I need to rewrite equation so that i can use gamma function. Below are assignment text and my steps and reasoning so far:
The probability of finding a translational energy in the range $$E_{tr}, E_{tr} + dE_{tr}$$ is given by: $$P(E_{tr})dE_{tr} = 2\cdot \pi \cdot(\frac{1}{\pi \cdot K_{B} \cdot T})^{\frac{3}{2}}\sqrt{E_{tr}} \cdot exp(\frac{-E_{tr}}{K_B\cdot T})dE_{tr}$$
Calculate the average translational energy using equation above
The average value or mean value of a variable x, which can take any value between $-\infty \ and\ \infty$ $$\left\langle x\right\rangle= \intop_{-\infty}^{\infty}xP(x)dx$$
So to find the average translational energy:
$$\left\langle E_{tr}\right\rangle= \intop_{0}^{\infty}E_{tr}\cdot P(E_{tr})dx =\intop_{0}^{\infty}E_{tr}\cdot2\cdot \pi \cdot(\frac{1}{\pi \cdot K_{B} \cdot T})^{\frac{3}{2}}\sqrt{E_{tr}} \cdot exp(\frac{-E_{tr}}{K_B\cdot T})dE_{tr}$$
rearranging : $$\left\langle E_{tr}\right\rangle=2\cdot \pi \cdot(\frac{1}{\pi \cdot K_{B} \cdot T})^{\frac{3}{2}}\intop_{0}^{\infty}E_{tr}^{\frac{3}{2}} \cdot exp(\frac{-E_{tr}}{K_B\cdot T})dE_{tr}$$
I was given a hint: in connection with the evaluation of the integrals the gamma function $\Gamma(n)$ is useful. it is defined as $$\Gamma(n) = \intop_{0}^{\infty}x^{n-1} \cdot exp(-x)dx, n>0$$
How do I get rid / rearrange the equation so $K_{B}\cdot T$ no longer is in exp() ? I want to get rid of them so the integral will be in the same form as the Gamma $\Gamma(n)$ function.
Thank you for your time reading/helping me with my problem!
from $$\left\langle E_{tr}\right\rangle=2\cdot \pi \cdot(\frac{1}{\pi \cdot K_{B} \cdot T})^{\frac{3}{2}}\intop_{0}^{\infty}E_{tr}^{\frac{3}{2}} \cdot exp(\frac{-E_{tr}}{K_B\cdot T})dE_{tr}$$
using $y=\frac{E_{tr}}{k_B T}$ in the integration you get
$$\left\langle E_{tr}\right\rangle=\frac{2}{\pi^{1 / 2}} \cdot k_B T\intop_{0}^{\infty}y^{\frac{3}{2}} \cdot exp(-y)dy = \frac{2}{\pi^{1 / 2}} \cdot k_B T \space\ \Gamma(5 / 2)$$
which is reasonable being $k_B T$ an energy