Why doesn't $x$ reach a constant for a block experiencing $v^n$ resistive force? I am stuck on the Exercise 3.5 of Newtonian Dynamics by R. Fitzpatrick:

A block of mass $m$ slides along a horizontal surface which is lubricated with heavy oil such that the block suffers a viscous retarding force of the form
$$F = - c\,v^n,$$
where $c>0$ is a constant, and $v$ is the block's instantaneous velocity. If the initial speed is $v_0 $ at time $t=0$, find $v$ and the displacement $x$ as functions of time $t$. Also find $v$ as a function of $x$. Show that for $n=1/2$ the block does not travel further than $2\,m\,v_0^{3/2}/(3\,c)$.

The last part of the question asks to show that for $n=1/2$ the block does not travel further than $2mv_0^{3/2}/(3c)$.
We start from Newton's second law
$$ m \frac{d^2x}{dt^2} = m \frac{dv}{dt} = m v \frac{dv}{dx}= -cv^n. $$
Separating variables gives
$$ \int_{v_0}^{v} \frac{dv'}{(v')^{n-1}} = -\frac{c}{m} \int_0^x dx', $$
$$ v^{-n+2} = v_0^{-n+2} - \frac{(-n+2)cx}{m}. $$
Plugging $n=1/2$,
$$ v^{3/2} = v_0^{3/2} - \frac{3cx}{2m}. $$
Setting the velocity to zero (this must be the case if the block stops moving),
$$ x =\frac{2m v_0^{3/2}}{3c}, $$
which is the desired result.
The problem arises when I try to solve for $x$ in terms of $t$. Now,
$$ m \frac{dv}{dt} = -cv^n, $$
$$ \int_{v_0}^{v} \frac{dv'}{(v')^n} = -\int_0^t \frac{c}{m} dt', $$
$$ \frac{1}{v^{n-1}} = \frac{1}{v_0^{n-1}} - \frac{(-n+1)c}{m} t. $$
Rising everything to $1/(1-n)$ power (of course, assuming that $n \ne 1$),
$$ v = \left( \frac{1}{v_0^{n-1}} - \frac{(-n+1)c}{m} t \right)^\frac{1}{1-n}.$$
Plugging $n=1/2$ gives:
$$ \frac{dx}{dt} = \left( v_0^{1/2} -\frac{c}{2m} t \right)^2. $$
Let's separate the variables and try to integrate,
$$ \int_0^x dx = \int_0^t \left( v_0^{1/2} - \frac{c}{2m} t' \right)^2 dt', $$
$$ x_{\mathrm{f}} = \int_0^{\infty} \left( v_0^{1/2} - \frac{c}{2m} t' \right)^2 dt'. $$
I've plugged $t = \infty$ because it seems to me that the block must stop to this time if it's going to stop at all. The problem is that the integral on the right hand side won't converge! So $x$ has no finishing point, which contradicts the first part of the solution. What's going on here?
 A: From
$$\dfrac{dx}{dt}=\left(v_0^{1/2}-\dfrac{c}{2m}{t}\right)^2$$
and
$$v(t_f)=\left.\dfrac{dx}{dt}\right|_{t=t_f}=0$$
you should be able to get a finite bound on your last integral.

EDIT: one possible reason for which your final integral doesn't properly converge comes from an earlier step. Indeed, you moved from:
$$m\dfrac{dv}{dt}=-cv^n$$
to:
$$\dfrac{dv}{v^n}=-\dfrac{c}{m}dt$$
The big caveat here is of course that this is only valid for $v\neq 0$. And in fact, the physical solution tells us that $v=0$ forever when $t_f$ is reached!
A: As you said,
$$ \int_0^x dx = \int_0^t \left( v_0^{1/2} - \frac{c}{2m} t' \right)^2 dt', $$
which means, after the integration, that:
$$
x(t) = x_1(t)=\frac{c^2 t^3}{12m^2} - \frac{ct^2v_0^{1/2}}{2m}+tv_0
$$ 
The point in time for which $x_1(t_f) = x_f = \frac{2mv_0^{3/2}}{3c}$ is $t_f=\frac{2mv_0^{1/2}}{c}$ which is not infinite. At this point, the velocity is zero, so the force is zero, so the acceleration is zero, so the velocity remains zero and the position remains constant. More specifically:
$$
x(t)= \begin{cases} 
      x_1(t)  & t\leq t_f \\
      x_f & t \gt t_f \\
   \end{cases}
$$
This function is not analytic, though it's continuous and has a first derivative. This is induced by the (unrealistic) $v^{1/2}$ force. Non-analytic solutions don't typically appear in realistic scenarios, but they can pop up in toy scenarios like this one or Norton's dome.
