I am having trouble understanding the derivation of the integral formula, most papers just assume the resistance to bending is:
$$\frac{A}{2}\left|\frac{dt(s)}{ds}\right|^2.$$
I have found some derivation but they are missing a step, how is the bottom summation formula, $$E=-\frac{B}{b}\sum_{i=2}^N\cos(\theta_i) - Fb\sum_{i=1}^N \cos(\gamma_i)$$ transformed to the integral formula: $$ \frac{E}{k_BT}=\frac{\xi_T}{2}\int_0^L\left(\frac{d\vec{t}(s)}{ds}\right)^2ds - \frac{F}{k_BT}\int_0^L \cos(\gamma(s))ds$$
Where $\xi_T=B/k_BT$ and $\cos(\theta_i)=\vec{t}_i\cdot\vec{t}_{i+1}$, when $b \to 0$, $\ t_i=t(s)$ and $t_{i+1}=t(s+ds)$
The question might be answer in some paper, but I can't find it so if you now where I could look this up or you know how to do this. I would greatly appreciate it.
