Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I asked about the value of an integral here:
Hard integral that standard CAS get totally wrong

The question got downvoted and voted to close because I didn't understand (and wasn't able to answer) the following question:

In which sense is the integral supposed to exist?

So in what senses can integrals exist? What are the options here?

share|improve this question
2  
It would help to know something of your background. For example, if you are just-now studying elementary calculus, a different sort of answer might help you, while if you are studying Fourier analysis or functional analysis a more substantial answer would make sense. –  paul garrett Jul 31 '13 at 15:58
    
@paulgarrett: I would put it like this: Please give the most elementary answer possible that could do justice to the question. The more intuitive the better :-) –  vonjd Jul 31 '13 at 16:03

4 Answers 4

up vote 10 down vote accepted

In addition to other useful answers... Before giving the standard example of an integral whose convergence is problemmatical, but which does have some sense... we should ask ourselves what it is we are expecting "integrals" to do, what properties the process should have. For example, "integration" should not just produce random numerical outcomes. And which functions, on which intervals, should be expected to be acceptable inputs? At the very least, if $f,g$ are acceptable, then linear combinations $af+bg$ should be, and, letting $I$ denote the integration procedure, $I(af+bg)=aI(f)+bI(g)$. ("Linearity".)

Riemann integration works best on finite intervals with nearly-continuous functions, while Lebesgue integration accommodates very-discontinuous functions, etc. In both cases, the integral of $f$ on a set or interval is a limit of finite sums, and the set-up for the game consists of proving these limits will exist (and be finite numbers!) under various assumptions on $f$.

A simpler standard example similar to one in your other post is $\int_0^\infty \sin(x^2)\;dx$. This has the disturbing feature that the function doesn't go to $0$ at infinity, so if we think of Cauchy's criterion for convergence of a series (rather than integral), we might conclude that this would diverge, meaning that $\lim_N \int_0^N \sin(x^2)\;dx$ might be $\pm\infty$? Or not exist? However, the oscillation produces enough self-cancellation so that this doesn't happen. In fact, changing variables, replacing $x$ by $\sqrt{x}$, gives integral $\int_0^\infty {\sin(x)\over \sqrt{x}}\;dx$. Now, at least it looks like it decays at infinity, and there is still cancellation due to the oscillation. In fact, the limit can be evaluated by various tricks: I think it is $\pi/2$ or something similar.

In fancier circumstances, it often happens that "an integral" is not meant to be taken literally, but only to indicate the structure of some operation on functions. The basic case is with Fourier or Laplace transforms on the real line. Fourier transforms expressed as integrals $\hat{f}(\xi)=\int_{\mathbb R} e^{-i\xi x}\,f(x)\;dx$ make best sense for $\int_{\mathbb R} |f(x)|\;dx<\infty$, but, in fact, via the Plancherel theorem for Fourier transforms, we know that $\int_{\mathbb R} \hat{f}(\xi)\;\hat{g}(\xi)\;dx=\int_{\mathbb R}f(x)\,g(x)\;dx$ (maybe up to a constant multiple), so there is a unique extension of Fourier transform to square-integrable functions $f$, that is, such that $\int_{\mathbb R}|f(x)|^2\;dx<\infty$. This extension has the same properties as the Fourier transform that is literally an integral, but is not quite given by that integral.

Similarly, Fourier transforms can be extended to ("tempered") "generalized functions" (="distributions", not in the probabilistic sense), in a way that is completely sensible structurally, but in which the integrals are wildly not-convergent. For example, $\int_{\mathbb R} x^n\cdot e^{-i\xi x}\;dx$ doesn't converge at all, but by other means we can conclude that it is (a constant multiple of) the $n$th derivative of the Dirac delta distribution.

And, in case there were any doubt, computer algebra systems have their limitations, especially in dealing with "divergent" (not numerically docile!) integrals.

share|improve this answer
    
+1: Great explanation - Thank you! –  vonjd Jul 31 '13 at 16:23
1  
Good! Hope it hit the mark, but also with pointers to things to think about later, perhaps. –  paul garrett Jul 31 '13 at 16:25
1  
I don't think the substitution ($x \mapsto \sqrt x$) for $\sin x^2$ is exactly right –  Argon Jul 31 '13 at 17:25
1  
@Argon, Oops, thanks! Repaired! :) –  paul garrett Jul 31 '13 at 17:29

Basically, the definite (Riemann) integral $\int_a^b f(x)\,\mathrm dx$ exists if, well, $f$ is Riemann integrable. The most important class of Riemann integrable functions are the continuous functions. Your example is an improper integral, i.e. one or (in your case both) ends are infinite, and that can only be evalutaed in the sense of $$\tag1\int_{-\infty}^\infty f(x)\,\mathrm dx:=\lim_{a\to-\infty\atop b\to+\infty}\int_a^bf(x)\,\mathrm dx$$ (There are other cases of improperness, e.g. when $f$ is not continuous at $a$ and/or $b$ and again you have to take a limit of definite integrals). Although $f$ is continuous and hence all $\int_a^b$ exist, the (double) limit is a completely different question.

You may also want to have a look at Lebesgue integral, a different integration theory altoether that handles some problems of Riemann integral with a different systematic. Lebesgue treats the intregal "all at once" but requires some other conditions to treat oscillating functions (as yours), namely that $\int_{-\infty}^\infty|f(x)|\,\mathrm dx$ be finite - which is not the case with your example.

share|improve this answer
    
+1: This gives great insight - Thank you! –  vonjd Jul 31 '13 at 16:26

The function you are considering, $$ f(x):=\exp \left ( i\left ( ax^3+bx^2 \right ) \right ), $$ is not Lebesgue integrable on $(-\infty,\infty)$, as $|f(x)|=1$ for all $x\in(-\infty,\infty)$. However, the limit $$ \lim_{M,N\to\infty}\int_{-M}^Nf(x)\,dx $$ does exist. This is called an improper integral.

share|improve this answer
1  
I'm still confused. His original question was "what is this integral's value?", and you claim that the value exists, which would mean his question can be answered. So what does the fact that it's not Lebesgue integrable have to do with anything? And why would that fact make his question a bad one? –  BlueRaja - Danny Pflughoeft Jul 31 '13 at 16:53
1  
@BlueRaja-DannyPflughoeft I mention Lebesgue integrability to address the question "in what sense can integrals exist?". I just wanted to point out that Lebesgue integrability is one possible sense. I certainly didn't mean to imply the question was bad. In fact, I think this integral is quite interesting. –  Julian Rosen Jul 31 '13 at 16:57

Well, there is a simple definition: an Integral $\int_{a}^{\infty} f(x)\;dx$ exists $\Leftrightarrow$ the series $\sum_{n=a}^{\infty} f(n)$ converges. The hard part might be proving this. Use a convergence test to proof, if it converges or not.

For Integrals with both bounds of integration infinite you should look up the Residue Theorem.

share|improve this answer
5  
This is not quite true, in either direction. –  paul garrett Jul 31 '13 at 15:57
    
@paulgarrett: Could you please expound on this? Thank you –  vonjd Jul 31 '13 at 15:57
4  
The integral can be absolutely convergent while the sum diverges, and vice-versa. For example, for $f$ with graph consisting of very narrow but tall spikes at integers, the sum of $f(n)$ at integers can be made to diverge, while the integral is finite. Conversely, there are non-negative functions with value $0$ at all integers whose integral blows up. –  paul garrett Jul 31 '13 at 16:01
1  
... but, to be fair, under sufficiently strong hypotheses on $f$ there are comparisons between integral and sum. For example, if $f$ is monotone decreasing and positive real-valued, there is indeed the "integral test", and such. But the range of validity of the integral test is pretty far away from the posted question, I think. –  paul garrett Jul 31 '13 at 17:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.