Is there an intuitive reason why $\int_{-\infty}^{\infty} x dx$ doesn't exist but $\lim\limits_{N\to \infty} \int_{-N}^N xdx$ exists? Is there some intuitive reason why $\int_{-\infty}^{\infty} x dx$ doesn't exist but $\lim\limits_{N\to \infty} \int_{-N}^N xdx$ does? It would seem they represent the same thing (area beneath $f$ everywhere), but apparently not.
Consider the following problem from Ch. 14 "The Fundamental Theorem of Calculus" from Spivak's Calculus

27 (b) The improper integral $\int_{-\infty}^a f$ is defined in the obvious
way, as $\lim\limits_{N\to -\infty} \int_N^a f$. But another kind of
improper integral $\int_{-\infty}^{\infty} f$ is defined in a
nonobvious way: it is $\int_0^{\infty} f + \int_{-\infty}^0 f$,
provided these improper integrals both exist.
(b) Explain why $\int_{-\infty}^{\infty} x dx$ does not exist. (But
notice that $\lim\limits_{N\to \infty} \int_{-N}^N xdx$ does exist)

My calculations are as follows
$$\int_{-N}^N xdx = \left . \frac{x^2}{2} \right |_{-N}^N=\frac{N^2-(-N)^2}{2}=0$$
On the other hand
$$\int_{-\infty}^{\infty} xdx=\int_{-\infty}^0xdx+\int_0^{\infty}xdx$$
$$=\lim\limits_{N \to -\infty} \int_{N}^0 xdx+\lim\limits_{N\to\infty}\int_0^{N} xdx$$
$$=\lim\limits_{N\to -\infty} \left . \frac{x^2}{2}\right |_N^0+\lim\limits_{N\to\infty} \left . \frac{x^2}{2} \right |_0^N$$
$$\lim\limits_{N\to -\infty} \frac{-N^2}{2}+\lim\limits_{N\to \infty} \frac{N^2}{2}$$
Neither of these limits exists.
This seems like a weird result.
Furthermore, if we had defined $\int_{-\infty}^a f$ as $\lim\limits_{N\to \infty} \int_{-N}^a f$ then in the calculation above we would have
$$\int_{-\infty}^{\infty} xdx=\int_{-\infty}^0xdx+\int_0^{\infty}xdx$$
$$=\lim\limits_{N \to \infty} \int_{-N}^0 xdx+\lim\limits_{N\to\infty}\int_0^{N} xdx$$
$$=\lim\limits_{N \to \infty} \left ( -\frac{(-N)^2}{2}+\frac{N^2}{2} \right )$$
$$=\lim\limits_{N \to \infty} 0$$
$$=0$$
 A: “Filling” the line with intervals of the form $[-N,N]$ is just a particular way of doing it. Yes, it's symmetric and indeed this particular way gives rise to what's called “Cauchy principal value”.
But we'd like the integral, if it exists, to be independent of the way we fill the line. Why not
$$
\lim_{N\to\infty}\int_{-N}^{N^2} x\,dx
$$
or some other way? Note that
$$
\int_{-N}^{N^2} x\,dx=\Bigl[x^2/2\Bigr]_{-N}^{N^2}=\frac{N^4-N^2}{2}
$$
and the limit is $\infty$.
Your final argument is flawed: you can do
$$
\lim_{x\to c}(f(x)+g(x))=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)
$$
provided, in the right hand side,

*

*both limits exist finite or

*one limit is finite and the other is infinite or

*both limits are $\infty$ or

*both limits are $-\infty$
with obvious meaning of $+$ in cases 2, 3 and 4.
Your case does not fit, because
$$
\lim_{N\to\infty}\int_0^N x\,dx=\infty
$$
and
$$
\lim_{N\to\infty}\int_{-N}^0 x\,dx=-\infty
$$
A: The problem here is that "real infinity" (i.e., $\infty$) is a concept, not a number.  We tend to think of infinity as a number, but it is really an entire group of numbers.  That's why you can't perform a lot of arithmetic operations using $\infty$.  It would be like doing arithmetic operations on "odd" or "even". Sure, there are some things we can do, but not a lot.
With hyperreal numbers, however, you can be much more exact, since hyperreal numbers refer to specific values of infinity, not just the concept in general.  There are a lot of different notations for hyperreal numbers, but the one I prefer is to use $\omega$ (greek lowercase omega) as a "unit" infinity.
Then, your limit becomes much more reasonable, because you have specified which infinity you are talking about.  That's why your limit works when your integral doesn't (since $N$ can only take one value, you are forcing both parts to be the same infinite value).  If you want to do your integral using hyperreal, it is straightforward enough:
$$ \int_{-\omega}^{\omega} x\,dx = \frac{x^2}{2}\biggr|_{-\omega}^{\omega} = \frac{\omega^2}{2} - \frac{(-\omega)^2}{2} = 0 $$
Note that this is the same reason that the Cauchy rearrangement theorem works - by treating $\infty$ as a number when it is really a concept you get weird results.  The rearrangement theorem goes away if you use specific hyperreal values rather than a general "infinity".
