difference between two improper integrals I can't grasp the difference between
$\int_{-\infty}^{\infty}\,f(t)\,dt$  and    $\lim\limits_{x \to \infty} \int_{-x}^x\,f(t)\,dt$ for example if $ f(t)=t $ then the first one will give a divergent improper integral whereas the second one gives us 0.
Thanks for your help.
 A: Note that $\int_{-\infty}^{\infty} f(t)dt$ should be interpreted as
$$\lim_{\overset{x \to -\infty}{y \to \infty}}\int_{x}^y f(t)dt$$
which is different from $\lim_{x \to \infty} \int_{-x}^x f(t)dt$.

As an example consider
$$f(x) =
\begin{cases}
1 & \text{ if }\vert x \vert \leq 1\\
\dfrac1{x} & \text{ if } \vert x \vert \geq 1
\end{cases}$$
Now the integral
$$\lim_{x \to \infty}\int_{-x}^x f(t) dt = \lim_{x \to \infty}\left(\int_{-x}^{-1} \dfrac{dt}t + \int_{-1}^1 f(t)dt + \int_1^x \dfrac{dt}t \right) = 2$$
However, one way to interpret $\int_{-\infty}^{\infty} f(t)dt$ is
$$\lim_{x \to \infty}\int_{-x}^{x^2} f(t) dt = \lim_{x \to \infty}\left(\int_{-x}^{-1} \dfrac{dt}t + \int_{-1}^1 f(t)dt + \int_1^{x^2} \dfrac{dt}t \right) = \lim_{x \to \infty} \left(2 + \log(x^2) - \log(x)\right) = \infty$$
A: In the Lebesgue integral, the real line is just another domain of integration, not treated any differently from domains of finite measure.  In a common approach (Rudin Real and Complex Analysis 1.23, 1.31), the integral $\int_{-\infty}^\infty f(x)\,dx$ is defined by splitting real functions $f$ into positive and negative parts, calculating the integrals of the positive and negative parts separately (as a supremum of simple functions, i.e., functions which take a finite number of values each on a measurable set, which are $\le$ the integrand), and then subtracting the two separate integrals.
Now, when the integral exists in this sense, the limit you've indicated also exists, but not vice versa.  The integral is not defined as the limit, but rather by the supremum of simple functions as indicated.  Perhaps there is a different approach to integration where the integral is defined as the limit. 
