Integrating $x^3$ perplexes me. $\int_{-1}^1 x^3dx=0$ and $\int_{-2}^2 x^3dx=0$ and even $\int_{-1000}^{1000} x^3dx=0$. Intuitively, it would make sense that $\int_{-\infty}^\infty x^3dx=0$. However, this integral is undefined.
Why does my intuition fail? The integral from $-a$ to $a$ of $x^3$ is always $0$, so why doesn't this hold for bounds of $-\infty$ to $\infty$? I understand how the math checks out (improper integrals), but instinctively this result makes no sense. Can anyone give me an intuitive explanation of this solution?
Going into details of Rene Schipperus' comment:
$\int_{a}^b x^3 dx = \frac {x^4}4|_{a}^b = \frac {b^4 - a^4}4$.
So $\int_{-\infty}^{\infty}x^3 dx = \lim\limits_{a\to -\infty,b\to \infty}\frac {b^4 - a^4}4$ which is undefined as the nature of $a\to -\infty,b\to \infty$ may vary.
We could say $\lim\limits_{a\to \infty}\int_{-a}^a x^3 dx = 0$.
But we cold also say $\lim\limits_{a\to \infty}\int_{-a}^{2a} x^3 dx = \lim\limits_{a\to \infty} \frac {(2a)^4 - a^4}4 = \lim\limits_{a\to \infty} a^4\frac {16-1}4 = \infty$.
or $\lim\limits_{a\to \infty}\int_{-2a}^{a} x^3 dx = \lim\limits_{a\to \infty} \frac {a^4 - (2a)^4}4 = \lim\limits_{a\to \infty} a^4\frac {1-15}4 = -\infty$