Improper Integrals

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 7.8 – Improper Integrals

Expected Educational Results

• Objective 09–01: I can identify an improper integral.

• Objective 09–02: I can use limits to evaluate an improper integral.

• Objective 09–03: I can use the Comparison Theorem to determine if an improper integral converges or diverges.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Improper Integrals

Definition: Improper Integral, Type I

• If ${\int }_a^{t}f(x)dx$$\int_a^t{f(x)\,dx}$ exists for every $t\ge a$$t\geq a$, then ${\displaystyle {\int }_a^{\mathrm{\infty }}f(x)dx=\underset{t\to \mathrm{\infty }}{lim}{\int }_a^{t}f(x)dx}$$\displaystyle\int_a^{\infty}{f(x)\,dx}=\lim_{t\to\infty}\int_a^t{f(x)\,dx}$ provided the limit exists.

• If ${\int }_t^{b}f(x)dx$$\int_t^b{f(x)\,dx}$ exists for every $t\le b$$t\leq b$, then ${\displaystyle {\int }_{-\mathrm{\infty }}^{b}f(x)dx=\underset{t\to -\mathrm{\infty }}{lim}{\int }_t^{b}f(x)dx}$$\displaystyle\int_{-\infty}^b{f(x)\,dx}=\lim_{t\to -\infty}\int_t^b{f(x)\,dx}$ provided the limit exists.

Definition: Improper Integral, Type II

• If $f(x)$$f(x)$ is continuous on $[a,b)$$\big[a,b\big)$ and is discontinuous at $b$$b$, then ${\displaystyle {\int }_a^{b}f(x)dx=\underset{t\to b^{-}}{lim}{\int }_a^{t}f(x)dx}$$\displaystyle\int_a^b{f(x)\,dx}=\lim_{t\to b^{-}}\int_a^t{f(x)\,dx}$ provided the limit exists.

• If $f(x)$$f(x)$ is continuous on $(a,b]$$\big(a,b\big]$ and is discontinuous at $a$$a$, then ${\displaystyle {\int }_a^{b}f(x)dx=\underset{t\to a^{+}}{lim}{\int }_t^{b}f(x)dx}$$\displaystyle\int_a^b{f(x)\,dx}=\lim_{t\to a^{+}}\int_t^b{f(x)\,dx}$ provided the limit exists.

• If $f(x)$$f(x)$ is continuous on $[a,b]$$\big[a,b\big]$ and is discontinuous at $x=c$$x=c$, where $a<c<b$$a<c<b$, then ${\displaystyle {\int }_a^{b}f(x)dx=\underset{x\to c^{-}}{lim}{\int }_a^{c}f(x)dx+\underset{x\to c^{+}}{lim}{\int }_c^{b}f(x)dx}$$\displaystyle\int_a^b{f(x)\,dx}=\lim_{x\to c^{-}}\int_a^c{f(x)\,dx}+\lim_{x\to c^{+}}\int_c^b{f(x)\,dx}$ provided the limit exists.

Convergence and Divergence

Definition: Convergence

An improper integral is convergent if the corresponding limit exists.

Definition: Divergence

An improper integral is divergent if the corresponding limit does not exist.

Convergence and Divergence of Improper Integrals

• If both ${\int }_a^{\mathrm{\infty }}f(x)dx$$\int_a^{\infty}{f(x)\,dx}$ and ${\int }_{-\mathrm{\infty }}^{a}f(x)dx$$\int_{-\infty}^a{f(x)\,dx}$ are convergent, then ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)dx={\int }_{-\mathrm{\infty }}^{a}f(x)dx+{\int }_a^{\mathrm{\infty }}f(x)dx$$\int_{-\infty}^{\infty}{f(x)\,dx}=\int_{-\infty}^a{f(x)\,dx}+\int_a^{\infty}{f(x)\,dx}$ for any real number $a$$a$.

• If $f$$f$ has a discontinuity at $c$$c$, where $a\le c\le b$$a\leq c\leq b$, and both ${\int }_a^{c}f(x)dx$$\int_a^c{f(x)\,dx}$ and ${\int }_c^{b}f(x)dx$$\int_c^b{f(x)\,dx}$ are convergent, then ${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$$\int_a^b{f(x)\,dx}=\int_a^c{f(x)\,dx} + \int_c^b{f(x)\,dx}$

Theorem

${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{1}{x^p}dx}$$\displaystyle\int_1^{\infty}{\frac{1}{x^p}\,dx}$ is convergent for $p>1$$p>1$ and divergent for $p\le 1$$p\leq 1$.

Identifying Improper Integrals

Investigation 05

Explain why the following integrals are improper. HINT: Graph the function in the embedded DESMOS graph, if needed.

1. ${\displaystyle {\int }_0^1\frac{1}{x}dx}$$\displaystyle\int_0^{1}{\frac{1}{x}\,dx}$

2. ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{1}{x}dx}$$\displaystyle\int_1^{\infty}{\frac{1}{x}\,dx}$

3. ${\displaystyle {\int }_{-2}^3\frac{1}{x-1}dx}$$\displaystyle\int_{-2}^{3}{\frac{1}{x-1}\,dx}$

4. ${\displaystyle {\int }_2^{\mathrm{\infty }}\frac{1}{x^2}dx}$$\displaystyle\int_{2}^{\infty}{\frac{1}{x^2}\,dx}$

5. ${\displaystyle {\int }_4^{\mathrm{\infty }}\frac{1}{\sqrt{x}}dx}$$\displaystyle\int_{4}^{\infty}{\frac{1}{\sqrt{x}}\,dx}$

6. ${\displaystyle {\int }_3^{\mathrm{\infty }}{e}^{-x}dx}$$\displaystyle\int_3^{\infty}{e^{-x}\,dx}$

7. ${\displaystyle {\int }_{-\mathrm{\infty }}^0{e}^{-x}dx}$$\displaystyle\int_{-\infty}^{0}{e^{-x}\,dx}$

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Created: Thursday, 17 September 2020 2:48 EDT Last Modified: Sunday, 12 June 2022 - 17:31 (EDT)