Question on a problem regarding improper integral

Question

I'm studying improper integrals with Paul's Online Notes as a reference. Sorry if I'm quoting it here, but the website has the following problem:

Determine if the following integral is convergent or divergent. If it is convergent find its value. $$\int_{-2}^{3} \frac{1}{x^3} \, dx$$

And the solution provided in the website is:

This integrand is not continuous at $x=0$ and so we'll need to split the integral up at that point. $$\int_{-2}^{3} \frac{1}{x^3} \, dx=\int_{-2}^0\frac{1}{x^3}\,dx+\int_{0}^3\frac{1}{x^3}\,dx$$ Now we need to look at each of these integrals and see if they are convergent. $$\int_{-2}^0\frac{1}{x^3}\,dx=\lim_{t \to 0^-}\int_{-2}^t\frac{1}{x^3}\,dx$$ $$=\lim_{t \to 0^-}(-\frac{1}{2t^2}+\frac{1}{8})$$ $$=-\infty$$ At this point we're done. One of the integrals is divergent that means the integral that we were asked to look at is divergent. We don't even need to bother with the second integral.

Question is is the solution correct? If I use my intuition, the integral $$\int_{-2}^{3} \frac{1}{x^3} \, dx$$ should be equal to $$\int_{2}^{3} \frac{1}{x^3} \, dx$$ because $$\int_{-2}^{2} \frac{1}{x^3} \, dx=0$$ If it's true, then the integral should be convergent and its value should be $\frac{5}{72}$.

I checked an online integral calculator https://www.integral-calculator.com/ and it seemed to confirm my answer. So which solution and reasoning is correct and why?

Answer 1

You have to be very, very careful when it comes to concluding results like

$${\int_{-2}^{2}\frac{1}{x^3}dx=0}$$

Because this simply isn't true by the standard definition of convergence of improper integrals. The Fundamental Theorem of Calculus requires our function to be continuous over the domain we are integrating - and as you can see, ${\frac{1}{x^3}}$ is not continuous over ${(-2,2)}$ Namely, it is discontinuous at ${x=0}$.

Now, it is true if we do the following:

$${\lim_{\epsilon \rightarrow 0^+}\left[\int_{-2}^{-\epsilon}\frac{1}{x^3}dx + \int_{\epsilon}^{2}\frac{1}{x^3}dx\right]=0}$$

So you may wonder - "why don't we just call the integral $0$ then"? Well, in this case it's ${0}$, because both integrals approach the point $0$ at the same rate. But there is absolutely no reason they should have to do this, for example, something like

$${\lim_{\epsilon\rightarrow 0^+}\left[\int_{-2}^{-\epsilon^2}\frac{1}{x^3}dx + \int_{\epsilon}^{2}\frac{1}{x^3}dx\right]}$$

would also be completely valid! And from this principle, you can arrive at different answers (although as others have pointed out - the "principle value" for the integral is $0$. This is just a way of assigning a number to the integral, but it's NOT what it "converges" to in the sense of convergence that we usually care/talk about).

The website does exactly how we define how to handle improper integrals of this type (ones that are discontinuous at a point over our domain of integration). In general, say we are integrating the function ${f(x)}$ over ${[a,c]}$ (${a<c}$), and ${f}$ has a discontinuity at ${b\ |\ a < b < c}$. Then we always must do the following

$${\int_{a}^{c}f(x)dx = \lim_{l\rightarrow b^-}\int_{a}^{l}f(x)dx + \lim_{m\rightarrow b^+}\int_{m}^{c}f(x)dx}$$

And you notice in this case, the limits are actually decoupled. So indeed if one diverges, you can conclude right away that the integral ${\int_{a}^{c}f(x)dx}$ is divergent, because of the fact the limits are decoupled. Otherwise, you could end up with nonsense like ${\infty-\infty}$, which is not something you can evaluate (and as I said, taking a "combined" limit is out of the question since our definition on how to handle improper integrals require the limits to be decoupled).