# Improper integrals debate

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 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?

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**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”

1. I think that the CPV is beautiful.

That’s the only justification I need.