Show that the mean kinetic energy for a one-dimensional particle equals $k_BT/2$. That is, show
$$E[p^2/2] =\frac{1}{Z} \int_{\mathbb{R}^2} \frac{p^2}{2} e^{-\beta(V(x) + p^2/2)} \mathop{dxdp},$$
where $\beta = (k_BT)^{-1}$. I also know that
$$Z = \int_\mathbb{R^2} e^{-\beta(V(x) + p^2/2)} \mathop{dx dp}.$$
I just tried to expand the integral
$$E(p^2/2) = \frac{1}{Z} \int_{-\infty}^{\infty} \frac{p^2 e^{-\beta p^2/2}}{2} \int_{-\infty}^{\infty} e^{-\beta V(x)} \mathop{dx} \mathop{dp},$$
but the innermost integral works out to be $\exp(-\beta V(x)/(-\beta V'(x))$ evaluated from $x = -\infty$ to $x = \infty$, and I'm not entirely sure how to do that. I was thinking maybe to somehow cancel out the $Z$ in the denominator with an integral but I couldn't come up with anything.
Any help is appreciated.
We can use Fubini's theorem to write
$$\left\langle \frac{p^2}{2} \right\rangle = \frac{\int_{\Bbb{R^2}}\frac{p^2}{2}\exp\left(-\beta(V(x)+p^2/2)\right)dxdp}{\int_{\Bbb{R^2}}\exp\left(-\beta(V(x)+p^2/2)\right)dxdp} = \frac{\left(\int_\Bbb{R}\exp\left(-\beta V(x)\right)dx\right)\cdot\left(\int_\Bbb{R}\frac{p^2}{2}\exp\left(- \frac{\beta p^2}{2}\right)dp\right)}{\left(\int_\Bbb{R}\exp\left(-\beta V(x)\right)dx\right)\cdot\left(\int_\Bbb{R}\exp\left(-\frac{\beta p^2}{2}\right)dp\right)}$$
$$= \frac{\int_\Bbb{R}\frac{p^2}{2}\exp\left(- \frac{\beta p^2}{2}\right)dp}{\int_\Bbb{R}\exp\left(-\frac{\beta p^2}{2}\right)dp} = \frac{\frac{1}{4}\sqrt{\frac{8\pi}{\beta^3}}}{\sqrt{\frac{2\pi}{\beta}}} = \frac{1}{2\beta} = \frac{1}{2}k_BT$$
from Feynman's trick and the result of the Gaussian integral.