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If we divide it into two parts such that $$I=\int_{-2}^0\frac{1}{x^3}dx+\int_0^3\frac{1}{x^3}dx$$

And then use substitution $x=-t$ we get $$I=\int_2^3\frac1{x^3}dx=\frac5{72}$$

However, If we use limits on both part separately, they both diverge, so the integral diverges too.

Which explanation is correct?

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    $\begingroup$ See Cauchy principal value. $\endgroup$
    – Chen Wang
    Jun 16, 2018 at 12:03
  • $\begingroup$ @ChenWang: expand this into an answer. $\endgroup$
    – robjohn
    Jun 16, 2018 at 12:14

2 Answers 2

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You are right that the integral is not defined (because of the singularity), so you can't trust what the anti-derivative tells you. It's even worse with $$\int_{-1}^{1}\frac{1}{x^2}dx=-2\tag{1}$$

The area is clearly positive ($+\infty$). If you were to break up $(1)$ you would get $$\int_{-1}^{1}\frac{1}{x^2}dx=2\int_{0}^{1}\frac{1}{x^2}dx\rightarrow\infty$$ So it's not always true that you can break up integrals.

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  • $\begingroup$ Indeed, the area is positive, the problem is that it is infinite $\endgroup$ Jun 16, 2018 at 12:36
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    $\begingroup$ The OP is not blindly applying the anti-derivative, so this seems to miss the point. The point is that the integral of an odd, integrable function over an interval symmetric about the origin is zero. Can this be extended to functions with a singularity at the origin? In the strict sense, no; but in the Cauchy Principal Value sense, yes. This is why I encouraged Chen Wang to expand their comment into an answer. $\endgroup$
    – robjohn
    Jun 16, 2018 at 12:49
  • $\begingroup$ @robjohn Integral of odd integrable symmetric function is equal to zero in strict sense or Cauchy Princicple Sense? $\endgroup$
    – Anvit
    Jun 16, 2018 at 12:59
  • $\begingroup$ Also, what about $$\int_{-\infty}^\infty x^3dx$$, $0$ or not defined? $\endgroup$
    – Anvit
    Jun 16, 2018 at 13:01
  • $\begingroup$ @AFalseName: "... the integral of an odd, integrable function over an interval symmetric about the origin is zero. Can this be extended to functions with a singularity at the origin? In the strict sense, no; but in the Cauchy Principal Value sense, yes." Does this not cover the first of your comments? The Cauchy Principal Value might cover the second case if you consider the one point compactification of the reals. $\endgroup$
    – robjohn
    Jun 16, 2018 at 17:21
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On account of the singularity at $x=0$, the integral you give is undefined. The only way you could get close is with: $$\int_{-2}^{3}{[x^{-3}dx]}\approx\int_{-2}^{-\epsilon}{[x^{-3}dx]}+\int_{\epsilon}^{3}{[x^{-3}dx]}$$ With $\epsilon$ as a very small positive constant.

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  • $\begingroup$ With a bit more rigor, what you are suggesting here is called Cauchy Principal Value. $\endgroup$
    – robjohn
    Jun 16, 2018 at 12:51

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