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?

  • $\begingroup$ Because $\infty$ is not a number is a concept. $\endgroup$ – valer Oct 9 '17 at 22:41
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    $\begingroup$ Your top and bottom limits are equal and this gives the cauchy principal value, and it is zero as you say. But the definition of the improper integral is that both limits must vary independently, and under this definition there is no well defined improper limit. $\endgroup$ – Rene Schipperus Oct 9 '17 at 22:43

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 $\lim\limits_{a\to \infty}\int_{-2a}^{a} x^3 dx = \lim\limits_{a\to \infty} \frac {a^4 - (2a)^4}4 = \lim\limits_{a\to \infty} a^4\frac {1-15}4 = -\infty$


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