Subsections

• Purpose
• Getting Started
• Background
  • Unbounded integrands
    • Examples
  • Unbounded intervals of integration
    • Examples
  
  
• Exercises

Improper Integrals

Purpose

The purpose of this lab is to use Maple to introduce you to the notion of improper integral and to give you practice with this concept by using it to prove convergence or divergence of integrals involving unbounded integrands or unbounded intervals or both.

Getting Started

To assist you, there is a worksheet associated with this lab that contains functions entered associated with each exercise. On your Maple screen go to File - Open then type the following in the white rectangle:

\\storage\academics\math\calclab\MA1023\Improper_int_start_C19.mw

Background

Our basic theorem for $\int_{a}^{b} f(x) dx$ is that the integral exists if $f(x)$ is continuous on the closed interval $[a,b]$. We have actually gone beyond this theorem a few times, and integrated functions that were bounded and had a finite number of jump discontinuities on $[a,b]$. However, we don't have any theory to help us deal with integrals $\int_{a}^{b} f(x) dx$ involving one or more of the following.

1. Functions $f(x)$, for example rational functions, that have vertical asymptotes in $[a,b]$ (or are not bounded on $[a,b]$).
2. Integrals where the interval $[a,b]$ is unbounded, for example intervals like $[a, \infty)$, $(-\infty, b]$, or $(-\infty, \infty)$.

We have already seen at least one example of the problems you can run into if the function is unbounded. Recall the clearly absurd result

$$\int_{-1}^{1} \frac{1}{x^2} dx = -2$$

that is obtained by blindly applying the FTOC. The second type of problem, where the interval of integration is unbounded, occurs often in applications of calculus, such as the Laplace and Fourier transforms used to solve differential equations. It also occurs in testing certain kinds of infinite series for convergence or divergence, as we will learn later.

We start with the following definition.

Definition 1   We say that the integral

$$\int_{a}^{b} f(x) dx$$

is improper if one or both of the following conditions is satisfied.

1. The interval of integration is unbounded.
2. The function $f(x)$ has an infinite discontinuity at some point $c$ in $[a,b]$. That is, $$\lim_{x \rightarrow c} f(x) = \pm \infty$$.

Unbounded integrands

To see how to handle the problem of an unbounded integrand, we start with the following special cases.

Definition 2   Suppose that $f(x)$ is continuous on $[a,b)$, but $$\lim_{x \rightarrow b^{-}} f(x) = \pm \infty$$. Then we define

$$\int_{a}^{b} f(x) dx = \lim_{t \rightarrow b^{-}} \int_{a}^{t} f(x) dx ,$$

provided that the limit on the right-hand side exists and is finite, in which case we say the integral converges and is equal to the value of the limit. If the limit is infinite or doesn't exist, we say the integral diverges or fails to exist and we cannot compute it.

Definition 3   Suppose that $f(x)$ is continuous on $(a,b]$, but $$\lim_{x \rightarrow a^{+}} f(x) = \pm \infty$$. Then we define

$$\int_{a}^{b} f(x) dx = \lim_{t \rightarrow a^{+}} \int_{t}^{b} f(x) dx ,$$

provided that the limit on the right-hand side exists and is finite, in which case we say the integral converges and is equal to the value of the limit. If the limit is infinite or doesn't exist, we say the integral diverges or fails to exist and we cannot compute it.

Cases where $f(x)$ has an infinite discontinuity only at an interior point $c, a <c < b$ are handled by writing

$$\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$$

and using the definitions to see if the integrals on the right-hand side exist. If both exist then the integral on the left-hand side exists. If either of the integrals on the right-hand side diverges, then $\int_{a}^{b} f(x) dx$ does not exist.

Examples

Here is a simple example using Maple to show that $$\int_{0}^{2} \frac{1}{x} dx$$ doesn't exist.

> limit(int(1/x,x=a..2),a=0,right);

The example above used the right option to limit because the right-hand limit was needed. If you need a left-hand limit, use the left option in the limit command. Maple can usually do the limit within the int command.

> int(1/x,x=0..2);

Unbounded intervals of integration

These are handled in a similar fashion by using limits. The definition we need the most is given below.

Definition 4   Suppose $f(x)$ is continuous on the unbounded interval $[a, \infty)$. Then we define

$$\int_{a}^{\infty} f(x) dx = \lim_{t \rightarrow \infty} \int_{a}^{t} f(x) dx ,$$

provided the limit on the right-hand side exists and is finite, in which case we say the integral converges and and is equal to the value of the limit. If the limit is infinite or fails to exist we say the integral diverges or fails to exist.

The other two cases are handled similarly. You are asked to provide suitable definitions for them in one of the exercises.

Examples

Using the definition for $$\int_{2}^{\infty} \frac{1}{x^2}$$.

> limit(int(1/x^2,x=2..a),a=infinity);

This command shows that Maple takes the limit definition into account in the int command.

> int(1/x^2,x=2..infinity);

Exercises

1. Determine which of the regions described below have finite area by evaluating the limit of an imporper integral. If Maple is unable to calculate the limit of the integral, use a comparison test (either by plotting for direct comparison or by limit comparison if applicable).

   A

   The region below $$f1(x)=\frac{2x}{x^3+1}$$, above the $x$-axis, over the interval $[2,\infty)$.
   The region below $$f2(x)=\frac{\cos(x)+x}{x^3-x \sin(x)}$$, above the $x$-axis, over the interval $[2,\infty)$.

   B

   The region below $$f3(x)=\frac{1}{(x^2+1)^{\frac{1}{3}}}$$, above the $x$-axis, over the interval $[1,\infty)$.
   The region below $$f4(x)=\frac{2x}{(2x^4+x^3+1)^{\frac{1}{3}}}$$, above the $x$-axis, over the interval $[1,\infty)$.

2. Determine, by calculating a limit, whether each of the following improper integrals converge or diverge and if the integral converges, to what does it converge. Begin each exercise by first solving in Maple for what values of $x$ the denominator of the integrand is zero.

   A

   $$\int_{-1}^1 \frac{1}{(x^3+1)} dx$$

   B

   $$\int_0^\infty \frac{1}{\sqrt{x^4+x}} dx$$

   C

   $$\int_0^1 \frac{1}{(-x^4+x)} dx$$

   D

   $$\int_0^1 \frac{1}{(-x^4+x)^{\frac{1}{4}}} dx$$