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

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

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

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

exists. Then we say $f(x,y)$ is integrable on $\Omega$ and define

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

Understanding

If a function $z = f(x,y)$ is not bounded on a region, then we say the double integral is improper integral of the 1st kind. If the region is not bounded, then we say the double integral is improper integral of the 2nd kind. If a function is not bounded on unbounded region, then we say the double integral is improper integral of the 3rd kind.

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

$0 \leq x \leq 1, 0 \leq y \leq 1, x \leq y$.

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

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

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

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

Create $\Omega _n$ so that $\Omega \subset \Omega_n$ as $n$ goes to infinity.

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

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

The figure of $\{\Omega_{n}\}$ is given by the figure 5.19.

Figure 5.19: Sequences

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

Check

$x^2 + y^2 = r^2 \leq n^2$ implies that $r \leq n$. Since $x,y \geq 0$, $0 \leq \theta \leq \frac{\pi}{2}$.

$$\iint_{\Omega_{n}}e^{-(x^2 + y^2)}dxdy = \int_{0}^{\frac{\pi}{2}}d\theta\int_{0}^{n}e^{-r^{2}}rdr$$

$$= \frac{\pi}{2}\left[-\frac{e^{-r^2}}{2}\right ]_{0}^{n} = \frac{\pi}{4}(1 - e^{-n^2})$$

Thus, by letting $n \rightarrow \infty$, we find $I^2$ and $I^2 = \pi/4$ $\blacksquare$

$\int_0^{\infty}e^{-x^2}\:dx = \frac{\sqrt{\pi}}{2}$

$I = \int_0^\infty e^{-x^2}dx$, $I = \int_0^\infty e^{-y^2}dy$. Then $I^2 = \int_0^\infty e^{-x^2}dx \int_0^\infty e^{-y^2}dy = \int_0^\infty \int_0^\infty e^{-x^2-y^2}dydx$implies $I = \frac{\sqrt{\pi}}{2}$.

Consider a function $f(x,y)$ is not bounded on $\Omega$.

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

Exercise5-5-2.

Create $\Omega _n$ so that $\Omega$ is a subset of $\Omega _n$ as $n$ goes to infinity.

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 $\Omega _n$.

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

Then we get figure 5.21.

Figure 5.20: Sequences of $\Omega _n$

Thus,

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

$$= 2\int_{1/n}^{1}[(x+\frac{1}{n})^{-\frac{1}{2}} - (x+1)^{-\frac{1}{2}} ]dx$$

$$= 2\left[2(x+\frac{1}{n})^{\frac{1}{2}} - 2(x+1)^{\frac{1}{2}}\right ]_{\frac{1}{n}}^{1}$$

$$= 4[2(1+\frac{1}{n})^{\frac{1}{2}} - \sqrt{2} - (\frac{2}{n})^{\frac{1}{2}}]$$

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

Check

$\int_{1/n}^{1} \frac{dy}{(x+y)^{3/2}} = \\ \int_{1/n}^{1} (x+y)^{-3/2}\:dy$.
This is the integral with respect to $y$. Thus we treat $x$ as a constant. Now $\int_{1/n}^{1} (x+y)^{-3/2}\:dy = \left[-2(x+y)^{-\frac{1}{2}}\right]_{\frac{1}{n}}^1$

Exercise5-5-2.

$\tan^{-1}{\frac{y}{x}}$ is not bounded at $x = 0$. Using polar coordinate transformation $x = r\cos{\theta}, y = r\sin{\theta}$, the region $\Omega$ can be covered by taking $\theta$ goes from 0 to $\frac{\pi}{2}$ and the distance from the pole ranges from 0 to $1$. Thus, $\Gamma = \{(r,\theta):0 \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq 1\}$.

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

Figure 5.21: Sequence of Regions

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

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

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

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

and is bounded on $\Gamma_n$. Thus,

$$I = \iint_{\Omega}\tan^{-1}{\frac{y}{x}}dxdy = \lim_{n \to \infty}\iint_{\Gamma_n}\theta rdrd\theta$$

$$= \lim_{n \to \infty}\int_{\theta = 0}^{\frac{\pi}{2}-\frac{1}{n}}\... ...{\frac{\pi}{2}-\frac{1}{n}}\big(\left[\frac{r^2}{2}\right]_{\frac{1}{n}}^1\big)$$

$$= \frac{\pi^2}{8}\cdot \frac{1}{2} = \frac{\pi^2}{16}\ensuremath{ \blacksquare}$$

Exercise A

1.

Evaluate the following improper integrals．

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

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

(c) $${\iint_{\Omega}\log(x^{2} + y^{2})dxdy, \Omega = \{(x,y) : x,y \geq 0, x^2 + y^2 \leq 1 \}}$$

Exercise B

1.

Evaluate the following improper integrals．

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

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

(c) $${\iint_{\Omega}\tan^{-1}(\frac{y}{x})dxdy, \Omega = \{(x,y) : x,y \geq 0, x^2 + y^2 \leq 1 \}}$$

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

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

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

2.

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