Here’s a simple Calc 1 problem:
Evaluate
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]](https://s0.wp.com/latex.php?latex=%5Cint_%7B-1%7D%5E1+%5Cfrac%7B1%7D%7Bx%7Ddx+%3D+%5Clim_%7Ba%5Cto+0%5E%2B%7D%5Cleft%5B%5Cint_%7B-1%7D%5E%7B-a%7D%5Cfrac%7B1%7D%7Bx%7Ddx%2B%5Cint_a%5E%7B1%7D%5Cfrac%7B1%7D%7Bx%7Ddx%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
![= \lim_{a\to 0^+}\left[\ln(a)-\ln(a)\right]=\boxed{0}](https://s0.wp.com/latex.php?latex=%3D+%5Clim_%7Ba%5Cto+0%5E%2B%7D%5Cleft%5B%5Cln%28a%29-%5Cln%28a%29%5Cright%5D%3D%5Cboxed%7B0%7D&bg=ffffff&fg=333333&s=0&c=20201002)
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]](https://s0.wp.com/latex.php?latex=%5Cint_%7B-1%7D%5E1+%5Cfrac%7B1%7D%7Bx%7Ddx+%3D+%5Clim_%7Ba%5Cto+0%5E%2B%7D%5Cleft%5B%5Cint_%7B-1%7D%5E%7B-a%7D%5Cfrac%7B1%7D%7Bx%7Ddx%2B%5Cint_%7B2a%7D%5E%7B1%7D%5Cfrac%7B1%7D%7Bx%7Ddx%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
![= \lim_{a\to 0^+}\left[\ln(a)-\ln(2a)\right]=\boxed{\ln{\frac{1}{2}}}](https://s0.wp.com/latex.php?latex=%3D+%5Clim_%7Ba%5Cto+0%5E%2B%7D%5Cleft%5B%5Cln%28a%29-%5Cln%282a%29%5Cright%5D%3D%5Cboxed%7B%5Cln%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D&bg=ffffff&fg=333333&s=0&c=20201002)
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]](https://s0.wp.com/latex.php?latex=%5Cint_%7B-1%7D%5E1+%5Cfrac%7B1%7D%7Bx%7Ddx+%3D+%5Clim_%7Ba%5Cto+0%5E-%7D%5Cleft%5B%5Cint_%7B-1%7D%5E%7Ba%7D%5Cfrac%7B1%7D%7Bx%7Ddx%5Cright%5D%2B%5Clim_%7Bb%5Cto+0%5E%2B%7D%5Cleft%5B%5Cint_b%5E%7B1%7D%5Cfrac%7B1%7D%7Bx%7Ddx%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
![= \lim_{a\to 0^-}\left[\ln|a|\right]+\lim_{b\to 0^+}\left[-\ln|b|\right]](https://s0.wp.com/latex.php?latex=%3D+%5Clim_%7Ba%5Cto+0%5E-%7D%5Cleft%5B%5Cln%7Ca%7C%5Cright%5D%2B%5Clim_%7Bb%5Cto+0%5E%2B%7D%5Cleft%5B-%5Cln%7Cb%7C%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
But notice that we can’t actually resolve this last expression, since the first limit is 
All that being said, we still intuitively feel like this integral should have the value 0 rather than something else like 
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
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.
Random Walks 
I think that the CPV is beautiful.
That’s the only justification I need.