Why is the rotation period finite for nonrotating object? This question is conceptual in nature as I am obviously failing to grasp something simple here. Say you have a rotating astronomical object (say a pulsar) with rotation period T. The rotation rate is slowing at the constant rate $\frac{dT}{dt} = C$.
At some time $t$, the pulsar will have come to a complete stop and its frequency will be $f = \frac{1}{T} = 0$. So far, so good.
However, if we separate the original differential equation, and integrate, we get $T(t) = ct+T_0$. Shouldn't this give the pulsar's period at any time $t$? However, there's no $t$ for which $\frac{1}{T} = 0$. In other words, the pulsar should never come to a stop--it should only approach zero rotation asymptotically. What is causing this contradiction?
 A: I'm going to come at this from a maths perspective. You're confusing finite, and infinite.
I'll start with an example, take a sequence $(a_n)_{n=1}^\infty$ where $a_n:=\frac{1}{n}$.
It is easy to see that: $$\lim_{n\rightarrow\infty}(a_n)=0$$
I call it "A-level logic" to understand limits as 'fancy substitution' so let us be formal. Because we cannot define the sequence at $n=\infty$ (or really define any notion of infinity) we sidestep the issue entirely, by saying $a_n\rightarrow L$ if:

$\forall\epsilon > 0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies |a_n-L|<\epsilon]$

Pick any $\epsilon$ you like, there exists some $N$ such that all points of the sequence after $N$ are within $\epsilon$ of the limit.

Lets apply this. A sequence is a special case of a function, $f:\mathbb{N}\rightarrow\mathbb{R}$ by $f:n\rightarrow a_n$ so you should have little trouble following along.
Let $f:\mathbb{R_{\ge 0}}\rightarrow\mathbb{R}$ - this is your frequency.
Nice example

Let us suppose the frequency is decreasing slowly - as in the ODE example above - but is initially spinning
  Given a $t\in\mathbb{R}_{\ge 0}$ $f(t)>0$ so $\frac{1}{f(t)}$ is defined.
  There is no time where this is not the case, if $$\lim_{t\rightarrow+\infty}(f(t))=0$$ then this means:
  $$\forall\epsilon>0\exists C\in\mathbb{R}_{\ge0}\forall x\in\mathbb{R}_{\ge0}[x>C\implies |f(t)-L|<\epsilon]$$
  Notice that this basically says 'there exists a point, $C$ such that after $C$ $f(x)$ is always within $\epsilon$ of the limit' - so again, we never actually get to $t=\infty$, we sidestep it.

Nasty example

Suppose that at some time, $t'$ - for an instant or more - the frequency is truly $0$ then the period function: $p(t):=\frac{1}{f(t)}$ is undefined at this instant.
  There are a few ways we can play this. If you consider the real line, with a single $\infty$ added, you're dealing with something that is (topologically) a circle. Like $\tan$, $-\infty=+\infty$ in this case. This makes sense if something is changing the direction by which it spins (if you've opted to deal with negative frequencies)
  The circle homeomorphism should make sense now. If not I can draw some pictures.  
Suppose however that it is not changing direction, it just stopped and is starting up again. Here period would head up to $+\infty$ then come back down from $+\infty$, which is distinct from going to $-\infty$ then coming back from $-\infty$
Let us suppose that you went for 'both infinities are the same' - you can now say $f(t')=\infty$ accurately, as the limit is the same from both sides of $t'$ however this does not tell you if it came up or down from there. (where as something tending to $+\infty$ tells you it is getting bigger) which isn't a good way.
Let us suppose you went for extended real values, that is adding $\pm\infty$ into the mix. Then if something changes direction, and has no period at $t'$ the limit $\lim_{t\rightarrow t'}(\frac{1}{f(t)})$ doesn't exist, as if you come in from the left, it's different to coming at it from the right! So $f(t')$ is truly undefined. 

As you can see we cannot have it both ways. As such we simply say 'undefined'
A: There's no contradiction, because your assumption that at some finite time the pulsar will stop is false. We can solve the equation for $f$:
$$\frac{dT}{dt} = \frac{d(1/f)}{dt} = -\frac1{f^2}\frac{df}{dt} = C$$
Which is equivalent to
$$\frac{df}{dt} = -Cf^2$$
The solution is 
$$f(t) = \frac{1}{Ct+1/f_0}$$
Of course, this is the same as taking the expression for $T(t)$ and replacing $T=1/f$. You can see from this that $f$ approaches zero asymptotically. Your confusing probably comes from assuming that if $T$ increases linearly, then $f$ decreases linearly; as the math shows, this is not true.
