Applications of Double Integrals: Learn It 3

Giselle Miller

Applications of Double Integrals: Learn It 3

Improper Double Integrals

In single-variable calculus, an improper integral arises when attempting to integrate a function on an unbounded region, or attempting to integrate a function on an interval where that function is discontinuous. We briefly recall both types of improper integrals below.

Recall: Improper Integrals
[latex]\\[/latex]
A type I improper integral arises when integrating a function [latex]f(x)[/latex] on an unbounded interval, either [latex][a,\infty)[/latex], [latex][-\infty,b)[/latex], or [latex](-\infty,\infty)[/latex]. Each of these integrals is defined in terms of a limit.

[latex]\displaystyle \int_a^\infty f(x)dx = \displaystyle \lim_{t\rightarrow\infty} \int_a^t f(x)dx[/latex]
[latex]\displaystyle \int_{-\infty}^b f(x)dx = \displaystyle \lim_{t\rightarrow-\infty} \int_t^b f(x)dx[/latex]
[latex]\displaystyle \int_{-\infty}^\infty f(x)dx = \displaystyle \int_{-\infty}^c f(x)dx + \displaystyle \int_c^\infty f(x)dx[/latex] (where [latex]c[/latex] is a real number)

If the limits above exist, the improper integrals are said to be convergent. Otherwise, the integrals are divergent.

A type II improper integral arises when integrating a function [latex]f(x)[/latex] on a half-open interval [latex][a, b)[/latex] or [latex](a,b][/latex]. In these respective cases, the improper integrals are defined as follows:

[latex]\displaystyle \int_a^b f(x)dx = \displaystyle \lim_{t\rightarrow b^-} \int_a^t f(x)dx[/latex]
[latex]\displaystyle \int_a^b f(x)dx = \displaystyle \lim_{t\rightarrow a^+} \int_t^b f(x)dx[/latex]

The two definitions above can be used to define an improper integral on a closed interval [latex][a,b][/latex] where there exists a [latex]c \in [a,b][/latex] such that [latex]f(x)[/latex] is discontinuous at [latex]c[/latex].

[latex]\displaystyle \int_a^b f(x)dx = \displaystyle \int_a^c f(x)dx + \int_c^b f(x)dx[/latex]

An improper double integral is an integral [latex]\underset{D}{\displaystyle\iint}fdA[/latex] where either [latex]D[/latex] is an unbounded region or [latex]f[/latex] is an unbounded function. For example, [latex]D = {(x,y)\mid |x-y| \geq 2 }[/latex] is an unbounded region, and the function [latex]f(x,y) = \frac{1}{1-x^2-2y^2}[/latex] over the ellipse [latex]x^2 + 3y^2 \leq 1[/latex] is an unbounded function. Hence, both of the following integrals are improper integrals:

[latex]\underset{D}{\displaystyle\iint}{xy\,dA}[/latex] where [latex]D = \{(x,y)| \mid{x}-y\mid\geq2\}[/latex];
[latex]\underset{D}{\displaystyle\iint}{\dfrac{1}{1-x^2-2y^2}}{dA}[/latex] where [latex]\large{D} = {\left \{{(x,y)}{\mid}{x^2} + {3y^2} \ {\leq} \ {1} \right \}}[/latex].

In this section we would like to deal with improper integrals of functions over rectangles or simple regions such that [latex]f[/latex] has only finitely many discontinuities. Not all such improper integrals can be evaluated; however, a form of Fubini’s theorem does apply for some types of improper integrals.

Fubini’s theorem for improper integrals

If [latex]D[/latex] is a bounded rectangle or simple region in the plane defined by [latex]{\left \{{(x,y)}{:} \ {a} \ {\leq} \ {x} \ {\leq} \ {b,g(x)} \ {\leq} \ {y} \ {\leq} \ {h(x)} \right \}}[/latex] and also by [latex]{\left \{{(x,y)}{:} \ {c} \ {\leq} \ {y} \ {\leq} \ {d,j(y)} \ {\leq} \ {x} \ {\leq} \ {k(y)} \right \}}[/latex] and [latex]f[/latex] is a nonnegative function on [latex]D[/latex] with finitely many discontinuities in the interior of [latex]D[/latex], then:

[latex]\large{\underset{D}{\displaystyle\iint}f\,dA=\displaystyle\int_{x=a}^{x=b}\displaystyle\int_{y=g(x)}^{y=h(x)}f(x,y)dydx=\displaystyle\int_{y=c}^{y=d}\displaystyle\int_{x=j(y)}^{x=k(y)}f(x,y)dxdy.}[/latex]

It is very important to note that we required that the function be nonnegative on [latex]D[/latex] for the theorem to work. We consider only the case where the function has finitely many discontinuities inside [latex]D[/latex].

Consider the function [latex]{f(x,y)}={\frac{e^y}{y}}[/latex] over the region [latex]{D} = {\left \{{(x,y)}{:} \ {0} \ {\leq} \ {x} \ {\leq} \ {1,x} \ {\leq} \ {y} \ {\leq} \ {\sqrt{x}} \right \}}[/latex].

Notice that the function is nonnegative and continuous at all points on [latex]D[/latex] except [latex](0, 0)[/latex]. Use Fubini’s theorem to evaluate the improper integral.

As mentioned before, we also have an improper integral if the region of integration is unbounded. Suppose now that the function [latex]f[/latex] is continuous in an unbounded rectangle [latex]R[/latex].

improper integrals on an unbounded region

If [latex]R[/latex] is an unbounded rectangle such as [latex]{R} = {\left \{{(x,y)}{:} \ {a} \ {\leq} \ {x} \ {\leq} \ {\infty, c} \ {\leq} \ {y} \ {\leq} \ {c} \ {\leq} \ {\infty} \right \}}[/latex], then when the limit exists, we have:

[latex]\underset{R}{\displaystyle\iint}f(x,y)dA=\displaystyle\lim_{(b,d)\to(\infty,\infty)}\displaystyle\int_a^b\left(\displaystyle\int_c^d{f}(x,y)dy\right)dx=\displaystyle\lim_{(b,d)\to(\infty,\infty)}\displaystyle\int_c^d\left(\displaystyle\int_a^b{f}(x,y)dx\right)dy[/latex]

The following example shows how this theorem can be used in certain cases of improper integrals.

Evaluate the integral [latex]\underset{R}{\displaystyle\iint}{xy}{e^{{-x^2}{-y^2}}}{dA}[/latex] where [latex]R[/latex] is the first quadrant of the plane.

Applications to Probability

In some situations in probability theory, we can gain insight into a problem when we are able to use double integrals over general regions. Before we go over an example with a double integral, we need to set a few definitions and become familiar with some important properties.

joint density function

Consider a pair of continuous random variables [latex]X[/latex] and [latex]Y[/latex], such as the birthdays of two people or the number of sunny and rainy days in a month. The joint density function [latex]f[/latex] of [latex]X[/latex] and [latex]Y[/latex] satisfies the probability that [latex](X, Y)[/latex] lies in a certain region [latex]D[/latex]:

[latex]{P}{((X,Y)\in{D})} =\underset{D}{\displaystyle\iint}{f(x,y)}{dA}[/latex].

Since the probabilities can never be negative and must lie between [latex]0[/latex] and [latex]1[/latex], the joint density function satisfies the following inequality and equation:

[latex]{f(x,y)} \ {\geq} \ {0}[/latex] and [latex]\underset{R^2}{\displaystyle\iint}{f(x,y)}{dA} = {1}[/latex].

independent random variables

The variables [latex]X[/latex] and [latex]Y[/latex] are said to be independent random variables if their joint density function is the product of their individual density functions:

[latex]{f(x,y)} = {f_1}{(x)}{f_2}{(y)}[/latex].

At Sydney’s Restaurant, customers must wait an average of [latex]15[/latex] minutes for a table. From the time they are seated until they have finished their meal requires an additional [latex]40[/latex] minutes, on average. What is the probability that a customer spends less than an hour and a half at the diner, assuming that waiting for a table and completing the meal are independent events?

Waiting times are mathematically modeled by exponential density functions, with [latex]m[/latex] being the average waiting time, as

[latex]{f(t)}\left\{\begin{matrix} 0 & \text{if } t < 0,\\ {\frac{1}{m}}{e^{-t/m}} & \text{if } t \geq 0. \end{matrix}\right.[/latex]

If [latex]X[/latex] and [latex]Y[/latex] are random variables for ‘waiting for a table’ and ‘completing the meal,’ then the probability density functions are, respectively,

[latex]{{f_1}(x)} = \left\{\begin{matrix} 0 & \text{if } x < 0,\\ {\frac{1}{15}}{e^{-x/15}} & \text{if }x \geq 0. \end{matrix}\right. \text{and} \ {{f_2}(y)} = \left\{\begin{matrix} 0 & \text{if } y < 0,\\ {\frac{1}{40}}{e^{-y/40}} & \text{if }y \geq 0. \end{matrix}\right.[/latex]

Clearly, the events are independent and hence the joint density function is the product of the individual functions

[latex]{f(x,y)} = {f_1}{(x)}{f_2}{(y)} = \left\{\begin{matrix} 0 & \ \ \ \ \ \ \ \ \ \text{if } x < 0 \ \text{or} \ y < 0, \\ {\frac{1}{600}}{e^{-x/15}}{e^{-y/60}} & \text{if } x,y \geq 0. \end{matrix}\right.[/latex]

We want to find the probability that the combined time [latex]X+Y[/latex] is less than 90 minutes. In terms of geometry, it means that the region [latex]D[/latex] is in the first quadrant bounded by the line [latex]x+y=90[/latex] (Figure 2).

The line x + y = 90 is shown. — Figure 2. The region of integration for a joint probability density function.

Hence, the probability that [latex](X, Y)[/latex] is in the region [latex]D[/latex] is

[latex]{P} = {({X+Y} \ {\leq } \ 90)} = {P}{((X,Y)\in{D})} = {\iint\limits_D}{f(x,y)}{dA} = {\iint\limits_D}{\frac{1}{600}}{e^{-x/15}}{e^{-y/40}}{dA}.[/latex]

Since [latex]x+y=90[/latex] is the same as [latex]y=90-x[/latex], we have a region of Type I, so
[latex]\hspace{6cm}\begin{align} {D} &= {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {90,0} \ {\leq} \ {y} \ {\leq} \ {90-x} \right \}}, \\ P(X + Y\leq90)&=\frac1{600}\displaystyle\int_{x=0}^{x=90}\displaystyle\int_{y=0}^{y=90-x}e^{-x/15}e^{-y/40}dxdy \\ &=\frac1{600}\displaystyle\int_{x=0}^{x=90}\displaystyle\int_{y=0}^{y=90-x}e^{-(x/15+y/40)}dxdy=0.8328. \end{align}[/latex]

Thus, there is an [latex]83.2 \%[/latex] chance that a customer spends less than an hour and a half at the restaurant.

Another important application in probability that can involve improper double integrals is the calculation of expected values. First we define this concept and then show an example of a calculation.

expected values

In probability theory, we denote the expected values [latex]E(X)[/latex] and [latex]E(Y)[/latex], respectively, as the most likely outcomes of the events. The expected values [latex]E(X)[/latex] and [latex]E(Y)[/latex] are given by:

[latex]{E(X)} = \underset{S}{\displaystyle\iint}{xf}{(x,y)}{dA}[/latex] and [latex]{E(Y)} = \underset{S}{\displaystyle\iint}{yf}{(x,y)}{dA}[/latex],

where [latex]S[/latex] is the sample space of the random variables [latex]X[/latex] and [latex]Y[/latex].

Find the expected time for the events ‘waiting for a table’ and ‘completing the meal’ in Example “Application to Probability”.