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If $\frac{1}{x}$ is symmetric about the x and y axis, why isn't $\int_{-1}^{1} \frac{1}{x}dx = 0$?

Visually, it's pretty easy to see that the area we get calculating $\int_{0}^{1} \frac{1}{x}dx$ is the exact same as the negative area we get calculating $\int_{-1}^{0} \frac{1}{x}dx$. Yes, they're both infinity but they're the exact opposite infinities of each other. So why do we say $\int_{-1}^{1} \frac{1}{x}dx$ diverges?

To clarify, I understand that when dealing with infinities it's improper to do a lot of calculations with infinity because we don't know which infinities we're dealing with or how fast things are approaching infinity (the reason we need L'Hospital's Rule for indeterminate form when evaluating certain limits).

But in a case like this, I feel like we could make some sort of argument that if we remove the point x = 0 for a moment, then if $f(x) = \frac{1}{x}$ for $-1 \leq x < 0$, and g(x) = $\frac{1}{x}$, for $0 < x \leq 1$, then we know that f(x) = -g(x) in their respective intervals and so f(x) + g(x) = 0 in their respective intervals. This would inherently mean that f(x)dx + g(x)dx = 0 in their respective intervals.

Obviously there's something mathematically fishy with the non-removable discontinuity at x = 0 but the natural intuition for me is that these integrals should perfectly cancel. Why do we say "divergent?"

PS: I read something about a Cauchy principal value being 0 for this integral, but I think that's beyond my scope of understanding. (think first-time integral calculus student)

J. W. Tanner
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The Math Potato
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  • What makes you believe that (a) that integral is even defined, and (b) nonzero? Given that you're a first-time integral calculus student, I assume you're talking about the Riemann integral, right? – John Hughes Mar 16 '25 at 20:18
  • Right. From the perspective of thinking about it as area under the curve, it would seem that the areas are perfectly symmetric, so asking to evaluate that definite integral is asking "what's the area under the curve" and I would think it's 0 since they perfectly, symmetrically cancel out. Even if the areas are both infinity and negative infinity, they are the exact same (well, exact opposites), right? – The Math Potato Mar 16 '25 at 20:21
  • @DietrichBurde the first answer in that thread simply says something along the lines of "because the Stewart book defines each of those integrals to be divergent so we say their sum is also divergent" with no explanation addressing the area argument, and the next answer refers to imaginary numbers and complex math which I don't have any exposure to. I was hoping for an intuitive way of making sense of this. – The Math Potato Mar 16 '25 at 20:25
  • In general, the fact that two things both blow up ("go to infinity") doesn't make them equal. But in this particular case, the Cauchy principal value is zero. We just don't tend to use that particular definition in most calculus textbooks. – David K Mar 16 '25 at 20:33
  • I do understand that in general two things going to infinity don't make them equal (and therefore able to cancel out). But in this case they are the exact opposites of each other. No argument could be made (as far as I can tell) that the area from -1 to 0 is any different in size than the area from 0 to 1 (due to symmetry). So how could we not conclude that the areas cancel each other out?? – The Math Potato Mar 17 '25 at 21:05

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