Improper integrals debate

Posted on by

Here’s a simple Calc 1 problem:

Evaluate  \int_{-1}^1 \frac{1}{x}dx

Before you read any of my own commentary, what do you think? Does this integral converge or diverge?

image from illuminations.nctm.org

Many textbooks would say that it diverges, and I claim this is true as well. But where’s the error in this work?

\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^+}\left[\int_{-1}^{-a}\frac{1}{x}dx+\int_a^{1}\frac{1}{x}dx\right]

= \lim_{a\to 0^+}\left[\ln(a)-\ln(a)\right]=\boxed{0}

Did you catch any shady math? Here’s another equally wrong way of doing it:

\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^+}\left[\int_{-1}^{-a}\frac{1}{x}dx+\int_{2a}^{1}\frac{1}{x}dx\right]

= \lim_{a\to 0^+}\left[\ln(a)-\ln(2a)\right]=\boxed{\ln{\frac{1}{2}}}

This isn’t any more shady than the last example. The change in the bottom limit of integration in the second piece of the integral from a to 2a is not a problem, since 2a approaches zero if a does. So why do we get two values that disagree? (In fact, we could concoct an example that evaluates to ANY number you like.)

Okay, finally, here’s the “correct” work:

\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^-}\left[\int_{-1}^{a}\frac{1}{x}dx\right]+\lim_{b\to 0^+}\left[\int_b^{1}\frac{1}{x}dx\right]

= \lim_{a\to 0^-}\left[\ln|a|\right]+\lim_{b\to 0^+}\left[-\ln|b|\right]

But notice that we can’t actually resolve this last expression, since the first limit is \infty and the second is -\infty and the overall expression has the indeterminate form \infty - \infty. In our very first approach, we assumed the limit variables a and b were the same. In the second approach, we let b=2a. But one assumption isn’t necessarily better than another. So we claim the integral diverges.

All that being said, we still intuitively feel like this integral should have the value 0 rather than something else like \ln\frac{1}{2}. For goodness sake, it’s symmetric about the origin!

In fact, that intuition is formalized by Cauchy in what is called the “Cauchy Principal Value,” which for this integral, is 0. [my above example is stolen from this wikipedia article as well]

I’ve been debating about this with my math teacher colleague, Matt Davis, and I’m not sure we’ve come to a satisfying conclusion. Here’s an example we were considering:

If you were to color in under the infinite graph of y=\frac{1}{x} between -1 and 1, and then throw darts at  the graph uniformly, wouldn’t you bet on there being an equal number of darts to the left and right of the y-axis?

Don’t you feel that way too?

(Now there might be another post entirely about measure-theoretic probability!)

What do you think? Anyone want to weigh in? And what should we tell high school students?

.

**For a more in depth treatment of the problem, including a discussion of the construction of Reimann sums, visit this nice thread on physicsforums.com.

1 thought on “Improper integrals debate”

Leave a comment