Calculate the time it takes a point particle of mass $m$ to roll down a log potential, $ V(x) = \frac{V_0}{2}ln\frac{x}{x_0}$ using incomplete gamma function. The particle starts at position $x_0$ at time $t=0$ with energy $E=0$, and falls all the way into the singularity at $x=0$.
ok so I tried solving using conservation of energy:
$E = T + V$
$E(x) = \frac{1}{2}m\ddot r$ + $\frac{V_0}{2}ln\frac{x}{x_0}$
with initial conditions:
$r(0) = x_0$; $E = 0$
Plug in:
$0 = \frac{1}{2}m x_0$ + $\frac{V_0}{2}ln\frac{x}{x_0}$
$-mx_0$ = $v_0 ln\frac{x}{x_0}$
From here how do you proceed using a gamma function? to be honest I have no idea how to solve this problem with using a gamma function.. any help will be appreciated.
From conservation of energy, we have
$$\frac12m \dot x^2+V_0\log(x/x_0)=0 $$
Solving $(1)$, we find
$$\frac{dx}{dt}=\sqrt{-\frac{2V_0}{m}\log(x/x_0)}\tag 1$$
Proceeding to integrate $(1)$ reveals
$$\int_0^{x(t)} \left(-\frac{2V_0}{m}\log(x'/x_0)\right)^{-1/2}\,dx'=t$$
Then, enforcing the substitutuion $x'= x_0e^{-s}$ yields
$$\begin{align} t&=\sqrt{\frac{m}{2V_0}}x_0\int_{-\log(x(t)/x_0)}^\infty s^{-1/2}e^{-s}\,ds\\\\ &=\Gamma\left(\frac12,-\log(x(t)/x_0)\right) \end{align}$$
where $\Gamma(\cdot,\cdot)$ is the Upper Incomplete Gamma Function.