Evaluate the following expression: ׬ 4𝜋ሺ ௠
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మ 𝑢 ଶ 𝑒 ି
೘ೠమ
మೖಳ೅ 𝑑𝑢
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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.

Evaluate the following expression: ׬ 4π( m / 2πk_BT )^(3/2) u^2 e^(-mu^2 / 2k_BT) du from 0 to ∞

Step-by-step evaluation of the Maxwell-Boltzmann distribution integral to determine the speed of particles in a gas. This problem covers Gaussian integrals, exponential functions, and the role of the Boltzmann constant. Ideal for university-level students studying statistical mechanics.

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