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

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?

• It is only zero if the integral is defined. The function $x \mapsto x^3$ is not integrable on the real line, so the integral is not defined. You can define an improper integral as the limit, but the integral per so is not defined. – copper.hat Aug 9 '18 at 15:31
• In general when you have improper integrals you consider the bounds as limits (ie $\int_{0}^{\infty}$ actually represents $\lim_{n \to \infty} \int_{0}^{n}$)(same principle if both bounds are infinite) – aidangallagher4 Aug 9 '18 at 15:32
• @copper.hat Not sure what you are implying about $x^3$ not being integratable. – Don Thousand Aug 9 '18 at 15:32
• @RushabhMehta: Integrable usually means that the integral is finite. – Clayton Aug 9 '18 at 15:32
• @aidangallagher4 that's correct, which is why what TheSimpliFire did is quite incorrect. – Don Thousand Aug 9 '18 at 15:32

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 $0$.
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.
• There is a small defect in your notation. You have written that the two limits are taken sequentially. Correctly, they are taken simultaneously and independently, that is, as $\lim_{(\alpha, \beta) \rightarrow (-\infty, \infty)} \dots$. – Eric Towers Aug 9 '18 at 20:56