The improper integral of $\int_{-1}^1 \frac{x-1}{ \sqrt[3]{x^5} } dx$

Question

I have a kind request to check whether my solution is correct.

$$ \int \limits_{-1}^1 \frac{x-1}{ \sqrt[3]{x^5} } dx = \int \limits_{-1}^0 \frac{x-1}{ \sqrt[3]{x^5} } dx + \int \limits_{0}^1 \frac{x-1}{ \sqrt[3]{x^5} } dx = $$ $$ =\lim \limits_{A \to 0^+} \int \limits_{-1}^A \frac{x-1}{ \sqrt[3]{x^5} } dx + \lim \limits_{A \to 0^-} \int \limits_{A}^1 \frac{x-1}{ \sqrt[3]{x^5} } dx $$

$$ \int \frac{x-1}{\sqrt[3]{x^5}} dx = 3 x^{\frac{1}{3}} - \frac{3}{2} \frac{1}{x^{\frac{2}{3}}} + C $$

$$ \int \limits_{-1}^1 \frac{x-1}{ \sqrt[3]{x^5} } dx = \lim \limits_{A \to 0^+} \Big{[}3 x^{\frac{1}{3}} - \frac{3}{2} \frac{1}{x^{\frac{2}{3}}} \Big{]}^{A}_{-1} + \lim \limits_{A \to 0^-} \Big{[}3 x^{\frac{1}{3}} - \frac{3}{2} \frac{1}{x^{\frac{2}{3}}} \Big{]}^{1}_{A} $$ $$ \lim \limits_{A \to 0^+} 3 A^{\frac{1}{3}} - \frac{3}{2} \frac{1}{A^{\frac{2}{3}}} - \Big{(}3 (-1)^{\frac{1}{3}} - \frac{3}{2} \frac{1}{ (-1)^{\frac{2}{3}} } \Big{)} = -\infty $$

$$ \lim \limits_{A \to 0^-} \frac{3}{2} - 3 A^{\frac{1}{3}} + \frac{3}{2} \frac{1}{A^{\frac{2}{3}}} = - \infty $$

Since: $$ \lim \limits_{A \to 0+} \frac{1}{A^{\frac{2}{3}}} = \infty $$ $$ \lim \limits_{A \to 0-} \frac{1}{A^{\frac{2}{3}}} = - \infty $$

Therefore $\int \limits_{-1}^1 \frac{x-1}{ \sqrt[3]{x^5} } dx$ does not converge.

Answer 1

Your work is correct. The oddness of $x^{-5/3}$ does not matter, because $$\lim_{A \to 0+0} \int_A^1 x^{5/3} \mathrm{d}x$$ is not finite. The same problem appears when you are trying to compute $$\int_{-1}^1 \frac{1}{x} \mathrm{d}x$$

Answer 2

In$$x^{-2/3}-x^{-5/3},$$ the second term is odd and vanishes by symmetry. The first term integrates with exponent $\frac13$, and this converges.

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Alternatively, set $x:=t^3$, which is invertible.

$$\int \limits_{-1}^1 \frac{x-1}{ \sqrt[3]{x^5} } dx = 3\int \limits_{-1}^1 \left(1-\frac1{t^3}\right)dt.$$