I would like to compare these two integrals
$$\int_{-\infty}^{+\infty} f(x) dx \quad\text{ and }\quad \lim_{t\rightarrow +\infty} \int_{-t}^{t} f(x) dx$$
My Thoughts:
Let $\displaystyle I=\int_{-\infty}^{+\infty} f(x)dx$, and $\displaystyle J=\lim_{t \to +\infty} \int_{-t}^{+t} f(x)dx.$
If $I$ exist, then $J$ exists, and $I=J$ :
If $I$ exists, Then $\displaystyle I = \lim_{u \to -\infty} \lim_{v \to +\infty} \int_{u}^{v} f(x)dx = \lim_{u \to +\infty} \lim_{v \to +\infty} \int_{-u}^{v} f(x)dx =\lim_{t \to +\infty} \lim_{t \to +\infty} \int_{-t}^{t} f(x)dx = J.$
The converse is false.
Conter-example : $\displaystyle f(x) = x.$
$\displaystyle J=\lim_{t \to +\infty} \int_{-t}^{+t} f(x)dx = \lim_{t \to +\infty} \int_{-t}^{+t} xdx = \lim_{t \to +\infty} \frac{x^2}{2} \mid_{-t}^{+t} =\lim_{t \to +\infty} \ 0=0 .$ then $J$ exists.
$\displaystyle \lim_{v \to +\infty} \int_{u}^{v} f(x)dx = \lim_{v \to +\infty} \int_{u}^{v} xdx =\lim_{v \to +\infty} \frac{x^2}{2} \mid_{u}^{v} = \lim_{v \to +\infty} (\frac{v^2}{2}-\frac{u^2}{2}) = +\infty.$ then $I$ doesn't exists.
Is my proof correct also I'm interested in more ways of proving it