Improper Integrals

Exam 2 Wed Mar 1: 7pm-7:50pm in ??
Today: 7.8 Improper Integrals
Monday - president's day holiday (and almost my bday)
Next -- 11.1 sequences

Example 5.7.1   Make sense of $\int_{0}^{\infty} e^{-x} dx$. The integrals

$$\int_0^t e^{-x} dx$$

make sense for each real number $t$. So consider

$$\lim_{t\to\infty} \int_0^t e^{-x} dx = \lim_{t\to\infty} [-e^{-x}]_0^t = 1.$$

Geometrically the area under the whole curve is the limit of the areas for finite values of $t$.

Figure 5.7.1: Graph of $e^{-x}$

Example 5.7.2   Consider $\int_0^1\frac{1}{\sqrt{1-x^2}} dx$ (see Figure 5.7.2).

Figure 5.7.2: Graph of $\frac{1}{\sqrt{1-x^2}}$

Problem: The denominator of the integrand tends to 0 as $x$ approaches the upper endpoint. Define

$$\int_0^1\frac{1}{\sqrt{1-x^2}} dx = \lim_{t\to 1^-} \int_0^t\frac{1}{\sqrt{1-x^2}} dx$$

$$= \lim_{t\to 1^-} \Bigl( \sin^{-1}(t) - \sin^{-1}(0) \Bigr) = \sin^{-1}(1) = \frac{\pi}{2}$$

Here $t\to 1^-$ means the limit as $t$ tends to $1$ from the left.

Example 5.7.3   There can be multiple points at which the integral is improper. For example, consider

$$\int_{-\infty}^{\infty} \frac{1}{1+x^2} dx.$$

A crucial point is that we take the limit for the left and right endpoints independently. We use the point 0 (for convenience only!) to break the integral in half.

$$\int_{-\infty}^{\infty} \frac{1}{1+x^2} dx = \int_{-\infty}^{0} \frac{1}{1+x^2} dx + \int_{0}^{\infty} \frac{1}{1+x^2} dx$$

$$= \lim_{s\to -\infty} \int_{s}^{0} \frac{1}{1+x^2} dx + \lim_{t\to\infty} \int_{0}^{t} \frac{1}{1+x^2} dx$$

$$= \lim_{s\to-\infty} (\tan^{-1}(0) - \tan^{-1}(s)) + \lim_{t\to\infty} (\tan^{-1}(t) - \tan^{-1}(0))$$

$$= \lim_{s\to-\infty} (-\tan^{-1}(s)) + \lim_{t\to\infty} (\tan^{-1}(t))$$

$$= -\frac{-\pi}{2} + \frac{\pi}{2} = \pi.$$

The graph of $\tan^{-1}(x)$ is in Figure 5.7.3.

Figure: Graph of $\tan^{-1}(x)$

Example 5.7.4   Brian Conrad's paper on impossibility theorems for elementary integration begins: ``The Central Limit Theorem in probability theory assigns a special significance to the cumulative area function

$$\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-u^2}{u} du$$

under the Gaussian bell curve

$$y=\frac{1}{\sqrt{2\pi}} \cdot e^{-u^2/2}.$$

It is known that $\Phi(\infty) = 1$.''

What does this last statement mean? It means that

$$\lim_{t\to\infty} \frac{1}{\sqrt{2\pi}} \int_{-t}^0 e^{-u^2}{u} du + \lim_{x\to\infty} \frac{1}{\sqrt{2\pi}} \int_{0}^x e^{-u^2}{u} du = 1.$$

Example 5.7.5   Consider $\int_{-\infty}^{\infty} x dx$. Notice that

$$\int_{-\infty}^{\infty} x dx = \lim_{s\to-\infty} \int_{s}^{0} x dx + \lim_{t\to \infty} \int_{0}^{t} x dx.$$

This diverges since each factor diverges independtly. But notice that

$$\lim_{t\to\infty} \int_{-t}^t x dx = 0.$$

This is not what $\int_{-\infty}^{\infty} x dx$ means (in this course - in a later course it could be interpreted this way)! This illustrates the importance of treating each bad point separately (since Example 5.7.3) doesn't.

Example 5.7.6   Consider $\int_{-1}^1 \frac{1}{\sqrt[3]{x}} dx$. We have

$$\int_{-1}^1 \frac{1}{\sqrt[3]{x}} dx = \lim_{s \to 0^-} \int_{-1}^s x^{-\frac{1}{3}} dx + \lim_{t\to 0^+} \int_t^1 x^{-\frac{1}{3}} dx$$

$$= \lim_{s \to 0^-} \left(\frac{3}{2} s^{\frac{2}{3}} - \frac{3}{2... ...+ \lim_{t \to 0^+} \left(\frac{3}{2} - \frac{3}{2} t^{\frac{2}{3}} \right) = 0.$$

This illustrates how to be careful and break the function up into two pieces when there is a discontinuity.

NOTES for 2006-02-22
Midterm 2: Wednesday, March 1, 2006, at 7pm in Pepper Canyon 109
Today: 7.8: Comparison of Improper integrals
11.1: Sequences
Next 11.2 Series

Example 5.7.7   Compute $\int_{-1}^3 \frac{1}{x-2} dx$. A few weeks ago you might have done this:

$$\int_{-1}^3 \frac{1}{x-2} dx = [\ln\vert x-2\vert]_{-1}^{3} = \ln(3) - \ln(1) \qquad{\text{(totally wrong!)}}$$

This is not valid because the function we are integrating has a pole at $x=2$ (see Figure 5.7.4). The integral is improper, and is only defined if both the following limits exists:

$$\lim_{t\to 2^-} \int_{-1}^t \frac{1}{x-2} dx$$   and$$\qquad \lim_{t\to 2^+} \int_{t}^3 \frac{1}{x-2} dx.$$

However, the limits diverge, e.g.,

$$\lim_{t\to 2^+} \int_{t}^3 \frac{1}{x-2} dx = \lim_{t\to 2^+} (\l... ...t 1\vert - \ln\vert t-2\vert) = - \lim_{t\to 2^+} \ln\vert t-2\vert = -\infty.$$

Thus $\int_{-1}^3 \frac{1}{x-2} dx$ is divergent.

Figure 5.7.4: Graph of $\frac {1}{x-2}$