Why is it that
$$\int_{-\infty}^\infty x \,dx$$
does not exist, but
$$\lim_{N \to \infty} \int_{-N}^{N} x\,dx$$
does exist?
I was thinking that it involves the fact that in the second case, the limit's behavior at the actual value does not matter, but in the first case it would.
If this is the case, is there any way to show this rigorously?
Let's show that this just goes catastrophically wrong if we were to guess that the first integral does exist, and that you think it should match up with the second--which is $0$, because $x$ is an odd function. If that's the "right" value, then we should be able to get this every time, but I claim we can really get whatever we want.
Pick your favorite real number, $s\ge 0$. Then we seek to define functions $f(x), g(x)$ so that
$1$) $$\lim_{x\to \infty} f(x)=-\lim_{x\to\infty}g(x)=\infty$$
$2$) $$\int_{g(t)}^{f(t)}x\,dx=s$$
for every $t\ge 0$. The point is that these two functions, $f$ and $g$ are our new ways of letting the limits go to infinity. If such functions exist, then clearly we can make
$$\int_{-\infty}^\infty x\,dx\;``="\;s$$
so that the integral is really seen to make no sense. The quotes are on there to indicate how little sense it makes.
In face many such function pairs exist, allow me to demonstrate one such:
Let $g(t)=-t$. Clearly it satisfies condition ($1$), now we need an $f$ satisfying ($1$) so that we get condition ($2$). However, note the fact that integrals are continuous, that is:
$$\lim_{x\to b}\int_a^x f(t)\,dt=\int_a^bf(t)\,dt$$
so that for any $T>0$, we have
$$\int_{-T}^t x\,dx={1\over 2}\left(x^2-T^2\right)$$
Now if we set this equal to $s$ we see that
$${1\over 2}\left(x^2-T^2\right)=s\iff x=\sqrt{T^2+2s}.$$
Now we just define
$$f(t)=\sqrt{t^2+2s}.$$
Then we see that $f(x)$ satisfies condition ($1$), and this choice along with $g$ satisfy condition ($2$), hence
$$\lim_{t\to\infty} \int_{g(t)}^{f(t)} x\,dx$$
can be made to "take on" the value $s$. You can do the same thing with $s\le 0$ or even $s=\pm \infty$ by appropriate modifications, so it goes to show that how you go to infinity really matters!
The reason that the first does not exist is because we define doubly improper limits to be given by
$$\int_{-\infty}^{\infty} x\,dx = \lim_{r\to-\infty}\lim_{s\to\infty}\int_r^s x\,dx.$$
If we try to evaluate this integral (say fix $r$ and evaluate the limit for $s\to\infty$ and vice versa), we would get all sorts of contradictory results. Sometimes we would get $-\infty$, sometimes $+\infty$; we could even concoct it to give any real number we please (by relating $r$ and $s$ in some way)!
The other limit you posit, however, is well-defined. Particularly, it evaluates to
$$\lim_{N\to\infty}\int_{-N}^N x\,dx = \lim_{N\to\infty} \frac{x^2}{2}\bigg|_{-N}^N = 0.$$
This kind of limit is akin to a principal value and it is used to evaluate otherwise ill-defined integrals.
Since $\displaystyle \int_{a}^\infty xdx = +\infty$, the first improper integral does not exist while the second does because for any $N$, $\displaystyle \int_{-N}^N xdx = 0$
As I understand it, the convention is that $$\int_{-\infty}^{\infty} f(x) ~{\rm d}x$$ is notation for
$$\lim_{(a,b) \to (\infty, \infty)} \int_{-a}^{b} f(x) ~{\rm d}x$$
And that $\lim_{(a,b) \to Z} P(a,b)$ is only defined if all 2-dimensional paths that end at $Z$ which $(a,b)$ could take yield the same value. For example, consider the set of paths to $\infty, \infty$:
$$I = \lim_{t \to \infty} \int_{-tk}^{tj} x ~{\rm d}x$$
Notice how different choices of $k$ and $j$ give different values of the integral:
So the value that $\lim_{(a,b) \to (\infty, \infty)} \int_{-a}^{b} x ~{\rm d}x$ take depends on the path of $(a,b)$ and so isn't defined.
Think of it as an order of operations problem. In the first, the limit is 'cooked in' to the integral operation being performed and asks us to evaluate something like $\infty$ - $\infty$. In the second we are considering the limit of a (discrete, since $N$ is an integer) sequence of numbers -- each of which turns out to be zero.
You can see that the ways the two integral convey to "sum up" the integrand are very different from each other.
You can see that in the second integral the limits go to infinity in the SAME manner because they are both $N$. Therefore, at all points of consideration, the integral will take the sum from $-N$ to $N$ and will results in $0$.
However, in the first integral, the limits do not behave in the same manner but are rather independent of each other. As Cameron Williams has pointed out, the integral's result will be ill-defined because infinity is not definite but is just a notation. It is not quantifiable in an absolute manner, and $|\infty| \neq |-\infty|$. Therefore, the first integral does not exist.
