To solve this integral using the gamma function, let's analyze the integral step by step.

The given integral:

$\int_{0}^{\infty} e^{-\sqrt[3]{8x}} \, dx$

We rewrite it in a form suitable for substitution and relating to the gamma function.

Step 1: Substitution

Let: $t = \sqrt[3]{8x} \implies x = \frac{t^3}{8}$

Differentiating: $dx = \frac{3t^2}{8} \, dt$

Step 2: Update the limits and integrand

When $x = 0$, $t = 0$.
When $x = \infty$, $t = \infty$.

Substituting into the integral: $\int_{0}^{\infty} e^{-\sqrt[3]{8x}} \, dx = \int_{0}^{\infty} e^{-t} \cdot \frac{3t^2}{8} \, dt$

Simplify: $\int_{0}^{\infty} e^{-\sqrt[3]{8x}} \, dx = \frac{3}{8} \int_{0}^{\infty} t^2 e^{-t} \, dt$

Step 3: Express in terms of the gamma function

The gamma function is defined as: $\Gamma(n) = \int_{0}^{\infty} t^{n-1} e^{-t} \, dt$

In our case, $t^2 e^{-t}$ corresponds to $n - 1 = 2 \implies n = 3$. Therefore: $\int_{0}^{\infty} t^2 e^{-t} \, dt = \Gamma(3)$

Using the property of the gamma function, $\Gamma(n) = (n-1)!$: $\Gamma(3) = 2! = 2$

Step 4: Final result

Substitute $\Gamma(3) = 2$ back into the integral: $\int_{0}^{\infty} e^{-\sqrt[3]{8x}} \, dx = \frac{3}{8} \cdot 2 = \frac{6}{8} = \frac{3}{4}$

Final Answer:

$\int_{0}^{\infty} e^{-\sqrt[3]{8x}} \, dx = \frac{3}{4}$

Let me know if you want further clarification or detailed steps!

Here are 5 related questions to explore:

1. How is the gamma function derived and used in mathematical analysis?
2. What properties of the gamma function make it useful in integrals?
3. How would the solution differ if the limits of integration were finite instead of infinite?
4. Can this integral be evaluated using methods other than the gamma function?
5. How does substitution simplify complex integral forms into recognizable patterns?

Tip: Familiarizing yourself with common substitutions (like exponential forms) is crucial when working with improper integrals and gamma functions.