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One condition of integrability is that the function is bounded across the interval.

$1/x^3$ however has a pole at $x=0$ yet it's integral is defined over all real numbers.

Doesn't this violate the condition since it's integral is defined across a interval where it's unbounded?

asd
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The integral $$\int_{-\infty}^{\infty}\frac{1}{x^3} dx$$ does not exist. This is because the function blows up too rapidly near $x = 0$: the limits $$\lim_{a \rightarrow 0^+} \int_a^{1} \frac{1}{x^3} dx$$ and $$\lim_{a \rightarrow 0^-} \int_{-1}^{a} \frac{1}{x^3} dx$$ do not exist, and therefore $$\int_{-1}^{1} \frac{1}{x^3} dx$$ does not exist. I limited the integration interval to $[-1,1]$ to highlight that the problem is with the blow-up at the origin, not with the tails.

The principal value of the integral, defined as $$\lim_{a \rightarrow 0^+} \left[\int_{-1}^{-a} \frac{1}{x^3} dx + \int_{a}^{1} \frac{1}{x^3} dx\right]$$ does exist and equals zero. This is because the two integrals inside the brackets are finite and have opposite sign, so they cancel each other out. To distinguish the principal value from the ordinary integral, we sometimes write $$\text{PV}\int_{-1}^{1} \frac{1}{x^3} dx = 0$$

  • But the integral of the function is $\frac{-1}{2x^2}$. So why doesn't $\int_{-1}^{1}\frac{1}{x^3} = \frac{-1}{2(1)^2}? - \frac{-1}{2(-1)^2} = 0$ – asd Aug 19 '14 at 02:11
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    @asd It is not correct to say that "the integral of $1/x^3$" is $-1/(2x^2)$. You may be thinking of antiderivatives, which is quite a different topic. –  Aug 19 '14 at 02:14
  • @900sit-upsaday Oops yeah, you're right. I mean't antiderivative. – asd Aug 19 '14 at 02:17
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    @asd The distinction clarifies a few things. Under certain assumptions (like boundedness), we can say that a function is integrable on $[a,b]$, and we may be able to use its antiderivative to evaluate the integral (as the Fundamental Theorem of Calculus says). When those assumptions do not hold, an integral isn't defined and there is not much that we can use the antiderivative for. –  Aug 19 '14 at 02:19
  • Should that be $\int_{-\infty}^\infty$ rather than $\int_\infty^\infty$? – Tavian Barnes Aug 19 '14 at 03:27
  • @Tavian - Thanks, fixed. –  Aug 19 '14 at 03:34