# Why doesn't $\int_{-1}^{1}\frac{dx}{x} = \ln|x|\biggr\rvert_{-1}^{1} = 0$?

$1/x$ is an odd function, so it makes sense to me intuitively that the area would be $0$, and similarly I would expect that $\int_{-1}^{2}\frac{dx}{x} = \ln(2)$.

Proof Wiki seems to confirm my intuition, but with the exception of functions that don't have a primitive (i.e. integral?), which I guess this one doesn't, because of the discontinuity at $x=0$.

Nonetheless, it seems to me that the area under $1/x$ must be $0$ because:

$$\int_{-1}^{1}\frac{dx}{x} = \lim_{a\to0} \left[ \int_{-1}^{a}{x^{-1} + \int_{a}^{1}{x^{-1}}} \right] = 0$$

I just can't shake the intuitive feeling that the area is $0$. Bonus points if you can explain why it is not $0$ in an intuitive way.

• Although the two integrals are symmetric and opposite, they are divergent; that is, the areas are infinite. – Tim Thayer Feb 3 '16 at 1:44
• @TimThayer: But $\lim_{x\to∞}(x-x) = 0$, right? Why is it not the same here? – Zaz Feb 3 '16 at 1:46
• You could then say something like this: $\int_{-1}^1 x^{-1} = \lim_{a\to 0} \left[\int_{-1}^{-a/2} x^{-1} + \int_{a}^{1} x^{-1} \right] = \lim (\ln(a/2)-\ln(a))=\ln(1/2)$. Is $0=\ln(1/2)$? – user160738 Feb 3 '16 at 1:48
• Because the first equality is valid only if both limits inside of it exist (i.e., are finite). – B. Freitas Feb 3 '16 at 1:49
• Only the CPV of the integral is zero. You get the CPV when you take the limits symmetrically from either side of the discontinuity. See en.wikipedia.org/wiki/Cauchy_principal_value. – okrzysik Feb 3 '16 at 2:15

Your intuition (expressed in the comments below the OP) is that the integral in question is analogous to $$\lim_{x \to \infty}(x-x)$$ which is, of course, $0$. But a better way to think about it is to consider $$\lim_{(x,y)\to(\infty,\infty)}(x-y)$$ which informally can be thought of as $\infty - \infty$, but this cannot be simply expressed as $0$. More precisely, the value of that limit depends on how $x$ and $y$ go to infinity -- what path they follow, what rate they diverge relative to one another.
Likewise, when you look at an integral of an odd function over a symmetric interval with a pole in it, it may be very tempting to imagine approaching the vertical asymptote from the left and right sides at the same "rate"; if you do that then the areas always cancel out, and you end up with $0$. This is what other answers have referred to as the Cauchy principal value. But you could also take those two limits (approaching from the left and from the right) independently, at different rates, in which case the areas would not cancel, and almost anything is possible.
• Ahh. So an example of approaching the asymptote at different rates would be @user160738's comment: $$∫^1_{−1}x^{−1}=\lim_{a→0}\left[∫^{−a/2}_{−1}x^{−1}+∫^{1}_{a}x^{−1}=\ln\left( \frac {a}{2}\right)−\ln(a)\right]=\ln\left(\frac{1}{2}\right)$$ – Zaz Feb 3 '16 at 22:15
• Would you ever say that there is a sense in which the integral is $0$, or would you always talk about the Cauchy Principle Value as something distinct from the integral? – Zaz Feb 3 '16 at 22:21
• What we would say is either "The integral is $0$ in the sense of the Cauchy Principal Value" or "The Cauchy Principal Value of the integral is $0$." – mweiss Feb 3 '16 at 22:45
Simple: $\int_{-1}^{1} \frac{dx}{x}$ is not well defined. The area, if it should exist in whatever definition of it you choose, is certainly not given by this expression.