The given integral is:

$\int_{-\infty}^{\infty} x^3 e^{-x^2} \, dx$

To determine whether this integral is convergent or divergent, let's go through the following steps:

1. Check the integrand's behavior at infinity:

• The function inside the integral is $x^3 e^{-x^2}$.
• As $x \to \infty$ or $x \to -\infty$, the term $e^{-x^2}$ decays very rapidly to zero because it is an exponentially decaying function.
• However, $x^3$ grows for large values of $|x|$, but the decay of $e^{-x^2}$ is much faster than the growth of $x^3$.
• Therefore, the integrand tends to zero at both $+\infty$ and $-\infty$.

2. Symmetry of the integrand:

• The function $x^3$ is odd, meaning $x^3 = -(-x)^3$.
• The exponential function $e^{-x^2}$ is even, meaning $e^{-x^2} = e^{-(-x)^2}$.
• Therefore, the integrand $x^3 e^{-x^2}$ is odd because it is the product of an odd and an even function.
• Integrals of odd functions over symmetric limits (from $-\infty$ to $\infty$) are zero.

3. Conclusion:

• Since the integrand is odd and we are integrating over a symmetric interval (from $-\infty$ to $\infty$), the integral converges and evaluates to 0.

Thus, the integral is convergent and its value is:

$\int_{-\infty}^{\infty} x^3 e^{-x^2} \, dx = 0$

Let me know if you would like further details!

Here are five related questions for further practice:

1. What is the result of $\int_{-\infty}^{\infty} x^2 e^{-x^2} \, dx$?
2. How do you evaluate $\int_{0}^{\infty} e^{-x} \, dx$?
3. How would the integral change if the power of $x$ were even, e.g., $x^2$?
4. What is the value of $\int_{-\infty}^{\infty} e^{-x^2} \, dx$?
5. How do you handle improper integrals where the limits are infinite?

Tip: Integrals involving Gaussian functions like $e^{-x^2}$ often have closed-form solutions and are useful in probability and statistics!