Is $\int_{-2}^3\frac{1}{x^3}dx=\frac5{72}$ or not defined?

Question

If we divide it into two parts such that $$I=\int_{-2}^0\frac{1}{x^3}dx+\int_0^3\frac{1}{x^3}dx$$

And then use substitution $x=-t$ we get $$I=\int_2^3\frac1{x^3}dx=\frac5{72}$$

However, If we use limits on both part separately, they both diverge, so the integral diverges too.

Which explanation is correct?

Answer 1

You are right that the integral is not defined (because of the singularity), so you can't trust what the anti-derivative tells you. It's even worse with $$\int_{-1}^{1}\frac{1}{x^2}dx=-2\tag{1}$$

The area is clearly positive ($+\infty$). If you were to break up $(1)$ you would get $$\int_{-1}^{1}\frac{1}{x^2}dx=2\int_{0}^{1}\frac{1}{x^2}dx\rightarrow\infty$$ So it's not always true that you can break up integrals.

Answer 2

On account of the singularity at $x=0$, the integral you give is undefined. The only way you could get close is with: $$\int_{-2}^{3}{[x^{-3}dx]}\approx\int_{-2}^{-\epsilon}{[x^{-3}dx]}+\int_{\epsilon}^{3}{[x^{-3}dx]}$$ With $\epsilon$ as a very small positive constant.