To evaluate the given integral, let's rewrite it in a more standard mathematical notation:

$\int_{0}^{\infty} 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} u^2 e^{-\frac{mu^2}{2k_B T}} \, du$

This is a common integral encountered in statistical mechanics, specifically in deriving the Maxwell-Boltzmann distribution for the speed of particles in a gas. Let's break down and evaluate the integral step-by-step.

Step-by-Step Solution

1. Identify the integral: $\int_{0}^{\infty} 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} u^2 e^{-\frac{mu^2}{2k_B T}} \, du$

2. Simplify the constants: The constant $4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2}$ is independent of $u$ and can be factored out of the integral.

   Let: $A = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2}$ Thus, the integral becomes: $A \int_{0}^{\infty} u^2 e^{-\frac{mu^2}{2k_B T}} \, du$

3. Substitute to simplify the exponential: To simplify the integral, let's make the substitution: $x = \frac{mu^2}{2k_B T} \quad \text{so that} \quad dx = \frac{m}{k_B T} u \, du$ Therefore, $u \, du = \frac{k_B T}{m} \, dx$ and $u^2 = \frac{2k_B T}{m} x$.

   Substituting these into the integral: $\int_{0}^{\infty} u^2 e^{-\frac{mu^2}{2k_B T}} \, du = \int_{0}^{\infty} \frac{2k_B T}{m} x e^{-x} \frac{k_B T}{m} \, dx$ Simplifying further: $\int_{0}^{\infty} \frac{(k_B T)^2}{m^2} 2x e^{-x} \, dx$

4. Evaluate the resulting integral: The integral $\int_{0}^{\infty} x e^{-x} \, dx$ is a standard integral that evaluates to 1.

   So: $\int_{0}^{\infty} \frac{(k_B T)^2}{m^2} 2x e^{-x} \, dx = \frac{(k_B T)^2}{m^2} \times 2 \times 1 = \frac{2(k_B T)^2}{m^2}$

5. Combine with the constant $A$: Recall that $A = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2}$.

   Thus, the integral evaluates to: $4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} \cdot \frac{2(k_B T)^2}{m^2}$

6. Simplify the final expression: Simplifying further:

   $4\pi \cdot \frac{1}{\left(\frac{2\pi k_B T}{m}\right)^{3/2}} \cdot \frac{2(k_B T)^2}{m^2}$ $= 4\pi \cdot \frac{(2\pi k_B T)^{3/2} (2k_B T)^2}{m^{3/2} m^2}$ $= 4\pi \cdot 2\left(\frac{2k_B T}{m}\right)^{1/2}$ $= 4 \times 2 = 8$

Hence, the value of the integral is:

$8$

Would you like more details or explanations? Here are some related questions to explore:

1. What is the significance of this integral in statistical mechanics?
2. How does the Maxwell-Boltzmann distribution describe the speed of particles?
3. What are common substitutions for simplifying Gaussian integrals?
4. How does temperature affect the speed distribution of particles in a gas?
5. Can you explain the role of $k_B$ (Boltzmann constant) in thermodynamics?

Tip: When dealing with integrals involving exponential terms, try using substitution or parts to reduce the expression into a standard form that is easier to evaluate.