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What is the definite integral $\displaystyle\int_{-1}^1 \frac{1}{x} dx$ equal to?

Is it $0$ like it would be if you just integrated and plugged in the bounds, or does the discontinuity at $x=0$ invalidate that line of reasoning and make this integral undefined. If you split up the integral $$\int_{-1}^0 \frac{1}{x} dx+\int_{0}^1 \frac{1}{x} dx.$$ You get $\infty-\infty$ which is undefined. I find it difficult to prove either of these solutions.

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    $\begingroup$ Improper integral $\int\limits_{-1}^1 \dfrac{1}{x} dx$ is divergent. $\endgroup$ – M. Strochyk Oct 15 '13 at 21:40
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    $\begingroup$ Function $\ln|x|$ is not continuous on $[-1,1]$, so the "plug in the bounds" method (fundamental theorem of calculus) is not applicable. $\endgroup$ – GEdgar Oct 16 '13 at 0:25
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The improper integral

$$ \int^{1}_{-1} \frac{1}{x} \, dx $$

is indeed, as M.Strochyk says, divergent. At least in the sense of Riemann integrability, it doesn't make any sense to try and evaluate this integral because we require our integrand to be bounded over the region we are integrating, which doesn't occur in this case. Also, for this reason, statements such as

$$ \int^{1}_{-1} \frac{1}{x} \, dx = \int^{1}_{0} \frac{1}{x} \, dx + \int^{0}_{1} \frac{1}{x} \, dx$$

also make no sense, seeing as the statement is true if and only if the integrals on both sides exist. However, because the function $f : \mathbb{R} \setminus \{0\} \rightarrow \mathbb{R}$ with $f(x)=1/x$ is odd, and we often like to use that if $g(x)$ is an odd function on some interval $I$ with $0 \in I$ then

$$ \int^{a}_{-a} g = 0$$

provided $a,-a \in I$, in some sense we would like to give the original integrand (I'd mark it with a star if I knew how, but alas I do not) a similar value of $0$. One way in which we can do it is by defining the Cauchy principal value, in which case after reading the definition we see that

$$ \text{P.V.} \int^{1}_{-1} \frac{1}{x} \, dx = \lim_{a \rightarrow 0^{+}} \left( \int^{-a}_{-1} \frac{1}{x} \, dx + \int^{1}_{a} \frac{1}{x} \, dx \right) = 0 $$

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$$ \left(\int_\varepsilon^1 + \int_{-1}^{-2\varepsilon} \right) \frac{dx}{x} \to \log_e 2\ \text{ as }\ \varepsilon\downarrow0. $$ But if $-3\varepsilon$ had appeared where $-2\varepsilon$ is, then the limit would be a different number.

When both the positive part and the negative part of an integral or an infinite series diverges to infinity, then you can change the value of the sum or integral by "rearranging" it (which, in the case of infinite series, means changing the order in which the terms appear).

The "principal value", sometimes called the "Cauchy principal value", is the limit when $-\varepsilon$ appears above where I put $-2\varepsilon$. By symmetry, it's $0$.

In Lebesgue's theory of integration, $\displaystyle\int_{-1}^1\frac{dx}{x}$ is undefined. Lebesgue deals only with cases in which $\displaystyle\int_A |f|<\infty$.

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