Equipartition Thm: mv_x^2/2 = k_B T/2

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The discussion focuses on demonstrating the equipartition theorem, specifically showing that the average kinetic energy in one dimension, represented as mv_x^2/2, equals k_B T/2. Participants suggest substituting the expression for the velocity distribution function ρ(v_x) into the integral. Calculating this integral is emphasized, with the recommendation to consult a table of integrals if needed. The goal is to establish the relationship between kinetic energy and temperature as outlined by the equipartition theorem. Successful completion of this calculation confirms the theorem's validity in the context of statistical mechanics.
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< \frac{mv_{x}^{2}}{2}> \equiv s\int_{-\infty }^{\infty} \frac{mv_{x}^{2}}{2} \rho (v_{x}) dv_{x}=\frac{k_{B}T}{2}

im stuck
 
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Plug in the expression for ρ(vx) inside the integral and then actually calculate the integral, or look it up in a table of integrals.
 
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