5
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – copper.hat Aug 9 '18 at 15:31
  • $\begingroup$ 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) $\endgroup$ – aidangallagher4 Aug 9 '18 at 15:32
  • $\begingroup$ @copper.hat Not sure what you are implying about $x^3$ not being integratable. $\endgroup$ – Don Thousand Aug 9 '18 at 15:32
  • $\begingroup$ @RushabhMehta: Integrable usually means that the integral is finite. $\endgroup$ – Clayton Aug 9 '18 at 15:32
  • $\begingroup$ @aidangallagher4 that's correct, which is why what TheSimpliFire did is quite incorrect. $\endgroup$ – Don Thousand Aug 9 '18 at 15:32
17
$\begingroup$

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 $0$.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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 $\lim_{(\alpha, \beta) \rightarrow (-\infty, \infty)} \dots$. $\endgroup$ – Eric Towers Aug 9 '18 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.