Why is the definite integral of 1/x from -1 to 1 undefined?

[Saracen Rue]

TL;DR Summary

Why is $\int_{-1}^{1} \frac {1} {x} dx$ undefined?

I've always been taught that the indefinite integral of $\frac{1}{x}$ is $\ln(|x|)$. Extending this to definite integrals, particularly over limits involving negative values, should work just like any other integral:

$$\int_{-1}^{1} \frac {1} {x} dx = \ln(|-1|) - \ln(|1|) = \ln(1) - \ln(1) = 0 - 0 = 0$$

However, this doesn't seem to be the case as every online calculator/program I use states the answer as being undefined. What I'm struggling with is why it's undefined, as all the rules I know for integration and maths in general imply the answer should be $0$. I should note that I've never been the strongest with understanding the theory behind a lot of calculus. My teachers always dismissed my questions in high school or didn't know the answers themselves and insisted for me to just follow the formulae they give to get the answers they want. Any actual theory I know is self taught, so I apologies in advance if I struggle to follow explanations for this question if cover more complicated theory.

Thank you for your time.

 

[Gaussian97]

Well, for short, you are using the fundamental theorem of calculus, which assumes the integrand function $f(x)$ to be defined in the closed interval $[a,b]$. Then
$$\int_a^b f(x)\, dx = F(b)-F(a)$$
But $f(x)=x^{-1}$ is not defined in $[-1,1]$ since it's not defined for $x=0$ and FTC is not applicable. So the integral $$\int_{-1}^{1} x^{-1}\, dx$$ defined as the area under the curve is simply not defined. This integral is usually defined by the limits
$$\int_{-1}^{1} x^{-1}\, dx = \lim_{\varepsilon \to 0^-}\int_{-1}^{\varepsilon} x^{-1}\, dx + \lim_{\varepsilon \to 0^+}\int_{\varepsilon}^{1} x^{-1}\, dx$$
Since neither of the two limits exist, the integral cannot exist. Although if you want, you can use a different definition for the integral, usually known as its Principal Value:
$$\int_{-1}^{1} x^{-1}\, dx = \lim_{\varepsilon \to 0^-}\left(\int_{-1}^{\varepsilon} x^{-1}\, dx + \int_{\varepsilon}^{1} x^{-1}\, dx\right) = 0$$

 

[Delta2]

The definition of the Riemann Integral allows for countable many discontinuities or points where the function is not defined over the domain of integration. However the function must be bounded over the domain of integration and this function isn't bounded over (-1,1)(it becomes $\pm\infty$ at 0) hence the Riemann Integral doesn't exist.

 

[Anixx]

As other people said, the Riemann integral does nor exist, but this does not mean you cannot integrate it this or that way. For instance, the Cauchy principal value of this integral is 0.
Also, via Laplace transform and comparison to the harmonic series, $\int_0^1 1/x dx$ has regularized value of 0.

 

[jbunniii]

It's worth mentioning that the Lebesgue integral also does not exist. The Lebesgue integrals over the intervals $[-1,0]$ and $[0,1]$ exist (as extended real numbers):
$$\int_{-1}^{0}\frac{1}{x}\ dx = -\infty$$
$$\int_{0}^{1}\frac{1}{x}\ dx = +\infty$$
But $\int_{-1}^{1} \frac{1}{x}\ dx$ does not exist, because if it did, it would have to equal $\int_{-1}^{0}\frac{1}{x}\ dx + \int_{0}^{1}\frac{1}{x}\ dx = -\infty + \infty$, which is undefined.

Geometrically speaking, the graph of the part of the function to the right of the y-axis has positive infinite area, and the graph of the part to the left of the y-axis has negative infinite area. You can't "cancel" $+\infty$ and $-\infty$ in a well-defined way.