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.

  • 1
    $\begingroup$ What is $\gamma(t)$? Is it $\gamma_3(t)$? Likewise, are $\Gamma(t)$ and $\Gamma_3(t)$ the same? $\endgroup$
    – Blue
    Commented May 15, 2018 at 19:57
  • $\begingroup$ Looks like the intervals are of approximately fixed length and regularly spaced, the derivative is of size $1/t$ and $\Gamma(t)$ is nearly constant at $+\infty$, so if you are talking about $(0,+\infty)$, the sum diverges like a harmonic series, does not it? $\endgroup$
    – fedja
    Commented May 15, 2018 at 23:22
  • 1
    $\begingroup$ @ Blue, yes they are the same. I will edit it. $\endgroup$ Commented May 16, 2018 at 4:56
  • $\begingroup$ You appear to have a typo in the definition of $\gamma(t)$. The coefficient of the $\sin(\omega_0 t)$ term should be $2\omega_0 t$, not $2\omega_0 t^2$. $\endgroup$
    – Blue
    Commented May 16, 2018 at 5:18
  • $\begingroup$ @ Blue Yes, I corrected the typo. Thank you $\endgroup$ Commented May 16, 2018 at 13:49

1 Answer 1


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 $k^{-1}$.

  • $\begingroup$ Thank you. I will look at it tomorrow. My day is over (in germany) $\endgroup$ Commented May 16, 2018 at 14:26
  • $\begingroup$ Thank you. I am not very familiar with the big O notation. Would you mind explaining a bit more where the $O(k^{-1}$ is coming from in the $k$th root? Also, how does it work for $\Gamma(r_k)$? I guess this is simply a question of big O notation. Then the last step, I would have $= e^{-\omega_0^2/2}(e^{\frac{\omega_0^2}{2 \pi k}}-e^{-\frac{\omega_0^2}{2 \pi k}})$ I feel you are missing an exponential there. Thank you for your help $\endgroup$ Commented May 17, 2018 at 9:38
  • 1
    $\begingroup$ Suppose we have $\cos t_0 + \phi(t_0)/t_0 = 0$, and, for large $t$ and for some constant $M$, we have $|\phi(t)| < M$. Then $|\cos t_0| < M/t_0$. From this we know how far $t_0$ can be from zeros of cosine. Next step is expanding $\exp(-\Gamma(t))$ around that point. The $O$ estimates are for $k \to \infty$; expand $\exp(\omega_0^2 / (2 \pi k))$ around $k = \infty$, and you'll get the same result. $\endgroup$
    – Maxim
    Commented May 17, 2018 at 11:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .