Improper Double Integrals

The double integrals treated so far are the case where a function is bounded on the bounded region. Now consider the case where $\Omega$ is not bounded.

The sequence of bounded closed regions $\Omega_{1},\Omega_{2},\cdots,\Omega_{n},\cdots$ in $\Omega$ satisfy

$\displaystyle \Omega_{1} \subset \Omega_{2} \subset \cdots \subset \Omega_{n} \subset \cdots$

Furthermore, every subset of $\Omega$ is a subset of $\Omega_{n}$ . Then if $f(x,y)$

is integrable on the region $\Omega_{n}$ ,

$\displaystyle \lim_{n \rightarrow \infty}\iint_{\Omega_{n}} f(x,y)dxdy$

exists. Then we say $f(x,y)$

is integrable on $\Omega$ and define

$\displaystyle \iint_{\Omega} f(x,y)dxdy = \lim_{n \rightarrow \infty}\iint_{\Omega_{n}} f(x,y)dxdy$

Example 5..5 Evaluate the following double integrals.
1. $\displaystyle{\textcolor{red}{I} = \iint_{\Omega}\frac{1}{\sqrt{y^2 - x^2}}dxdy , \Omega = \{(x,y) : 0 \leq x \leq y \leq 1\} }$
2. $\displaystyle{I^2 = \iint_{\Omega}e^{-(x^2 + y^2)}dxdy , \Omega = \{(x,y) : x,y \geq 0\} }$

SOLUTION 1. Using horizontally simple region, we have $\Omega = \{(x,y): 0 \leq y \leq 1, 0 \leq x \leq y\}$ . Then $f(x,y) = \frac{1}{\sqrt{y^2 - x^2}}$ is discontinuous at $y = x$ . Thus let $\Omega_n = \{(x,y): \frac{1}{n} \leq y \leq 1, 0 \leq x \leq y - \frac{1}{n}\}$ . Then

$\displaystyle I$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\int_{y=\frac{1}{n}}^{1}\big(\int_{x=0}^{y-\frac{1}{n}}\frac{1}{(y^2 - x^2)^{1/2}}dx\big) dy$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\int_{y=\frac{1}{n}}^{1}\big(\int_{0}^{y-\frac{1}{n}}\frac{1}{(y^2 - x^2)^{1/2}}dx\big)dy$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\int_{\frac{1}{n}}^1\big([\sin^{-1}{\frac{x}{y... ... \lim_{n \to \infty}(\sin^{-1}{\frac{y-1/n}{y}})(1-\frac{1}{n}) = \frac{\pi}{2}$

2. The region $\Omega$ is not bounded. So, consider the sequence of closed bounded regions $\{\Omega_{n}\}$ .

$\displaystyle \Omega_{n} = \{(x,y) : x^2 + y^2 \leq n^2, x,y \geq 0 \}$

The figure of $\{\Omega_{n}\}$ is given by the figure 5.7.

**Figure 5.7:** Sequences
$\includegraphics[width=4.5cm]{SOFTFIG-5/closedregion.eps}$

For $\Omega_{n}$ , use the polar coordinate $x = r\cos{\theta}, y = r\sin{\theta}$

$\displaystyle \iint_{\Omega_{n}}e^{-(x^2 + y^2)}dxdy$	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{2}}d\theta\int_{0}^{n}e^{-r^{2}}rdr$
	$\displaystyle =$	$\displaystyle \frac{\pi}{2}\left[-\frac{e^{-r^2}}{2}\right ]_{0}^{n} = \frac{\pi}{4}(1 - e^{-n^2})$

Thus, by letting $n \rightarrow \infty$ , we find $\textcolor{red}{I^2}$ and $I^2 = \pi/4$ $\blacksquare$ Consider a function $f(x,y)$

is not bounded on $\Omega$ .

Exercise 5..5 Evaluate the following double integrals.
1. $\displaystyle{I = \iint_{\Omega}\frac{dxdy}{(x+y)^{3/2}}, \Omega = \{(x,y) : 0 < x \leq 1, 0 < y \leq 1 \}}$
2. $\displaystyle{I = \iint_{\Omega}\tan^{-1}{\frac{y}{x}}dxdy, \Omega = \{(x,y) : x^2 + y^2 \leq 1, x \geq 0,y \geq 0 \}}$

SOLUTION $f(x,y) = \frac{1}{(x+y)^{3/2}}$ is discontinuous at $(0,0)$

. Then create $\Omega _n$ so that $(0,0)$

is not included in $\Omega _n$ .

$\displaystyle \Omega_{n} = \{(x,y) : \frac{1}{n} \leq x \leq 1, \frac{1}{n} \leq y \leq 1\}$

Then we get figure 5.9.

**Figure 5.8:** Sequences of $\Omega _n$
$% latex2html id marker 30249 \includegraphics[width=5cm]{SOFTFIG-5/enshu5-5-1.eps}$

Thus,

$\displaystyle \iint_{\Omega_{n}}\frac{dxdy}{(x+y)^{3/2}}$	$\displaystyle =$	$\displaystyle \int_{1/n}^{1}dx\int_{1/n}^{1} \frac{dy}{(x+y)^{3/2}} = \int_{1/n}^{1}\left[-\frac{2}{(x+y)^{1/2}}\right ]_{\frac{1}{n}}^{1}dx$
	$\displaystyle =$	$\displaystyle 2\int_{1/n}^{1}[(x+\frac{1}{n})^{-\frac{1}{2}} - (x+1)^{-\frac{1}{2}} ]dx$
	$\displaystyle =$	$\displaystyle 2\left[2(x+\frac{1}{n})^{\frac{1}{2}} - 2(x+1)^{\frac{1}{2}}\right ]_{\frac{1}{n}}^{1}$
	$\displaystyle =$	$\displaystyle 4[2(1+\frac{1}{n})^{\frac{1}{2}} - \sqrt{2} - (\frac{2}{n})^{\frac{1}{2}}]$

Therefore, as $n \rightarrow \infty$ we can find $I = 4(2 - \sqrt{2})$ $\blacksquare$

2. $f(x,y) = \tan^{-1}{\frac{y}{x}}$ is bounded except on $x$ -axis.

**Figure 5.9:** Sequence of Regions
$% latex2html id marker 30312 \includegraphics[width=5cm]{SOFTFIG-5/enshu5-5-2.eps}$

Using the polar coordinate, since $0 \leq x^2 + y^2 = r^2 \leq 1$ , $0 \leq r \leq 1$ . Since $x \geq 0, y \geq 0$ , $0 \leq \theta \leq \frac{\pi}{2}$ . Thus $\Omega$ maps to

$\displaystyle \Gamma = \{(r,\theta) : 0 \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq 1 \}$

Now let $\Gamma_n = \{(x,y): 0 \leq \theta \leq \frac{\pi}{2} - \frac{1}{n}, \frac{1}{n} \leq r \leq 1\}$ . Then,

$\displaystyle \tan^{-1}{\frac{y}{x}} = \tan^{-1}\left(\frac{r\sin{\theta}}{r\cos{\theta}}\right) = \tan^{-1}(\tan{\theta}) = \theta$

and is bounded on $\Gamma_n$ . Thus,

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Omega}\tan^{-1}{\frac{y}{x}}dxdy = \lim_{n \to \infty}\iint_{\Gamma_n}\theta rdrd\theta$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\int_{\theta = 0}^{\frac{\pi}{2}-\frac{1}{n}}\... ...{\frac{\pi}{2}-\frac{1}{n}}\big(\left[\frac{r^2}{2}\right]_{\frac{1}{n}}^1\big)$
	$\displaystyle =$	$\displaystyle \frac{\pi^2}{8}\cdot \frac{1}{2} = \frac{\pi^2}{16}\ensuremath{ \blacksquare}$

Exercise A

1.

Evaluate the following improper integrals．

(a) $\displaystyle{\iint_{\Omega}e^{y/x}dx dy, \Omega = \{(x,y) : 0 < x \leq 1, 0 \leq y \leq x \}}$

(b) $\displaystyle{\iint_{\Omega}\frac{x}{\sqrt{1 - x - y}}dx dy, \Omega = \{(x,y) : x+y < 1, x \geq 0, y \geq 0\}}$

Exercise B

1.

Evaluate the following improper integrals．

(a) $\displaystyle{\iint_{\Omega}\frac{dxdy}{(y^2 - x^2)^{1/2}}, \Omega = \{(x,y) : 0 \leq x < y \leq 1\}}$

(b) $\displaystyle{\iint_{\Omega}e^{-(x+y)}dxdy, \Omega = \{(x,y) : x \geq 0, y \geq 0\}}$

(d) $\displaystyle{\iint_{\Omega}\frac{1}{x^2y^{2}}dxdy, \Omega = \{(x,y) : x \geq 1, y \geq 1 \}}$

(e) $\displaystyle{\iint_{\Omega}\frac{dxdy}{\sqrt{x - y^2}}, \Omega = \{(x,y) : 0 \leq x \leq 1, y^2 \leq x \}}$

(f) $\displaystyle{\iint_{\Omega}\frac{dxdy}{1 + (x^2 + y^2)^2}, \Omega = \{(x,y) : -\infty < x,y < \infty \}}$

2.

Using thee example 5.5, show that $\displaystyle{\Gamma\left(\frac{1}{2}\right) = \int_{0}^{\infty} x^{-\frac{1}{2}}e^{-x} dx = \sqrt{\pi}}$ ．