For this improper integral: $$\int_{-1}^{1} \frac{1}{x^{3}}dx$$.
My way to solve it is: split this into two parts: $$\int_{-1}^{1} \frac{1}{x^{3}}dx=\lim_{B \rightarrow 0^{-}} \int_{-1}^{B} \frac{1}{x^3}dx+\lim_{A \rightarrow 0^{+}} \int_{A}^{1} \frac{1}{x^3}dx$$
After computing both parts, I got $$\lim_{B \rightarrow 0^{-}} \int_{-1}^{B} \frac{1}{x^3}dx=-\infty$$ and $$\lim_{A \rightarrow 0^{+}} \int_{A}^{1} \frac{1}{x^3}dx=\infty$$
I am confused about the final result, is it $$\int_{-1}^{1} \frac{1}{x^{3}}dx=0$$
Can someone illustrate why or why not? Thanks very much!
Since the part $\displaystyle\int_{0}^{1}\dfrac{1}{x^{3}}dx$ is not finite, then you are ready to conclude that the improper integral does not exist.
But if we take account to principal value sense that $\lim_{\epsilon\downarrow 0}\displaystyle\int_{\epsilon\leq|x|\leq 1}\dfrac{1}{x^{3}}$, then it is zero.