To evaluate the improper integral

$\int_{-\infty}^{-1} x^{-\frac{8}{3}} \, dx,$

let's go through the steps to determine if this integral converges and find its value if possible.

Step 1: Set up the Integral

The given function $x^{-\frac{8}{3}}$ has a negative exponent, making it undefined at $x = 0$, and potentially divergent as $x \to -\infty$. Since the interval of integration is $(-\infty, -1]$, we are dealing with an improper integral due to the unbounded lower limit.

Rewrite the integral with a limit to handle the improper bound:

$\int_{-\infty}^{-1} x^{-\frac{8}{3}} \, dx = \lim_{a \to -\infty} \int_{a}^{-1} x^{-\frac{8}{3}} \, dx.$

Step 2: Integrate $x^{-\frac{8}{3}}$

To integrate $x^{-\frac{8}{3}}$, recall the power rule for integration:

$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad \text{for } n \neq -1.$

In this case, $n = -\frac{8}{3}$, so:

$n + 1 = -\frac{8}{3} + 1 = -\frac{5}{3}.$

Thus,

$\int x^{-\frac{8}{3}} \, dx = \frac{x^{-\frac{5}{3}}}{-\frac{5}{3}} = -\frac{3}{5} x^{-\frac{5}{3}}.$

Step 3: Apply the Limits

Now, substitute back into the integral:

$\lim_{a \to -\infty} \int_{a}^{-1} x^{-\frac{8}{3}} \, dx = \lim_{a \to -\infty} \left[ -\frac{3}{5} x^{-\frac{5}{3}} \right]_{a}^{-1}.$

Evaluating this at the bounds:

[ = \lim_{a \to -\infty} \left( -\frac{3}{5} \cdot (-1)^{-\frac{5}{