Section 4.1 Definitions and Examples
Suppose \(f\) is a function defined on \([a,b)\) where \(b\) is finite or \(+\infty\text{.}\) If
\(f\) is integrable on \([a,t]\) for each \(a \lt t \lt b\text{;}\) and that
the limit \(\lim_{t \to b^{-}} \int_a^t f(x) dx\) exists in \(\mathbb{R} \cup \{\pm \infty\}\text{,}\)
then we define
\begin{equation}
\int_a^b f(x)\ dx = \lim_{t \to b^{-}} \int_a^t f(x)\ dx.\tag{4.1}
\end{equation}
This definition, according to the fundamental theorem of calculus, agrees with the original definition definite integral when
\(b \in \mathbb{R}\) and
\(f\) is integrable on
\([a,b]\text{.}\) If
\(b = \infty\) or
\(f\) is not integrable on
\([a,b]\text{,}\) for example
\(f\) is unbounded on
\([a,b]\text{,}\) then we call the integral in
(4.1) an
improper integral.
Likewise, if \(f\) is defined on \((a,b]\) where \(a \in \mathbb{R}
\cup \{-\infty\}\) and \(f\) is integrable on \([t,b]\) for all \(a
\lt t \lt b\) and then we define
\begin{equation}
\int_a^b f(x) dx = \lim_{t \to a^+} \int_t^b f(x) dx\tag{4.2}
\end{equation}
whenever the limit exists. Finally, if \(f\) is defined on \((a,b)\) where \(a\lt b\) and \(a,b \in \mathbb{R} \cup\{\pm \infty\}\) and integrable on all subintervals \([s,t]\text{,}\) then we fix \(c \in (a,b)\) and define
\begin{equation}
\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx\tag{4.3}
\end{equation}
provided that the integrals on the right-hand side exist, possibly as improper integrals, and that the sum is not of the form \(+\infty + (-\infty)\text{;}\) otherwise, we declare that the improper integral is undefined. The definition is independent of the choice of \(c\text{.}\) We say that an improper integral converges if it is has a finite value, if it is either \(+\infty\) or \(-\infty\text{,}\) then we say that the improper integral exists but diverges to \(+\infty\) or \(-\infty\text{,}\) respectively.
Let us work out a few examples to solidify these concepts.
Checkpoint 4.2.
Let \(t \gt 1\text{,}\) compute the following integrals
\begin{equation*}
\int_1^t \frac{1}{x}dx, \int_1^t \frac{1}{x^2}dx,\ \text{and}\
\int_1^t \frac{1}{\sqrt{x}}dx
\end{equation*}
Then decide the convergence of the corresponding improper integrals on \([0,\infty)\text{.}\)
Solution.
\begin{equation*}
\int_1^t \frac{1}{x}dx = \ln|t| - \ln|1| = \ln t.
\end{equation*}
\begin{equation*}
\int_1^t \frac{1}{x^2}dx = \left. -\frac{1}{x} \right|_1^{t} =
\left.\frac{1}{x} \right|_t^1 = 1 - \frac{1}{t}.
\end{equation*}
and
\begin{equation*}
\int_1^t \frac{1}{\sqrt{x}}dx = \left. 2\sqrt{x} \right|_1^{t} =
2\sqrt{t}-2.
\end{equation*}
Therefore, as \(t \to +\infty\text{,}\) we have
\begin{equation*}
\int_1^{\infty} \frac{1}{x} dx = +\infty, \int_1^{\infty} \frac{1}{x^2} dx=
1, \int_1^{\infty}\frac{1}{\sqrt{x}} dx = +\infty.
\end{equation*}
For improper integral of the form \(\int_a^{\infty}\frac{1}{x^p} dx\) (\(a \gt 0\)) the dividing line for convergence is at \(p=1\text{.}\) We record that in the following proposition.
Proposition 4.3. p-test for integrals.
For \(a \gt 0\text{,}\)
\begin{equation*}
\int_a^{\infty}\frac{1}{x^p} dx
\begin{cases}
\text{converges} & p \gt 1; \\
\text{diverges to}\ +\infty & p \le 1.
\end{cases}
\end{equation*}
Proof.
Since
\begin{equation*}
\int_a^t \frac{1}{x^p} dx = \left. \frac{1}{(1-p)x^{p-1}}\right|_a^t =
\frac{1}{(p-1)a^{p-1}} - \frac{1}{(p-1)t^{p-1}},
\end{equation*}
the integral converges to
\(\frac{1}{(p-1)a^{p-1}}\) when
\(p
\gt 1\) and diverges to
\(+\infty\) when
\(p \lt 1\text{.}\) We have already showed in
Checkpoint 4.2 that the integral diverges to
\(+\infty\) for
\(p=1\text{.}\)
Checkpoint 4.4.
Suppose \(a \gt 0\) and \(f\) continuous on \((0,a]\text{.}\) Show that
\begin{equation*}
\int_0^a f(x) dx = \int_{1/a}^{\infty} \frac{f(1/u)}{u^2} du
\end{equation*}
as improper integrals.
\begin{equation*}
\int_0^a \frac{1}{x^p} dx
\begin{cases}
\text{converges} & p \lt 1; \\
\text{diverges to}\ +\infty & p \ge 1.
\end{cases}
\end{equation*}
Hint.
Make the substitution \(u = 1/x\)
Solution.
For \(t > 0\text{,}\)
\begin{equation*}
\int_t^a f(x) dx \stackrel{u=1/x}{=} \int_{1/t}^{1/a}
f(1/u)\frac{-du}{u^2}
= \int_{1/a}^{1/t}\frac{f(1/u)}{u^2}du.
\end{equation*}
As \(t \to 0^+, 1/t \to \infty\text{,}\) the first assertion follows. Apply that to \(f(x) = 1/x^p\text{,}\) we have
\begin{equation*}
\int_0^a \frac{1}{x^p} dx = \int_{1/a}^{\infty} \frac{1}{u^{2-p}}
du.
\end{equation*}
Example 4.5.
Consider the improper integral
\(\int_0^{\infty} \frac{1}{x^2}
dx\text{.}\) According to
(4.3), we should pick a point
\(c \in (0,\infty)\text{,}\) say
\(c=1\) and investigate the two integrals
\begin{equation*}
\int_0^1 \frac{1}{x^2} dx\ \text{and}\ \int_1^{\infty} \frac{1}{x^2}dx
\end{equation*}
Even though
\(\int_1^{\infty} \frac{1}{x^2} dx =1\) (
Proposition 4.3)
\(\int_0^1 \frac{1}{x^2} dx = +\infty\) (
Checkpoint 4.4). Therefore,
\(\int_0^{\infty} \frac{1}{x^2} dx\) diverges to
\(\infty\) according to the definition.
Checkpoint 4.6.
Find the values of \(p\) for which the improper integral
\begin{equation*}
\int_{0}^{1} x^p\ln(x) dx
\end{equation*}
converges. Also, evaluate the integral when it is convergent.
Hint.
Use integration by parts with \(u=\ln(x)\) and \(dv = x^p dx\text{.}\)
Checkpoint 4.7.
Find the values of \(p\) for which the improper integral
\begin{equation*}
\int_{e}^{\infty} \frac{1}{x(\ln(x))^p} dx
\end{equation*}
converges. Also, evaluate the integral when it is convergent.
Checkpoint 4.8.
Find the area of the region
\begin{equation*}
S = \left\lbrace (x,y) \colon -2 \lt x \le 0, 0 \le y \le
\frac{1}{\sqrt{x+2}}
\right\rbrace.
\end{equation*}
Hint.
Recognize the area is given by the integral
\begin{equation*}
\int_{-2}^0 \frac{1}{\sqrt{x+2}} dx.
\end{equation*}
To find the integral, make the substitution
\(u=x+2\) then use the result in
Checkpoint 4.4.
Example 4.9.
Let us compute the improper integral \(\ds \int_{-\infty}^{\infty}
\frac{1}{1+x^2} dx\text{.}\) By definition, we should investigate, for some fixed \(c\text{,}\) the integrals
\begin{equation*}
\int_{-\infty}^c \frac{1}{1+x^2} dx \quad \text{and} \quad
\int_{c}^{\infty} \frac{1}{1+x^2} dx.
\end{equation*}
Since the graph \(y = 1/(1+x^2)\) is symmetric about the \(y\)-axis, the choice \(c=0\) is appropriate. For \(t \gt 0\text{,}\)
\begin{equation*}
\int_{0}^{t} \frac{1}{1+x^2} dx = \left. \arctan(x)\right|_0^t =
\arctan(t)
\end{equation*}
which tends to \(\pi/2\) as \(t \to \infty\text{.}\) Thus, the improper integral \(\ds \int_0^\infty \frac{1}{1+x^2} dx\) converges to \(\pi/2\text{.}\) Therefore
\begin{equation*}
\int_{-\infty}^{\infty} \frac{1}{1+x^2} dx = \int_{-\infty}^{0} \frac{1}{1+x^2} dx +
\int_0^{\infty} \frac{1}{1+x^2} dx = \frac{\pi}{2} + \frac{\pi}{2} = \pi.
\end{equation*}
