Improper Integrals - infinity The improper integral is defined as: $$\int_{-\infty}^{\infty} f(x)\,\mathrm{d}x= \int_{-\infty}^{a}f(x)\,\mathrm{d}x+\int_{a}^{\infty}f(x)\,\mathrm{d}x$$
And we also have $\lim_{N\to\infty} \int_{-N}^{a}f(x)\,\mathrm{d}x = \int_{-\infty}^{a}f(x)\,\mathrm{d}x$
The integral $\int_{-\infty}^{\infty} x\,\mathrm{d}x$ does not exist because $\int_{-\infty}^{a}x\,\mathrm{d}x$ does not exist. 
But the integral $\lim_{N\to\infty} \int_{-N}^{N} x\,\mathrm{d}x$ does exist since $\lim_{N \to \infty} [N^2/2 - N^2/2] = 0$
But can we not also define it as $$\lim_{N\to\infty} \int_{-N}^{N} x\,\mathrm{d}x = \lim_{N\to\infty} \int_{-N}^{a} x\,\mathrm{d}x+ \lim_{N\to\infty} \int_{a}^{N} x\,\mathrm{d}x=\int_{-\infty}^{a}x\,\mathrm{d}x + \int_{a}^{\infty}x\,\mathrm{d}x$$
Now doesn't this make the LHS does not exist?
 A: The main reason not to define $\int_{-\infty}^\infty$ as $\lim_{N\to\infty} \int_{-N}^{N}$ is that you'd expect:
$$\int_{-\infty}^{\infty} f(x)\;dx = \int_{-\infty}^{\infty} f(x+1)\;dx$$
Your definition, however, would not work. For example, under your definition, $$\int_{-\infty}^\infty x\; dx = 0$$ but $$\int_{-\infty}^{\infty} (x+1)\;dx=+\infty$$
The other problem with your definition is that the integral is defined often even if it is not true that $f(x)\to 0$ for $x\to\infty$.
If the integral exists by the usual definition, you get the same value with your definition, but there are lots of cases where the usual definition yields no value for the integral, but where your definition yields a value, and that value tends not to be useful very often.
Incidentally, $\lim_{N\to\infty}\int_{-N}^N f(x)\;dx$ is called the Cauchy Principal Value of the improper integral.
A: Your third statement is incorect,
 $\lim_{N\to\infty} \int_{-N}^{N} x$ does not exist.
And the reason is your last statement.
$\int_{-\infty}^{\infty} x = lim_{a\to\infty}lim_{b\to\infty} \int_{-a}^{b} x$
your last statement is also a way to define $\int_{-\infty}^{\infty} x$ and they both coreect 
and the integral does not Exist.
If you have to integrate  odd function between finite  symmetrical section your integral will be zero but in infinity section you cant know how your function will behave.
