TITLE: Can we represent an improper integral as $\int_{-\infty}^{\infty} f(x)\,dx = \lim_{a \to \infty} \int_{-a}^a f(x)\,dx$?
QUESTION [7 upvotes]: I was reading on improper integrals, and came across :

$$\int_{-\infty}^{\infty} f(x)\,dx = \lim_{A \to -\infty}
\int_A^Cf(x)\,dx + \lim_{B \to \infty} \int_C^B f(x)\,dx$$

My question is a rather silly one:
Is it any different if I write the above as :
$$\int_{-\infty}^{\infty} f(x)\,dx = \lim_{a \to \infty} \int_{-a}^a f(x)\,dx$$
If it's correct, is there any specific reason why it is written that way? Or is looking at it in that way any more beneficial than my way?
If it's not, why is it incorrect to write it this way? (Any counter-example also appreciated)

REPLY [2 votes]: The symmetric limit $$\int_{-\infty}^{\infty} f(x)\,dx = \lim_{a \to \infty} \int_{-a}^a f(x)\,dx$$
May exist even through the improper integral $\int_{-\infty}^{\infty} f(x)\,dx$ is divergent 
For example the integral $\int_{-\infty}^{\infty} x\,dx$ is divergent
Since $\lim_{a\to\infty} \int_{0}^{a}x \,dx=\lim \frac{a^2}{2}=\infty$
However $\lim_{a\to\infty} \int_{-a}^{a}x\,dx= \lim_{a\to\infty} \frac{[a^2-(-a)^2]}{2}=0$