Average kinetic energy of an ideal gas

For a gas of particles with mass $m$ in thermal equilibirum at temperature $T$, the distribution function for the velocity along an axis is \begin{align} f(v_i) &= \left(\frac{m}{2\pi k_B T}\right)^{1/2}e^{- \frac{m {v_i}^2}{2 k_B T}}\,. \end{align} This gives the distribution function for the velocity vector $\vec v = (v_x, v_y, v_z)$ \begin{align} F(\vec v) &= \prod_i f(v_i) = \left(\frac{m}{2\pi k_B T}\right)^{3/2} e^{- \frac{m {v}^2}{2 k_B T}}\,. \end{align} With this we can find the thermal average of any velocity-dependent quantity $s(\vec v)$ as \begin{align} \langle s(\vec v)\rangle &= \iiint_{\mathbb R^3} \! s(\vec v) F(\vec v)\,\mathrm d^3\vec v\,. \end{align} The kinetic energy for a single particle with mass $m$ and speed $v$ is \begin{align} E_{\text{kin}} &= \frac{1}{2} m v^2\,, \end{align} so the average kinetic energy is \begin{align} \langle E_{\text{kin}} \rangle &= \iiint_{\mathbb R^3} \! \frac{1}{2} m v^2 F(\vec v) \,\mathrm d^3\vec v\\ &= \frac{1}{2} m\left(\frac{m}{2\pi k_B T}\right)^{3/2} \iiint_{\mathbb R^3} \! v^2 e^{- \frac{m {v}^2}{2 k_B T}} \,\mathrm d^3\vec v\,. \end{align} The triple integral is converted to spherical coordinates. As the integrand depends only on the magnitude of $\vec v$ it is sufficient to consider the integral along the radial coordinate $v$. The differential becomes \begin{align} \mathrm d^3\vec v = 4\pi v^2 \mathrm dv\, \end{align} giving the integral \begin{align} \langle E_{\text{kin}} \rangle &= 2\pi m\left(\frac{m}{2\pi k_B T}\right)^{3/2} \int\limits_0^{\infty} \! v^4 e^{- \frac{m {v}^2}{2 k_B T}} \,\mathrm dv\,. \end{align} We use the substitution $x = \sqrt{\frac{m}{2 k_B T}} v$, so that \begin{align} \langle E_{\text{kin}} \rangle &= 2\pi m\left(\frac{m}{2\pi k_B T}\right)^{3/2}\left(\frac{2 k_B T}{m}\right)^{5/2} \int\limits_0^{\infty} \! x^4 e^{-x^2} \,\mathrm dx \\ &= \frac{4 k_B T}{\sqrt{\pi}} \int\limits_0^{\infty} \! x^4 e^{-x^2} \,\mathrm dx\,. \end{align} The integral is evaluated using the method shown in this post, yielding \begin{align} \int\limits_0^{\infty}\! x^4 e^{-x^2} \,\mathrm dx &= \frac{3}{2}\int\limits_0^{\infty}\! x^2 e^{-x^2}\,\mathrm dx = \frac{3\sqrt{\pi}}{8}\,. \end{align} Plugging into the above expression gives the average kinetic energy of an ideal gas: \begin{align} \langle E_{\text{kin}}\rangle &= \frac{3}{2} k_B T\,. \end{align}