Section 3.7 Improper Integrals
Some physical problems are naturally described in terms of an integral of all real values of a variable, or all positive values. For example, the variable might represent speed, with (in classical physics) no upper limit, leading to integrals like \(\ds E = \int_0^\infty v e^{-v^2} dv.\)
How do we make sense of the infinite value for the upper limit? The basic idea is to first approximate by integrating over “most” of the domain, up to some large but finite value \(M\text{:}\)
\begin{equation*}
E_M := \int_0^M v e^{-v^2} dv
\end{equation*}
and then define the exact value to be the limit of the ever-improving approximations given by increasing M:
\begin{equation*}
E = \int_0^\infty v e^{-v^2} dv := \lim_{M \to \infty} E_M = \lim_{M \to \infty} \int_0^M v e^{-v^2} dv.
\end{equation*}
Definition 3.7.1. Improper Integral, Type I: Infinitely Wide Domains.
-
The improper integral \(\ds\int_a^\infty f(x)\ dx\) is defined as
\begin{equation*}
\int_a^\infty f(x)\ dx := \lim_{M \to \infty} \int_a^M f(x)\ dx
\end{equation*}
if this limit exits; and then the integral is called convergent.
If the limit does not exit, this integral is called divergent. This includes the case of an infinite limit.
Likewise we define \(\ds \int_{-\infty}^a f(x)\ dx := \lim_{M \to -\infty} \int_M^a f(x)\ dx.\)
If both the above integrals exist, we can define
\begin{equation*}
\int_{-\infty}^\infty f(x)\ dx := \int_{-\infty}^a f(x)\ dx + \int_a^\infty f(x)\ dx
\end{equation*}
It can be checked that any choice of the value \(a\) gives the same answer.
Checkpoint 3.7.2.
Determine whether \(\ds \int_1^\infty \frac{dx}{x}\) or \(\ds \int_1^\infty \frac{dx}{x^2}\) is convergent, and find the values if they exist.
Checkpoint 3.7.3.
Verify that \(\ds \int_{-\infty}^0 x e^ x\ dx\) is convergent and evaluate it.
Checkpoint 3.7.4.
Evaluate \(\ds \int_0^\infty v e^{-v^2} \ dv\text{,}\) or show that it is divergent.
Checkpoint 3.7.5.
Evaluate \(\ds \int_{-\infty}^\infty \frac{dt}{1+t^2}\text{,}\) or show that it is divergent.
Checkpoint 3.7.6.
Determine for which values of \(p\) the improper integral \(\ds \int_1^\infty \frac{dx}{x^p}\) is convergent and for which it is divergent, and evaluate for all cases where it converges.
Improper Integrals of Type II: Regions of Infinite Height.
There is a second case where the definition of the “true” integral fails, but we can make sense of the area under the curve: regions that go to infinite height at some point, due to a vertical asymptote, like \(\ds \int_0^1 \frac{dx}{\sqrt{x}}\text{.}\)
A similar strategy to above works. First approximate the domain of integration by leaving out a bit near the bad point:
\begin{equation*}
I_t = \int_t^1 \frac{dx}{\sqrt{x}}, \; t>0.
\end{equation*}
Then take the limit as the modified end-point goes back to where it should be:
\begin{equation*}
I = \int_0^1 \frac{dx}{\sqrt{x}} = \lim_{t \to 0^+} I_t = \lim_{t \to 0^+} \int_t^1 \frac{dx}{\sqrt{x}}.
\end{equation*}
Note that this must be a one-sided limit, as only values \(t > 0\) avoid the vertical asymptote and give a proper integral.
Definition 3.7.7. Improper Integral, Type II: Vertical Asymptotes.
If
\(f\) is continuous on
\([a,b)\) but discontinuous (including undefined) at
\(b\text{,}\)
\begin{equation*}
\int_a^b f(x)\ dx := \lim_{t \to b^-} \int_a^t f(x)\ dx.
\end{equation*}
Likewise, if \(f\) is continuous on \((a,b]\) but discontinuous at \(a\text{,}\)
\begin{equation*}
\int_a^b f(x)\ dx := \lim_{t \to a^+} \int_t^b f(x)\ dx
\end{equation*}
Note that in each case, the modified endpoint \(t\) must always be “inside” the original interval, so as to exclude the vertical asymptote.
-
If the integrand is continuous everywhere on interval \([a,b]\) except at \(c\text{,}\) \(a < c < b\text{,}\) then we chop it into a sum of two improper integrals as above:
\begin{equation*}
\int_a^b f(x)\ dx := \int_a^c f(x)\ dx + \int_c^b f(x)\ dx
\end{equation*}
Checkpoint 3.7.8.
Evaluate \(\ds\int_{-1}^2 \frac{dx}{\sqrt[3]{x}}\) or show that it is divergent.
Do the same for \(\ds\int_{-1}^2 \frac{dx}{x^2}\text{.}\)
Checkpoint 3.7.9.
Evaluate \(\ds\int_0^1 \ln x\ dx\) or show that it is divergent.
A Comparison Theorem: When Showing Convergence is Enough.
Often with complicated integrands, it is not possible to find the indefinite integrals and limits needed to evaluate as above, but it is enough to show that an improper integral is convergent (or not). (The actual numerical value might then be approximated using methods like those in the previous section.)
One can show convergence by showing that the integrand is “smaller” than another function for which you know convergence, or conversely:
Theorem 3.7.10. Comparison of Improper Integrals of Non-Negative Functions.
Suppose that for \(a \geq 0\text{,}\) \(f\) and \(g\) are both non-negative and continuous and \(f\) is bigger than \(g\text{:}\) \(f(x) \geq g(x) \geq 0\text{.}\)
If the “larger” integral \(\ds\int_a^\infty f(x)\, dx\) converges, then the “smaller” one \(\ds\int_a^\infty g(x)\, dx\) also converges, and so
if the “smaller” integral \(\ds\int_a^\infty g(x)\, dx\) diverges, the “larger” one \(\ds\int_a^\infty f(x)\, dx\) also diverges.
In either case, when the improper integral of a non-negative function diverges, the limit in its definition is infinite, and allowing this infinite value for such an integral, we can always say that
\begin{equation*}
0 \leq \int_a^\infty g(x) dx \leq \int_a^\infty f(x) dx.
\end{equation*}
Checkpoint 3.7.11.
Show that \(\ds \int_0^\infty \frac{1+\sin(x^3)}{e^x} dx\) converges.
Do not try to evaluate it, but think about how you might approximate its value using numerical integration methods.
Section Study Guide.
The two Definitions and Theorem 3.7,
Examples 47, 48, 50, 51 and 52–56,
all Checkpoints,
and one or several exercises from each of the following groups: (348, 350, 351, 352), 347 & 349, 353 & 354, 355–371.
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