# Evaluating $\int_{-1}^2\frac{1}{x}dx$

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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You need to enclose your LaTeX code between $..$ :) – user2468 Mar 5 '12 at 21:43
Fixed - thanks. Thanks for centering the integral J.D. Mind showing how you did it? :) – Joe Mar 5 '12 at 21:44
You might take a look at the Wikipedia article on the Cauchy principal value. – Harald Hanche-Olsen Mar 5 '12 at 21:46
The Cauchy principal value is indeed a rather advanced tool, and needs to be used with great care. Until the day when you need it, you are better off thinking of such integrals as nonsensical. In fact, that is so for most applications. – Harald Hanche-Olsen Mar 5 '12 at 21:50
@Jay Electronics: The definition, and meaning of the (definite, Riemann) integral $\int_a^b f(x)\,dx$ come from the limiting behaviour of Riemann sums, not from antiderivatives. We choose very fine subdivisions of $[a,b]$ by points $x_i$, choose a $\xi_{i}$ between $x_i$ and $x_{i+1}$, and evaluate $\sum(x_{i+1}-x_i)f(\xi_i)$. With $a=-1$, $b=2$, $f(x)=1/x$ we can find arbitrarily fine subdivisions and points $\xi_i$ such that the Riemann sum is anything we like! So the Riemann integral does not exist. What an antiderivative manipulations says is not relevant. – André Nicolas Mar 5 '12 at 22:21

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}
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.