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I understand that when evaluating

$ \int_{-1}^{2} \frac{1}{x} \mathrm dx = \ln 2$

It's simple integration, I understand. I'm more focused on the theory behind if it even exists. I had a question from Larson, Edward's Calculus 9th edition that was a true or false relating to this earlier today.

In one sense, I thought it had to be $\ln 2$. But, when looking at it again, technically the area from $-1$ to $0$, and $0$ to $1$ are negative infinity and infinity, respectively. They should cancel out and our original integral from $-1$ to $2$ is the same as the integral from $1$ to $2$. However, technically the integral does not converge from those two endpoints. The areas from $-1$ to $0$ and $0$ to $1$ are only infinitesimally close as they both approach zero from one side.

I was looking for your guys' input here. I was debating myself most of today over this seamlessly simple true/false question.

Note: this is a Calc 1 class, but I suppose I'm just getting ideas from Calc 2 onward (I've self-studied a bit) intertwined with our knowledge we've learned so far.

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    @JayElectronica `$..$` will show inlined LaTeX, and `$..$` will show displaystyle centered TeX equations (similar to LaTeX's `\[..\]`).2012-03-05

2 Answers 2

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What you are getting is known as the Cauchy Principal Value for this Improper Integral. It is actually $ \begin{align} \lim_{\epsilon\to0^+}\left(\int_{-1}^{-\epsilon}\frac{1}{x}\mathrm{d}x+\int_{\epsilon}^{2}\frac{1}{x}\mathrm{d}x\right) &=\lim_{\epsilon\to0^+}\left([\log(\epsilon)-\log(1)]+[\log(2)-\log(\epsilon)]\vphantom{\int}\right)\\ &=\lim_{\epsilon\to0^+}\left(\log(2)-\log(1)\vphantom{\int}\right)\\ &=\log(2) \end{align} $

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Just a small remark here, usually in introductory calculus classes, when we consider an integral of the form $ \int_a^b f(x) dx$ where $f(x)$ is discontinuous at the point $c$ but is otherwise continuous on the interval $[a,b]$, then we say that the integral exists if and only if both the improper integrals $\int_a^c f(x) dx$ and $\int_c^b f(x) dx$ exist and then we define $\int_a^b f(x) dx = \int_a^c f(x)dx + \int_c^bf(x)dx$

As you have already observed, in your case with $f(x) = 1/x$, $a = -1$, $c = 0$, $b = 2$, neither of the improper integrals $\int_{-1}^0\frac{1}{x} dx$ or $\int_0^2 \frac{1}{x} dx$ are defined as their values would be "infinite". So by the usual Calculus definition, the integral you are considering doesn't exist!

I would suggest to you that this is a good way to think about this situation. In particular, the Fundamental Theorem of Calculus does not apply since $f(x)$ has a vertical asymptote (usually the FTC is only proved for either continuous functions or functions that have only a finite number of jump discontinuities). Therefore your initial calculation using FTC is incorrect.

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    Ah, nice explanation. Thanks. I wonder what my educator will have to say regarding this problem. She has a Ph.D, but I'd expect that she would follow a similar vein of thought you just presented - to not go over the heads of many students.2012-03-05