Take the improper integral
$$\int^1 _{-\infty} \cos \pi x \; dx $$
From which it is clear that:
$$\lim_{b \to -\infty} \int^1 _{b} \cos \pi x \; dx = -\frac{1}{\pi}\sin b \pi$$
The integral oscillates between $\frac{1}{\pi}\text{ and }-\frac{1}{\pi}$ as $b \to \infty$.
Now, my textbook, Calculus: One and Several Variables, by Salas et. al (10th ed.), says this integral "diverges"? Certainly, I agree it "does not converge", but I don't think the integral is diverging per se…
Is the diverge/converge distinction a mutually exclusive dichotomy? It seems like it should be possible to say an integral neither converges or diverges, in the situation that the function oscillates to infinity.
Best Answer
Yes, divergence and convergence are mutually exclusive; divergence means "does not converge" and since we have a very precise idea of what it should mean to converge, the integral you gave must diverge. Of course you can sub-classify types of divergence (e.g. whether the object is bounded), but then it's just a matter of nomenclature. And based on the definition of "diverge" as an everyday English word, I don't think that math has chosen a poor word for the oscillating case.