To determine if the given improper integral diverges or converges, let's evaluate:

$\int_{-\infty}^{\infty} \frac{1}{x^{4/5}} \, dx$

Step 1: Consider the Integral on $(0, \infty)$

First, we'll evaluate the integral from $0$ to $\infty$:

$\int_{0}^{\infty} \frac{1}{x^{4/5}} \, dx$

This can be rewritten as:

$\int_{0}^{\infty} x^{-4/5} \, dx$

To evaluate this, we find the antiderivative:

$\int x^{-4/5} \, dx = \frac{x^{1 - 4/5}}{1 - 4/5} = \frac{x^{1/5}}{1/5} = 5x^{1/5}$

Now, evaluate this at the limits $0$ and $\infty$:

$\lim_{b \to \infty} \left[ 5x^{1/5} \right]_{0}^{b} = \lim_{b \to \infty} \left( 5b^{1/5} - 5(0)^{1/5} \right)$

Since $b^{1/5}$ grows without bound as $b$ approaches infinity, the integral diverges on $(0, \infty)$.

Step 2: Consider the Integral on $(-\infty, 0)$

Similarly, consider the integral from $-\infty$ to $0$:

$\int_{-\infty}^{0} \frac{1}{x^{4/5}} \, dx$

For $x < 0$, $\frac{1}{x^{4/5}}$ is not well-defined in the real number system because it involves taking a fractional power of a negative number, which results in a complex number. Therefore, the integral diverges for $x < 0$.

Conclusion

The integral diverges on both $(0, \infty)$ and $(-\infty, 0)$. Hence, the integral

$\int_{-\infty}^{\infty} \frac{1}{x^{4/5}} \, dx$

diverges.

Would you like more details or have any questions? Here are some related questions you might find useful:

1. How do you evaluate improper integrals?
2. What conditions determine the convergence of an integral?
3. How do complex numbers relate to integrals over the real line?
4. What is the behavior of integrals with singularities at zero?
5. Can improper integrals over symmetric functions converge?

Tip: When dealing with improper integrals, always check the behavior at singularities and at the limits of integration.