int_a^2 \frac{1}{\sqrt{x}} dx
\end{equation*}
Then I can just calculate the integral with the bound \(a\) and take the limit.
\begin{equation*}
= \lim_{a \rightarrow 0^+} 2x^{\frac{1}{2}} \Bigg|_a^2
= \lim_{a \rightarrow 0^+} 2\sqrt{2} - 2 \sqrt{a} = 2\sqrt{2}
- 0 = 2\sqrt{2}
\end{equation*}
I get a nice, finite answer. The area in Figure 5.3.4, even though the asymptote grows to infinity, is finite.
Example 5.3.5.
Here is an example with an integral over an infinte domain. There is no veritcal asymptote, but the area extends out along the \(x\) axis to infinity, as show in Figure 5.3.6.
\begin{equation*}
\int_1^\infty \frac{1}{x^2} dx
\end{equation*}
It’s not at all obvious how much area there is under the integral as the bounds go off to infinity. The limit will indicate whether this in infinite or finite area. I set the upper bound to \(a\) and take the limit as \(a \rightarrow
\infty\text{.}\)
\begin{equation*}
\int_1^\infty \frac{1}{x^2} dx = \lim_{a \rightarrow \infty}
\int_1^a \frac{1}{x^2} dx = \lim_{a \rightarrow \infty}
\frac{-1}{x} \Bigg|_1^a = \lim_{a \rightarrow
\infty} \frac{-1}{a} + 1 = 1
\end{equation*}
In this way, I show that the unknown area under the curve of \(\frac{1}{x^2}\) is indeed finite, even when the bound goes to infinity.
Example 5.3.7.
\begin{equation*}
\int_1^\infty \frac{1}{x} dx
\end{equation*}
This is very similar to the previous example, but the antiderivative of \(\frac{1}{x}\) differs from that of \(\frac{1}{x^2}\) in an important way. I use a limit to evaluate the integral on the infinite interval.
\begin{equation*}
\int_1^\infty \frac{1}{x} dx = \lim_{a \rightarrow \infty}
\int_1^a \frac{1}{x} dx = \lim_{a \rightarrow \infty}
\ln |x| \Bigg|_1^a = \lim_{a \rightarrow \infty} \ln a - 0
= \infty
\end{equation*}
The shape of the graphs of \(\frac{1}{x^2}\) and \(\frac{1}{x}\) on the interval \([1,\infty)\) are very similar: they both start at \((1,1)\) and decay to zero. However, \(\frac{1}{x}\) decays slower. This difference in decay rate is enough that the area under \(\frac{1}{x^2}\) is finite and the area under \(\frac{1}{x}\) is infinite, which is somewhat surprising if you just look at the graph. These two graphs are shown in Figure 5.3.8. In the graph, it is clear to see that \(\frac{1}{x}\) has greater area, but it is hard to see from the image that the difference is enough to actually produce infinite area.
Example 5.3.9.
I want to look more carefully at the previous two examples. Both had integrands of the form \(\frac{1}{x^p}\) for some exponent \(p\text{.}\) I showed that \(p=1\) gives a divergent integral and \(p=2\) gives a convergent integral. If \(p
\lt 1\) then \(\frac{1}{x^p} > \frac{1}{x}\text{.}\) If one positive function is larger than another, then the area under its graph must be larger. Since the area under \(\frac{1}{x}\) between \(1\) and \(\infty\) is already \(\infty\text{,}\) the integral of \(\frac{1}{x^p}\) on \((1,\infty)\) must also be \(\infty\) when \(p \lt
1\)
Now, if \(p \gt 1\text{,}\) I want to do the improper integral in general.
\begin{align*}
\int_1^\infty \frac{1}{x^p} \amp = \lim_{a \rightarrow
\infty} \int_1^a \frac{1}{x^p} dx = \lim_{a \rightarrow
\infty} \frac{-1}{(p-1)x^{p-1}} \Bigg|_1^a \\
\amp = \lim_{a \rightarrow \infty} \frac{1}{p-1}
-\frac{1}{(p-1)a^{p-1}} = \frac{1}{p-1}
\end{align*}
For, for \(p \gt 1\text{,}\) these improper integrals all converge. However, as \(p\) gets close to \(1\text{,}\) the values of the integrals get very large. \(p=1\) is the crossover point, where the area under the curve becomes infinite.
Subsection 5.3.2 Improper Integrals and Comparisons
I’ve already implicitly used a comparison argument in Example 5.3.9. I said that \(\frac{1}{x^p} > \frac{1}{x}\) for \(p \lt 1\) and used that fact to argue convergence. This kind of comparsion is a general property of integrals: if \(f(x) \geq g(x)\) on \([a,b]\text{,}\) then
\begin{equation*}
\int_a^b f(x) dx \geq \int_a^b g(x)
\end{equation*}
The same inequality holds for improper integrals, at least adopting the convention that that \(\infty > a\) for any finite number \(a\text{.}\) (While this comparison is reasonable, infinity is still not a number and I will not do arithmetic with it). This type of comparison is useful for evaluating improper integrals. However, a more useful comparison makes use of asymptotic analysis. The following theorem is presented here wihtout proof.
Theorem 5.3.10.
Consider an improper integral with integrand \(f(x)\text{.}\) The asymptotic order of \(f(x)\) determines the convergence of an improper integral. That is, if \(f(x)\) is replaced with any other function with the same asymptotic order, then the integral will have the same convergence behaviour.
What does this mean? If means that if you only care about convergence/divergence (and not the exact value of the integral), then you only need to consider the asymptotic order of the integrand. This means you can deal with much easier integrals to determine convergence. Let me do an example which illustrates this.
Example 5.3.11.
\begin{equation*}
\int_1^\infty \frac{1}{4 + x + 18x^2}
\end{equation*}
The asymptotic order of the denominator is \(x^2\text{.}\) Therefore, the convengence behaviour is the same as if the integrand were \(\frac{1}{x^2}\text{.}\) This is \(p=2\text{,}\) which is \(p \gt 1\text{,}\) so the integral converges. This use of asymptotic order allows me to dramatically simply the situation: I don’t have to evaluate the complicated quadratic integral to determine convergence or divergence. (I would, of course, have to evaluate the complicated quadratic integral if I wanted to determine the exact value of the improper integral).
The previous example is an example of a relatively common occurance. Often, the important fact is convergence. If there is an improper integral that I want to work with, that I want to measure something, I need to make sure it converges. I might not need to know its value (at least not right away); I might just want to do some other proofs or calculations with it. Therefore, often using asymptotical analysis to quickly test convergence allows progress in some calculations, even though the value of the integral hasn’t be exactly determined.
