Why $\int_{-1}^1 \frac{1}{x^3}\, \mathrm{d} x \neq 0$? I was considering the integral $\int_{-1}^1 \frac{1}{x^3}\, \mathrm{d} x$. At first, I suspected that it diverged due to the singularity present at $x = 0$, and WolframAlpha verified my hypothesis. However, I attempted to prove this more rigorously, but was unable. This was my reasoning:
$$\int_{-1}^1 \frac{1}{x^3} \, \mathrm{d}x = \lim_{b \to 0} \int_{-1}^b \frac{1}{x^3} \, \mathrm{d}x + \int_{b}^1 \frac{1}{x^3} \, \mathrm{d}x = \lim_{b \to 0} -\frac{1}{2b^2}+1/2-1/2+\frac{1}{2b^2}=0$$
Because the function is odd, I suspect that the positive and negative infinities introduced by each half cancel to make 0. What was wrong with my reasoning above?
 A: The integral diverges. What you have discovered is the Cauchy Principal Value, which is occasionally useful. It is defined as you did, by finding 
$$\lim_{b\to 0^+}\left(\int_{-1}^{-b}\frac{dx}{x^3}+\int_b^1 \frac{dx}{x^3}\right).$$
Do not separate into two parts, just compute. Inside the big parentheses, we get $0$, so the limit is $0$. 
If we want to compute the integral, and not merely its Cauchy Principal Value, we need to evaluate 
$$\lim_{\substack{a\to 0^+ \\ b\to 0^+}}\left(\int_{-1}^{-a}\frac{dx}{x^3}+\int_b^1 \frac{dx}{x^3}\right)$$
as $a$ and $b$ approach $0$ independently from the right. (Or, equivalently, we need to do the standard splitting into two parts.) The double limit, in this case, does not exist. 
A: The Integral diverges, so it is not $0$. It is the same like saying 
$$\int_{-\infty}^\infty x \, \mathrm{d}x $$
is not convergent. The idea is that you can find partitions where the riemann-sums have different limits.
A: Let $f : [a, b] \rightarrow \mathbb R, a \leqq b$. $f$ is called Riemann integrable if and only if the following condition is satisfied: for every $\epsilon \gt 0$, there exists a tagged partition $Q(y, u)$ such that, for every refinement $P(x, t)$ of $Q(y, u)$, $|R[P(x, t), f] - I|\lt \epsilon$. Whenever $f$ is Riemann integrable, the corresponding real number $I$ is called the Riemann integral of $f$, and is denoted $\int_a^b f(x) dx$. This definition is just a very technical and rigorous way of saying that $f$ is Riemann integrable if the Riemann sums of $f$ converge to some real number $I$.
Why does this matter? Because this reminds us that, in order for any manipulation techniques and integral properties to be used as shortcuts for evaluating integrals, it is first necessary that the integral actually exist, which is to say, that the function actually be Riemann integrable. Otherwise, the properties of the function being integrated are irrelevant. And this is exatly the problem here: the function $f : [-1, 0)\cup(0, 1] \rightarrow \mathbb R, f(x) \equiv \frac{1}{x^3}$ has a non-removable singularity at $0$. Thus, even if we let $f(0) = s$ for some real number $s$, the Riemann sums will not converge, and so $f$ is not Riemann integrable. Since $f$ is not Riemann integrable, one cannot say the Riemann integral exists, let alone that it is equal to $0$. To show that it is not Riemann integrable, all you need to do is have two different partitions with one of the tags being at $0$, and then show how, even if you let the rectangles in those partitions decrease in width, the Riemann sums for those particular partitions will diverge to $\infty$ and $-\infty$, respectively.
I am aware that the Riemann integral is fairly weak, compared to other integrals. But the Lebesgue integral, the gauge integral, and the Khinchin integral, which is generally the strongest integral considered in the literature, all do not exist for $f$ either on $[-1, 1]$. Also, even if we considered it as an improper integral, the limit does not exist.
