Cfrgtkky

14 4 $begingroup$ This improper integral comes from a problem of periodic orbit. The integral evaluates one half of the period. In a special case, the integral is $$I=int_{r_1}^{r_2}frac{dr}{rsqrt{Phi^2(r,r_1)-1}}$$ where $$Phi(u,v)=frac{uexp{(-u)}}{vexp{(-v)}}$$ The interval follows $Phi(r_1,r_2)=1$ , $r_1<r_2$ . I have found a solution to a special case (by applying perturbation method to the original ODE), which is $$lim_{r_1rightarrow r_2} I =pi$$ When $r_1 rightarrow r_2$ , we have $r_1, r_2 rightarrow r_0$ , where $r_0$ is the peak position of $g(r)=rexp{(-r)}$ . The numerical verification is shown below: $uparrow$ The interval of the integral and the integrand $uparrow$ The integral as a function of $r_2$ My problem is to derive a closed form for $I(r_1)$ , or even just a Taylor e