# Integration question

1. Dec 5, 2009

### Zoe-b

1. The problem statement, all variables and given/known data
The integral of 1/(x^2) between -1 and 1 comes to -2. How is this possible when the graph is always above the x-axis?

3. The attempt at a solution
the integral of x^(-2) = -x^(-1) + c
in this range you get (-1/1 - 1/1) which does indeed equal 2.
My original solution was to say that since the graph is not continuous in this range it makes no sense to try to find the integral (which should be infinite surely?). Is there more to it than this?
Thanks

2. Dec 5, 2009

### HallsofIvy

Staff Emeritus
Short answer: the integral of $1/x^2$ between -1 and 1 does NOT "come to -2". In fact, $1/x^2$ is not differentiable between -1 and 1 because of the infinite discontinuity at 0. Specifically, you cannot just find the anti-derivative and evaluate at -1 and 1.

You could try to evaluate it using the definition of "improper integrals":
$$\int_{-1}^1 \frac{1}{x^2} dx= \int_{-1}^{-\alpha}\frac{1}{x^2}dx+ \int_\beta^1 \frac{1}{x^2}dx$$
and then take the limits as $\alpha$ and $\beta$ go to 0independently.
The two integrals are
$$-\frac{1}{x}\left|_{-1}^{-\alpha}+ \frac{1}{x}\right|_{\beta}^1= \frac{1}{\alpha}- 1- 1+ \frac{1}{\beta}$$
but that limit does not exist as $\alpha$ and $\beta$ go to 0 independently
.
If you were to assume that $\alpha= \beta$ you would be calcuating the "Cauchy Principle Value"- those two fractions would cancel and you would get "-2". If the integral itself exists, then the Cauchy Principle Value but if the integral does not exist, as here, you get the kind of ridiculous result you cite.

3. Dec 5, 2009

### Zoe-b

Ok thats fine, pretty much what I thought (though hadn't thought through the limits thing)
Thank you

4. Dec 5, 2009

### LCKurtz

. Actually, they wouldn't cancel; they both go to $\infty$. If the integrand was odd they would have cancelled.