Dear Uncle Colin,

I have a disagreement with my teacher about the integral $\int_{-1}^{1} x^{-1} \dx$. I understand you have to split the integral into two parts, which I’m happy with. They split it from -1 to $a$, letting $a \rightarrow 0_-$ and from $b$ to 1, letting $b \rightarrow 0_+$; both of those integrals diverge, so the answer is undefined.

I, however, think you can make use of the symmetry and let $a=b$ – the two integrals clearly have the same value with opposite signs so the answer should be zero. Is there a flaw with my method?

-- Limited Entertainment Involving Benignly Not Integrating Zero

Dear LEIBNIZ, ‘fraid so.

Sadly, your teacher is right — the integral, however much it looks symmetrical and cancelly-out, falls foul to Beveridge’s First Law Of Infinity ((Don’t “mess” about with infinity)). “Clearly” is an especially dangerous word when it comes to infinity.

Before getting on to the integral, here’s a taster of the sort of problem you’re running into, using a series: what is the sum of the integers? Obviously 0, right? You can write it as $0 + (1 + -1) + (2 + -2) + (3 + -3) + … = 0 + 0 + 0 + …$. It all “clearly” cancels out.

But it’s just as “clearly” positive infinity: $1 + (0 + 2) + (-1 + 3) + (-2 + 4) + … = 1 + 2 + 2 + 2 + …$

And just as “clearly” negative infinity: $-1 + (-2 + 0) + (-3 + 1) + (-4 + 2) + … = -1 + -2 + -2 + …$

Because you can arrange the series to give different sums (among other reasons), the sum doesn’t converge.

Still with me? Splendid.

You can do something pretty much identical with your integral — you can split the areas you’re interested in into nearly-trapezia of areas $\pm 1$, $\pm 2$ etc, getting narrower as you get close to the asymptote. Your way of matching them up is maybe the obvious one, but it’s not the only one — so the sum doesn’t converge.

There’s generally a problem with sums and integrals that go to infinity in both directions: they only converge if the two individual sums do. (For an example: $\int_{-1}^{1} x^{-\frac 13} \dx$ is 0 — each ‘half’ has a well-defined area of $\pm \frac{3}{2}$.)

As for why your method doesn’t work, it’s because no matter how small your $a$ is, there’s still an infinite area under the curve between 0 and $a$, and above the curve between $-a$ and 0. Even if they’re “clearly” the same area, you can’t simply cancel out integrals that don’t converge.

Hope that clears it up!

-- Uncle Colin