It's natural to define the integral $$\int_{0}^\infty f(x) dx := \lim_{M\to\infty} \int_{0}^M f(x) dx~~~~~~ ~~~(*)$$
But it's not obvious how we should define the integral $$\int_{-\infty}^\infty f(x) dx$$
One common approach is to say $$\int_{-\infty}^\infty f(x) dx := \int_{-\infty}^0 f(x) dx + \int_{0}^\infty f(x) dx~~~~~~~~~~~(A)$$ But we could as well define the integral as $$\int_{0}^\infty f(x) dx := \lim_{M\to\infty} \int_{-M}^M f(x) dx ~~~~~~~~ (B)$$
Now let's take the identity function $f(x) = x$. Then
- The definition (A) is helpless, since the expression is undefined.
- $$\lim_{M\to\infty}\int_{-M}^{M} x dx = 0$$
- $$\lim_{M\to\infty}\int_{-M}^{M+1} x dx = +\infty$$
- $$\lim_{M\to\infty}\int_{-M}^{\sqrt{M^2+2}} x dx = 1$$
Basically, we can get the integral to be anything. On the other hand, there's no reason why one of the given definitions to be better than the other, why definition A should be favored over B. (we used a limit in $(*)$, why shouldn't we here?)
How should we then define the extension of the Riemann integral? Is there any way to avoid this complication?
I really don't see why $\int_{-\infty}^{+\infty} x\mathrm{d}x$ should not exist. Since it is an odd function and we are integrating it over a symmetric interval it has to give zero, regardless of the fact that this interval is unlimited. Is is just a purely mathematical matter, if the real problem concerning this integral results from a symmetric context, that is if the symmetry of the interval is linked to actual situations and the limit comes after, saying that it cannot exist would be just an annoying obstacle.