To check whether improper integral converges or not $\int_{-1}^1 \frac{(x-1)}{x^{5/3}}dx$

Question

Given integral is

$$\int_{-1}^1 \frac{(x-1)}{x^{5/3}}dx$$

ATTEMPT

Since there are no problem spots here. so i evaluated integral directly and got some answer. But textbook says integral is divergent. Can you please let me know where i am going wrong?

Thanks

Answer 1

We see that at $x=0$ integrand is undefined. So, $$ \int_{-1}^1 \frac{x-1}{x^{5/3}}dx = \lim_{\substack{\epsilon_1\to+0\\\epsilon_2\to+0}} \left(\int_{-1}^{-\epsilon_1} \frac{x-1}{x^{5/3}}dx + \int_{-\epsilon_1}^{\epsilon_2} \frac{x-1}{x^{5/3}}dx + \int_{\epsilon_2}^1 \frac{x-1}{x^{5/3}}dx\right) $$ As you can see, second integral is diverges.

Answer 2

$$\int\left(\frac{x-1}{x^{\frac{5}{3}}}\right)\text{d}x=$$

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Substitute $u=\sqrt[3]{x}$ and $\text{d}u=\frac{1}{3x^{\frac{2}{3}}}\text{d}x$:

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$$3\int\left(\frac{u^3-1}{u^3}\right)\text{d}u=$$ $$3\int\left(1-\frac{1}{u^3}\right)\text{d}u=$$ $$3\left(\int\left(1\right)\text{d}u-\int\left(\frac{1}{u^3}\right)\text{d}u\right)=$$ $$3\left(u-\int\left(\frac{1}{u^3}\right)\text{d}u\right)=$$ $$3\left(u+\frac{1}{2u^2}\right)+C=$$ $$3\left(\sqrt[3]{x}+\frac{1}{2(\sqrt[3]{x})^2}\right)+C=$$ $$\frac{3(2x+1)}{2x^{\frac{2}{3}}}+C$$

If you now turn in your boundaries you'll see it won't work! It diverges