When asked to evaluate $\int_{a}^{\infty}f(x)dx$, you split the interval based on the improper points.

If there is another improper point other than $\infty$, at $b$, we will write: $\int_{a}^{\infty}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx+\int_{c}^{\infty}f(x)dx$ and ask whether each of the integrals on the right hand side converge. If they all converge, so does the original one.

But if at least one of them diverges, the original one doesn't. What is the justification for this conclusion?

I can see that if $\int_{c}^{b}f(x)dx$ diverges, we can assume that the original integral converges and move the integrals around like this: $\int_{a}^{\infty}f(x)dx-\int_{a}^{b}f(x)dx-\int_{c}^{\infty}f(x)dx=\int_{b}^{c}f(x)dx$ then we'll get a contradiction. But what if both $\int_{c}^{b}f(x)dx$ and $\int_{a}^{b}f(x)dx$ diverge? What is the argument then?


This is a good question. One way to think about the issue is that the "convergence" of your integral is not just about whether the value is finite, but also about whether the finite value is well defined. This is somewhat related to indeterminant forms where $+\infty + (-\infty)$ is not a well-defined quantity. (You can't say they cancel out, since, heuristically, $\infty - \infty = (1+\infty)-\infty = 1+\infty -\infty = 1 + (\infty - \infty)$...)

So as long as one of the integrals on the right hand side diverges, the entire algebraic expression becomes indeterminate, and hence we say the integral diverges. (Diverges doesn't necessarily mean that the value must run-off to infinity; it can just mean that the value does not converge to a definite number/expression.)

(A similar issue also crops up when summing infinite series that doesn't converge absolutely. Riemann rearrangement theorem tells you that, depending on "how" you sum the series you can get the final number to be anything you want.)

Sometimes, however, it is advantageous to try to make sense of an integral which can be split into two divergent improper integrals, but also where one can argue that there should be some natural cancellation. For example, one may want to argue that $\int_{-a}^a \frac{1}{x^3} dx$ evaluates to 0 since it is the integral of an odd function. For this kind of situations, the notion of Cauchy principal value is useful. But notice that the definition is taken in the sense of a limit that rely on some cancellation, and so much in the same way of Riemann rearrangement theorem, "how" you take the limit can affect what value you get as the end result. (This is compatible with the notion that the integral diverges; as I said above, divergence should be taken to mean the lack of a well-defined, unique convergence.)

Edit. Let me add another example of an integral that remains finite but does not converge. What is the value of $\int_{0}^\infty \sin(x) dx$? For every fixed $a > 0$, $\int_0^a\sin(x) dx = 1 - \cos(a) $ is a number between 0 and 2. But the limit as $a\to \infty$ doesn't exist! If you pick a certain way to approach $\infty$, say choose a sequence $a_n = 2\pi n$, then you'll come to the conclusion that the "limit" is 0; but if you choose $a_n = (2n + 1)\pi$, then you get the conclusion that the limit is $2$. The idea here is roughly similar: you take a left limit and a right limit approaching the improper point, and depending on how you choose your representative points (by an algebraic relation between the speed at which the left and right limits approach the improper point, say), you can get different answers.

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    $\begingroup$ It is interesting that you gave the $\int_0^\infty \sin(x)\mathrm dx$ example; if one uses the notion of Abel summability, for instance, we have that $$\lim_{\epsilon\to 0}\int_0^\infty\exp(-\epsilon x)\sin(x)\mathrm dx=1$$, the mean of the two boundary values. I suppose it's a bit like the summation of $1-1+1-1+1\dots$. $\endgroup$ Dec 10 '10 at 17:59
  • $\begingroup$ I think for any "stable" notion of summability, the answer has to be 1 for that case. $\endgroup$ Dec 10 '10 at 19:12
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    $\begingroup$ As for the usefulness of subtracting divergent quantities, the defining limit for the Euler-Mascheroni constant (which generalizes to the Stieltjes constants) also comes to mind... $\endgroup$ Dec 10 '10 at 19:19
  • $\begingroup$ @J.M. that's a very good point. Though I think it is getting a bit far from the topic at hand. :-) Though of course a preferred limit is taken in those cases. $\endgroup$ Dec 10 '10 at 19:25

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