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Why is the integral from negative inf to positive inf of x^3 divergent? It's an odd function from -a to a, and because its odd, its symmetric about y=-x, So shouldnt all the areas cancel and get 0?

Nawar Ismail
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  • Because it depends on how you approach $\infty$ and $-\infty$. For example, if you approach $\infty$ twice as fast as you approach $-\infty$, you won't have them cancel. Your notion is known as the Cauchy Principal Value of the improper integral. – Hayden Jul 09 '15 at 23:41
  • But we know it is symmetric, so we know +inf is approached just as fast as -inf. I agree with the last point – Nawar Ismail Jul 10 '15 at 00:23
  • what the function is shouldn't at all affect the rate at which you approach your limits, so I don't see how being symmetric means "we know $+\infty$ is approached just as fast as $-\infty$" – Hayden Jul 10 '15 at 00:28
  • Then I dont understand what you mean by "approach" – Nawar Ismail Jul 10 '15 at 01:09
  • I think that Bruce's answer reflects what I'm trying to say. You're pretending that $\int_{-\infty}^\infty{x^3dx}=\lim_{a\rightarrow \infty}{\int_{-a}^a{x^3dx}}$, when really it is $\int_{-\infty}^\infty{x^3dx}=\lim_{(a,b)\rightarrow (-\infty,\infty)}{\int_a^b{x^3dx}}$. If you take the path $t,2t$, then $b$ is "approaching" $\infty$ twice as fast as $a$ is "approaching" $-\infty$. The point is that it depends upon the path you take, and hence the limit doesn't exist. – Hayden Jul 10 '15 at 02:24

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See Why is it not true that $\int_{-\infty}^\infty{\rm} x\,dx=0 \, $ given that x is an odd function? for a similar discussion. Everything that is said there also applies to your function, because $f(-x) = -f(x)$ for your function. The basic consensus is that the integral is undefined for various reasons, unless you define your integral to be $\lim_{a \to \infty} \int_{-a}^a f(x) dx$, which is a non-traditional definition (although studied in some contexts).

Community
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user2566092
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  • But an integral is defined as the area under a curve, which we know is 0, just by looking at the graph! So, I understand how the definition of an improper integral gives a divergent answer. But the definition of an integral says it should be 0 – Nawar Ismail Jul 10 '15 at 00:26
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Following up on the comment of @user2566092: Would you really be satisfied with a definition of $\int_{-\infty}^{\infty} f(x)\,dx$ in which $\lim_{a \to \infty} \int_{-a}^a f(x)\, dx$ gives a different answer than $\lim_{a \to \infty} \int_{-2a}^a f(x)\, dx$?

BruceET
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