When is $\int_{-\infty}^\infty f(x) dx = \lim\limits_{n \to \infty} \int_{-n}^n f(x) dx$?
Normally, we would need to take two different limits, but I am wondering if there is necessary and/or sufficient condition that the integral can be evaluated by taking the limit simultaneously. If so, is $$\lim\limits_{n \to \infty} \int_{-n}^n f(x) dx = \lim\limits_{n \to \infty} \int_{-kn}^n f(x) dx$$ for any $k > 0$?
More specifically, if we know that the integral converges, why can we conclude that $\int_\infty^\infty f(x) dx = \lim\limits_{n \to \infty} \int_{-n}^n f(x) dx$?
Best Answer
Let $F'(x)=f(x).$ Then using definitions, and the FTC we can see
$$\int_{-\infty}^\infty f(x)dx= \lim_{s\to\infty}\int_a^s f(x)dx +\lim_{t\to-\infty}\int_t^a f(x)dx$$
$$=\lim_{s\to\infty}F(s)-F(a) +\lim_{t\to-\infty}F(a)-F(t)$$ $$=\lim_{s\to\infty}F(s)-\lim_{t\to-\infty}F(t).$$
If these two limits exist, then it means that the result is finite, and in particular, it is well defined, and we don't need to evaluate these limits separately. Thus
$$\lim_{s\to\infty}F(s)-\lim_{t\to-\infty}F(t)=\lim_{s\to\infty}F(s)-F(-s).$$
It is only in this situation that the principal value is equal to the result of the improper integral (equality can only be discussed because it converges, after all!).