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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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What, then, is your working definition of divergence, if you do not count oscillatory behavior? –  J. M. Nov 25 '11 at 15:10
    
@J. M. I have only an intuitive understanding - I don't know what the formal definition is. I have understood diverge to mean that the function diverges from the x-axis, with the implication that the area under the curve will be infinite? That was my understanding of divergence, but I hasten to add I'm not very knowledgeable on this topic... –  ptrcao Nov 25 '11 at 15:26
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"diverges" means, simply, that the appropriate limit does not exist. This can occur in one of two ways: the expression could oscillate, as you have, between two values $a\ne b$, or, the expression could tend to an infinity (and thus the limit does not exist, as we don't regard an infinity as a real number). Both types are classified as divergent. (In the latter case, you might say "diverges to infinity"). So, yes. "converge" and "diverge" are mutually exclusive and any improper integral is one or the other. –  David Mitra Nov 25 '11 at 15:29

2 Answers 2

up vote 7 down vote accepted

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.

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Quoting Quinn Culver: "Of course you can sub-classify types of divergence (e.g. whether the object is bounded)..." For divergence to occur, doesn't the object always have to be unbounded? –  ptrcao Nov 25 '11 at 15:40
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@ptrcao No, as in the example you gave where the integral diverges but is bounded between $\frac{1}{\pi}$ and $-\frac{1}{\pi}$. This is analogous to the sequence $0,1,0,1,\dots$, which does not converge but is bounded. –  Quinn Culver Nov 25 '11 at 16:28

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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Shouldn't it be $\int^1 _{b} \cos \pi x \quad dx = \left[ \frac{\sin(\pi x)}{\pi} \right]^{x=1} _{x=b} = 0 - \frac{\sin (\pi b)}{\pi} = -\frac{\sin (\pi b)}{\pi}$, not $-1 - \frac{\sin (\pi b)}{\pi}$? –  ptrcao Nov 25 '11 at 16:14

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