Why does $\int_{-\infty}^{\infty} f(x) dx \ne \lim\limits_{t \to \infty} \int_{-t}^{t} f(x) dx$? I just get this is in my head.
How come LHS is not equal to RHS.
 A: The reason is that the kind of convergence you are considering is very strong, akin to a Cauchy principal value. The improper integral is defined to exist if the following limit exists (and is defined to be that value):
$$\int_{-\infty}^{\infty}f(x)\,dx \stackrel{\text{def}}{=} \lim_{t\rightarrow\infty}\lim_{s\rightarrow-\infty}\int_s^t f(x)\,dx.$$
This fails for any odd function. Take for example $f(x) = x$. Then
$$\lim_{t\rightarrow\infty}\int_{-t}^t x\,dx = 0$$
however the proper limit I denoted above does not exist and it's not hard to see.
Basically the way you want to define an improper integral removes a lot of global behavior of your function which is actually important. What would you say the area between the function $f(x)=x$ and the $x$ axis is? Intuitively we want to say that it is zero just by looking at the overall behavior but what if we approach positive infinity faster than negative infinity? In this case, we don't get a well-defined result. Calculus is all about being careful with limits and how you view things.
