To analyze whether the given improper integral converges or diverges, we proceed as follows:

The integral is:

$\int_{1}^{\infty} x \cdot e^{-x^3} \cdot \left(2x^4 - 1\right) \, dx$

Step 1: Observe the behavior of the integrand

The function is: $f(x) = x \cdot e^{-x^3} \cdot \left(2x^4 - 1\right)$

1. The exponential decay:
   The term $e^{-x^3}$ decreases to 0 extremely fast as $x \to \infty$. This suggests that the integrand may converge because the exponential decay dominates growth in $x$-dependent terms.

2. Polynomial growth:
   The term $x \cdot \left(2x^4 - 1\right) = 2x^5 - x$ grows as a polynomial. However, since it is multiplied by $e^{-x^3}$, the fast decay of the exponential function can counteract this growth.

Step 2: Approximate behavior at infinity

At large $x$, the dominant term of the integrand is: $x \cdot e^{-x^3} \cdot 2x^4 \sim 2x^5 e^{-x^3}.$

• Use a substitution to analyze $\int_{1}^{\infty} x^5 e^{-x^3} dx$. Let: $u = x^3 \implies du = 3x^2 \, dx \quad \text{or} \quad x^2 \, dx = \frac{du}{3}.$

  Then: $\int x^5 e^{-x^3} dx = \int x^3 \cdot x^2 e^{-x^3} dx = \int u \cdot e^{-u} \cdot \frac{du}{3}.$

  Simplify: $\int x^5 e^{-x^3} dx = \frac{1}{3} \int u e^{-u} du.$

The integral $\int u e^{-u} du$ converges because $u e^{-u}$ decays rapidly as $u \to \infty$. Hence, this term does not diverge.

Step 3: Behavior at the lower limit ($x = 1$)

At $x = 1$, the integrand is finite: $f(1) = 1 \cdot e^{-1^3} \cdot (2 \cdot 1^4 - 1) = e^{-1} \cdot 1 = e^{-1}.$ Thus, there are no issues with convergence at the lower limit.

Conclusion:

Since the integral converges at both the upper limit ($x \to \infty$) and the lower limit ($x = 1$), the given improper integral converges.

Would you like me to evaluate the integral explicitly or explain any part of this further?

Here are five follow-up questions you might consider:

1. How do we explicitly compute the value of this integral?
2. Why does exponential decay dominate polynomial growth?
3. What are general tests for convergence of improper integrals?
4. How would the result change if the limits of integration were different?
5. Can substitution always simplify an integral involving $e^{-x^n}$?

Tip: When analyzing improper integrals, always check the behavior of the integrand both as $x \to \infty$ and near any singularities!