Finding roots of trigonometric function I have been given the following functions (where $\omega_0$ is some positive constant)
$$\begin{align}
\Gamma(t) &= \frac{1}{t^2} \left(\;  \frac{1}{2} \omega_0^2 t^2 - \cos(\omega_0 t) - \omega_0 t \sin(\omega_0 t) + 1 \;\right) \\[4pt]
\gamma(t) = \frac{d \Gamma(t)}{dt} &= \frac{1}{t^3} \left(\;  2 \omega_0 t \sin(\omega_0 t) - (\omega_0^2 t^2  - 2) \cos(\omega_0 t) - 2 \;\right) 
\end{align}$$
and need to find the intervals $(a_n, b_n)$ such that $\gamma(t)<0$ for $ t \in (a_n,b_n)$. 
I then need to calculate
$$N(\Phi)= \int_{\gamma(t)<0} - \gamma(t) e^{-\Gamma(t)}dt =\sum_n e^{-\Gamma(b_n)}-e^{-\Gamma(a_n)}$$
I have been working on this and can't get it to work. 
The question really only is if $N(\Phi)=\infty$ or if it is finite. Both are possible. I tried working on the roots but don't get anywhere for an explicit interval. I know that
$$\lim_{t \to \infty} \Gamma(t) = \frac12\omega_0^2$$
I am not sure though what this tells me about my sum. Clearly individually the sums are divergent. But if I take the difference I get that
$$\lim_{n \rightarrow \infty} e^{-\Gamma(b_n)} - e^{-\Gamma(a_n)}=0$$
and thus I have the possibility of convergence.
If it is not possible to calculate the roots explicitly but know that $N(\Phi)=\infty$ that is sufficient. Otherwise I guess I will have to use numerical methods.
I would really appreciate any kind of help or hint. Thank you.
Edit: What I gather so far from Maxims answer is this.
$\gamma(t)$ behaves like $\frac{-\omega_0^2 \cos(\omega_0 t)}{t}$. This I assumes is the case as $\sin(t)/t^2$ and $2/t^3$ are "faster" decreasing then my other term. So finding the roots of $\frac{-\omega_0^2 \cos(\omega_0 t)}{t}$ gives me
$$ r_k=\frac{\pi (2k +1)}{2 \omega_0} + O(k^{-3})$$
Shouldn't there be $O(k^{-3})$ as I have the term $2/t^3$? If not, why not? 
The calulations for $\Gamma(r_k)$ are clear for the roots. But then calculating 
$$ exp(-\Gamma(r_{2k}) - exp(-\Gamma(r_{2k-1}) = e^{\frac{-\omega_0^2}{2}}(e^{\frac{\omega_0^2}{2 \pi k}}-e^{-\frac{\omega_0^2}{2 \pi k}}) + O(k^{-2}) $$
Thus leaving me with the following sum
$$e^{\frac{-\omega_0^2}{2}} \sum_k (e^{\frac{\omega_0^2}{2 \pi k}}-e^{-\frac{\omega_0^2}{2 \pi k}}) + O(k^{-2}) $$ 
But then $\sum_k O(k^{-2}) \rightarrow \frac{\pi^2}{6}$.and what happens to the other sum? Thank you for your time and effort.  

The general idea goes like this. $\gamma(t)$ behaves like $-\omega_0^2 \cos(\omega_0 t)/t$, and the position of the $k$th root of $\gamma(t)$ is
$$r_k = \frac {\pi (2 k + 1)} {2 \omega_0} + O(k^{-1}).$$
(The next order term is $-2 / (\pi \omega_0 k)$, but we only need the $O$ estimate.) Then
$$\Gamma(r_{2 k}) =
 \frac {\omega_0^2} 2 - \frac {\omega_0^2} {2 \pi k} + O(k^{-2}), \\
\Gamma(r_{2 k - 1}) =
 \frac {\omega_0^2} 2 + \frac {\omega_0^2} {2 \pi k} + O(k^{-2}), \\
\exp(-\Gamma(r_{2 k})) - \exp(-\Gamma(r_{2 k - 1})) =
\frac {\omega_0^2 e^{-\omega_0^2 / 2}} {\pi k} + O(k^{-2})$$
(the remainder is in fact $O(k^{-3})$, but that isn't important either), and the sum over the first $n$ negative intervals grows as $\ln n$ times the coefficient at $k^{-1}$.
