Definitions and Examples

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

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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}

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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In this book, we will not work with the most general possible improper integrals. Instead, we will restrict our attention to improper integrals that can be broken into finitely many pieces, where each piece is either Type I, with an unbounded interval in the \(x\)-direction, or Type II, with an integrand that becomes unbounded in the \(y\)-direction. This covers the examples and applications that we need while keeping the main ideas transparent.

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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.

Graphs of one over sqrt of x, one over x and one over x square, for x from point one to two point five. — Figure 4.3. Graphs of \(1/\sqrt{x}, 1/x, 1/x^2\)
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\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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Proposition 4.4. 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*}

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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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A positive continuous decreasing function in the first quadrant with the region under the curve shaded from the point a on the x-axis to the right. — Figure 4.5. A Type I improper integral \(\int_a^{\infty} f(x)\,dx\text{,}\) represented by the shaded area under a positive continuous decreasing function \(f\text{.}\)
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Checkpoint 4.6.

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 4.4 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 \(t \gt 0\text{,}\)

\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 4.4.

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A positive decreasing continuous curve labeled f of x rises sharply near the y-axis, and the region under the curve is shaded from near zero to the point a on the x-axis. — Figure 4.7. A Type II improper integral \(\int_0^a f(x)\,dx\) represented by the shaded area under the graph of a positive decreasing continuous function \(f\text{.}\)
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Example 4.8.

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 4.4) \(\int_0^1 \frac{1}{x^2} dx = +\infty\) (Checkpoint 4.6). Therefore, \(\int_0^{\infty} \frac{1}{x^2} dx\) diverges to \(\infty\) according to the definition.

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Checkpoint 4.9.

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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Solution.

For \(a\in(0,1)\text{,}\) let

\begin{equation*} I(a)=\int_a^1 x^p\ln(x)\,dx. \end{equation*}

Using integration by parts with \(u=\ln(x)\) and \(dv=x^p\,dx\text{,}\) for \(p\neq -1\) we get

\begin{align*} I(a) &= \left.\frac{x^{p+1}}{p+1}\ln(x)\right|_a^1 -\frac{1}{p+1}\int_a^1 x^p\,dx\\ &= -\frac{a^{p+1}\ln(a)}{p+1} -\frac{1-a^{p+1}}{(p+1)^2}. \end{align*}

If \(p\gt -1\text{,}\) then \(a^{p+1}\to 0\) and \(a^{p+1}\ln(a)\to 0\) as \(a\to 0^+\text{.}\) Hence

\begin{equation*} \int_0^1 x^p\ln(x)\,dx =\lim_{a\to 0^+} I(a) =-\frac{1}{(p+1)^2}. \end{equation*}

If \(p\lt -1\text{,}\) then \(a^{p+1}\to\infty\text{,}\) so the expression above tends to \(-\infty\text{;}\) thus the integral diverges. For \(p=-1\text{,}\) \(\int_a^1 \frac{\ln(x)}{x}\,dx=-\frac{1}{2}(\ln(a))^2\to-\infty\text{,}\) so the original integral diverges (to \(-\infty\)).

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Therefore, the improper integral converges exactly when \(p\gt -1\text{,}\) and in that case

\begin{equation*} \int_0^1 x^p\ln(x)\,dx=-\frac{1}{(p+1)^2}. \end{equation*}

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Checkpoint 4.10.

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 4.4

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Checkpoint 4.11.

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.

First, sketch the region. Then 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 4.6.

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Example 4.12.

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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