Area under a curve of an odd function from negative infinity to positive infinity

Question

In integration, there is a property that says: If you're integrating from -a to a some odd function f(x), then the area under the curve between -a and a is zero.

I was listening to this in class , and then I thought about integrating some odd function, like x^3, from negative infinity to positive infinity.

But, if you integrate x^3 and then solve it from negative infinity to positive infinity, wouldn't you end up subtracting infinity from infinity, which is undefined?

Given this, which answer is the correct one: is the area 0 or is it undefined?

Answer 1

For improper integrals, you're correct: you have to be careful. Both limits need to exist independently of each other. In your case, $\int_{-\infty}^0 x^3\,dx$ is $-\infty$, hence the integral "doesn't exist" (except in the extended real numbers case). There is something called a principal value, where you take the limits simultaneously, e.g., $$\lim_{N\to\infty}\int_{-N}^N x^3\,dx=\lim_{N\to\infty}0=0,$$ and in this sense, the limit will always give $0$.

Answer 2

The improper integral is defined as $\int_{-\infty}^{\infty}x^3dx$=$\lim_{\alpha\to-\infty}\lim_{\beta\to\infty}\int_{\alpha}^{\beta}x^3 dx$, which as you said is undefined as the final calculation is $\infty-\infty$.

However, this integral can be given a value by something called the Cauchy Principal Value, which is a method of giving a value to double infinite integrals such as this. It is defined as $PV\int_{-\infty}^{\infty}x^3dx$$=\lim_{R\to\infty}\int_{-R}^{R}x^3dx$, which will converge to $0$ as your heuristic has you believe.