Can we represent an improper integral as $\int_{-\infty}^{\infty} f(x)\,dx = \lim_{a \to \infty} \int_{-a}^a f(x)\,dx$? 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) 
 A: The limit $\lim_{a \to \infty} \int_{-a}^a f(t) d t$ is called the principal value (PV) of the improper integral.
Now, if the inetgral 
$$\int_{-\infty}^{\infty} f(x)\,dx = \lim_{A \to -\infty}
\int_A^Cf(x)\,dx + \lim_{B \to \infty} \int_C^B f(x)\,dx$$
is convergent, then the PV exist and is equal to the integral.
BUT, there are many situations (for example any divergent integral of an odd function) where the PV is convergent but the integral is not convergent.
The main reason why people use the given definition and not the PV is because many of the properties of integral extend to the definition we use but not to the PV. For example, substitution rule does not work for PV (try the substitution $u=x+1$ for $f(x)=x$).
A: Counterexample  $f(x)=\frac{1}{x}$ for $|x|\gt 1$ and $f(x)=0$ for $-1\le x\le 1$.  The symmetric case $\lim_{a\to \infty}\int_{-a}^af(x)dx=0$  However the general case with different limits does not converge.
A: Your question is not silly; it's important to be clear on the definitions of these things.
It is indeed different if you write the above like that. Consider for instance an arbitrary odd function $f : \mathbb R \to \mathbb R$ (a function is called odd if $f(-x) = -f(x)$). Then, using your definition,
$$
\int_{-\infty}^{\infty} f(x) \, dx = \lim_{a \to \infty} \int_{-a}^{a} f(x) \, dx = \lim_{a \to \infty} \int_{-a}^0 f(x) \, dx + \int_0^a f(x) \, dx = 0,
$$
which implies that the integral from $-\infty$ to $\infty$ of $f$ is $0$, no matter how pathological it is. Clearly we don't want to consider arbitrary odd $f$ to integrate to $0$ over the entire real line. Take as an example $f(x) = x$. The concern becomes especially obvious if we consider
$$
\lim_{a \to \infty} \int_{-a}^{2a} x \, dx = \lim_{a \to \infty} \frac 3 2 a^2 = \infty \neq 0,
$$
so the "rate" at which the upper and lower bounds move changes the answer.
The method of splitting up the integral into two improper integrals is thus used as a convention; it doesn't have this problem. In fact, you can prove that if splitting up the integral yields a convergent result, then so will your method. In essence, the method of splitting up the integral prevents "too many" things from converging.
What you are suggesting does in fact have a name; it is the Cauchy principal value of the improper integral. This is useful in some special cases but certainly should not be used all the time for the reasons given above.
A: 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 $\lim_{a\to\infty} \int_{-a}^{a}x\,dx= \lim_{a\to\infty} \frac{[a^2-(-a)^2]}{2}=0$
