"Not converging" vs. diverging improper integral
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.
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$\begingroup$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.
$\endgroup$ 2 $\begingroup$It is not true that $\displaystyle\lim_{b \to -\infty} \int^1_b \cos \pi x \; dx = -\frac{1}{\pi}\sin b \pi$.
If you take the limit of anything as $b$ approaches anything, you will not see "$b$" appearing in the answer. At least not if it's correct.
Look at the integral, without the limit: $$ \int^1_b \cos \pi x \; dx = \left[\frac{\sin(\pi x)}{\pi}\right]_{x=b}^{x=1} = - \frac{\sin(\pi b)}{\pi}. $$ Now try to take the limit as $b\to-\infty$. As you say, it doesn't exist because $-\sin(\pi b)/\pi$ keeps oscillating between $-\frac1\pi$ and $\frac1\pi$.
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