To determine whether the given improper integral diverges or converges, let's analyze the integral:

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

The integrand is $\frac{1}{4x^5}$, and we can break this integral into two parts:

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

Since $\frac{1}{x^5}$ has a singularity at $x = 0$, we need to examine the behavior of this function near 0 and at infinity.

Analysis at infinity ($x \to +\infty$ and $x \to -\infty$):

As $x$ tends to infinity, $\frac{1}{x^5}$ decays rapidly to 0. Hence, the integral over the region $(1, \infty)$ and $(- \infty, -1)$ converges because the function decreases fast enough.

Analysis near $x = 0$:

For the behavior near $x = 0$, note that $\frac{1}{x^5}$ becomes extremely large as $x \to 0$. Let's analyze the integrals near $x = 0$.

For small positive $x$, say from $0$ to some small positive number $\epsilon$: $\int_{\epsilon}^{0} \frac{1}{x^5} dx$ This integral can be computed as: $\int_{\epsilon}^{0} x^{-5} dx = \left[ \frac{x^{-4}}{-4} \right]_{\epsilon}^{0}$ When evaluated, the term $\frac{1}{x^4}$ approaches infinity as $x \to 0$, which means the integral diverges.

Thus, the singularity at $x = 0$ causes the integral to diverge, as the function $\frac{1}{x^5}$ grows too fast near zero for the integral to remain finite.

Conclusion:

The integral diverges due to the behavior near $x = 0$.

Would you like a more detailed explanation or further clarification on any step?

Here are 5 related questions for deeper understanding:

1. How do we determine if an improper integral converges or diverges in general?
2. What other types of singularities can lead to the divergence of an integral?
3. How does the order of a pole affect the convergence or divergence of an integral?
4. Can you solve an improper integral that converges using comparison tests?
5. What are some examples of functions with singularities that still have convergent integrals?

Tip: Always check for singularities in the integrand, especially in improper integrals, as these often determine whether the integral diverges or converges.