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

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

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

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

$\displaystyle \int_0^1\frac{1}{\sqrt{1-x^2}} dx$ $\displaystyle = \lim_{t\to 1^-} \int_0^t\frac{1}{\sqrt{1-x^2}} dx$    
  $\displaystyle = \lim_{t\to 1^-} \Bigl( \sin^{-1}(t) - \sin^{-1}(0) \Bigr) = \sin^{-1}(1) = \frac{\pi}{2}$    

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

$\displaystyle \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.

$\displaystyle \int_{-\infty}^{\infty} \frac{1}{1+x^2} dx$ $\displaystyle = \int_{-\infty}^{0} \frac{1}{1+x^2} dx + \int_{0}^{\infty} \frac{1}{1+x^2} dx$    
  $\displaystyle = \lim_{s\to -\infty} \int_{s}^{0} \frac{1}{1+x^2} dx + \lim_{t\to\infty} \int_{0}^{t} \frac{1}{1+x^2} dx$    
  $\displaystyle = \lim_{s\to-\infty} (\tan^{-1}(0) - \tan^{-1}(s)) + \lim_{t\to\infty} (\tan^{-1}(t) - \tan^{-1}(0))$    
  $\displaystyle = \lim_{s\to-\infty} (-\tan^{-1}(s)) + \lim_{t\to\infty} (\tan^{-1}(t))$    
  $\displaystyle = -\frac{-\pi}{2} + \frac{\pi}{2} = \pi.$    

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

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

under the Gaussian bell curve

$\displaystyle 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

$\displaystyle \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

$\displaystyle \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

$\displaystyle \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

$\displaystyle \int_{-1}^1 \frac{1}{\sqrt[3]{x}} dx$ $\displaystyle = \lim_{s \to 0^-} \int_{-1}^s x^{-\frac{1}{3}} dx    +   \lim_{t\to 0^+} \int_t^1 x^{-\frac{1}{3}} dx$    
  $\displaystyle = \lim_{s \to 0^-} \left(\frac{3}{2} s^{\frac{2}{3}} - \frac{3}{2...
...+ \lim_{t \to 0^+} \left(\frac{3}{2} - \frac{3}{2} t^{\frac{2}{3}} \right) = 0.$    

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:

$\displaystyle \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:

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

However, the limits diverge, e.g.,

$\displaystyle \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}$
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William Stein 2006-03-15