We are tasked with evaluating the improper integral:

$I = \int_0^\infty e^{-x} x^3 \, dx$

This is a standard integral that can be solved using the Gamma function. The Gamma function is defined as:

$\Gamma(n) = \int_0^\infty e^{-x} x^{n-1} \, dx$

In this case, we can recognize that the given integral is a Gamma function with $n = 4$, because $x^3 = x^{4-1}$. So, we can rewrite the integral as:

$I = \int_0^\infty e^{-x} x^3 \, dx = \Gamma(4)$

Now, we know that:

$\Gamma(n) = (n-1)!$

For $n = 4$:

$\Gamma(4) = 3! = 6$

Thus, the value of the integral is:

$I = 6$

So, the value of the integral $\int_0^\infty e^{-x} x^3 \, dx$ is 6.

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Would you like a more detailed explanation or assistance with a similar problem? Here are some related questions:

1. What is the relationship between the Gamma function and factorials?
2. How does the Gamma function generalize the concept of factorials?
3. How do we handle improper integrals in calculus?
4. What other types of integrals can be solved using the Gamma function?
5. How would you solve an integral of the form $\int_0^\infty e^{-ax} x^n \, dx$?

Tip: When faced with integrals involving powers of $x$ multiplied by exponential functions, consider checking if they can be related to the Gamma function for a quick solution.