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Calculus An intuitive approach

Wai Yan Pong

Section 4.1 Definitions and Examples

Suppose \(f\) is a function defined on \([a,b)\) where \(b\) is finite or \(+\infty\text{.}\) If
  1. \(f\) is integrable on \([a,t]\) for each \(a \lt t \lt b\text{;}\) and that
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  2. the limit \(\lim_{t \to b^{-}} \int_a^t f(x) dx\) exists in \(\mathbb{R} \cup \{\pm \infty\}\text{,}\)
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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}
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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.
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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.
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Let us work out a few examples to solidify these concepts.
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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 \([0,\infty)\text{.}\)
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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*}
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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.
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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 \(p=1\text{.}\)
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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.
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Deduce from the above statement and Proposition 4.3 that
\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*}
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Hint.
Make the substitution \(u = 1/x\)
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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*}
As a result, the second assertion follows from Proposition 4.3.
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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.
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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.
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Hint.
Use integration by parts with \(u=\ln(x)\) and \(dv = x^p dx\text{.}\)
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Figure 4.7. Graphs of \(x^p\ln(x)\) for \(p=1,2\)
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Checkpoint 4.8.

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.
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Hint.
Use the substitution \(u=\ln(x)\) and then use Proposition 4.3
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Checkpoint 4.9.

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*}
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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.
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Example 4.10.

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*}
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