Closed form of an improper integral to solve the period of a dynamical system

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 $$I=\int_{r_1}^{r_2}\frac{dr}{r\sqrt{\Phi^2(r,r_1)-1}}$$ where $$\Phi(u,v)=\frac{u\exp{(-u)}}{v\exp{(-v)}}$$

The interval follows $$\Phi(r_1,r_2)=1$$, $$r_1.

I have found a solution to a special case (by applying perturbation method to the original ODE), which is $$\lim_{r_1\rightarrow r_2} I =\pi$$ When $$r_1 \rightarrow r_2$$, we have $$r_1, r_2 \rightarrow r_0$$, where $$r_0$$ is the peak position of $$g(r)=r\exp{(-r)}$$.

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 expansion about $$r_0$$. I appreciate any hint.

Thanks!

If you are interested, here is the general form of the integral: $$I=\int_{r_1}^{r_2}\frac{dr}{r\sqrt{\Phi^2(r,r_1)-1}}$$ where $$\Phi(u,v)=\frac{u\exp{(k(u))}}{v\exp{(k(v))}}$$ and $$k$$ is a decreasing function. The interval follows $$\Phi(r_1,r_2)=1$$, $$r_1.

By solving the original ODE using perturbation method, the solution to a special case is $$\lim_{r_1\rightarrow r_2} I =\frac{\pi}{\sqrt{1+r_0 k''(r_0)/k'(r_0)}}$$

When $$k(r)=-r$$, it reduces to $$\pi$$.

In fact, $$\lim_{r_1 \rightarrow r_2} I (k(r)=-C\cdot r^n) = \pi/\sqrt{n}$$.

Thanks to Fabian, the second derivative at $$r=1$$ matches $$I=\pi-\frac{\pi}{12}\epsilon^2+O(\epsilon^3)$$:

$$\uparrow$$ The above is the numerical second order derivative of figure 2.

I do not know how to obtain an explicit solution to the problem. However, it is possible to have a Taylor series of the integral $$I(r_1)$$ around $$r_1=1$$.

Let us first perform the substitution $$r= r_1 (1-x) + r_2 x$$ such that the boundaries of the integral do not depend on $$r_1$$. In particular, we obtain the expression $$I(r_1) =\int_0^1 \frac{r_2 -r_1}{(r_1 (1-x) + r_2 x) [\Phi(r_1 (1-x) + r_2 x,r_1)^2 -1 ]^{1/2}}\,dx\,.$$

Next, we need a relation between $$r_2$$ and $$r_1$$. If you look at the function $$r\exp(-r)$$ you see that it is monotonous on the interval $$r\in[0,1]$$ and $$r\in[1,\infty]$$. The inverse of this function is commonly called the Lambert W function. In particular, the inverse of the respective branches are denoted by $$r=- W(-x) \in [0,1], \qquad r=-W_{-1}(-x) \in [1,\infty]\,.$$ With this notation, we have $$r_2 =-W_{-1}(-r_1e^{-r_1}), \quad r_1 =-W(-r_2e^{-r_2})\,.$$

For $$r_1$$ close to $$1$$, we need the expansion of $$W$$ close to the branch point (see 􏰉􏰐􏰅􏰁􏰉􏰐􏰅􏰁(4.26) of this paper). We obtain $$r_2 = 1+ \epsilon + \frac{2}{3} \epsilon^2 + \frac{4}{9} \epsilon^3 + \frac{44}{135}\epsilon^4 + O(\epsilon^5) \tag{1}$$ with $$\epsilon = 1-r_1$$.

Investigating first the point $$r_1=r_2=1$$. We set $$r_1 = 1-\epsilon$$, $$r_2 = 1+\epsilon$$ (we know from (1) that $$r_1$$ and $$r_2$$ approach 1 from below and above at equal rate). To zeroth order in $$\epsilon$$, we obtain $$I(1) =\int_0^1 \frac{1}{\sqrt{x(1-x)}}\,dx = \pi \,$$ as you have already observed.

In a next step, we look at $$I(1) -I(r_1)$$ for $$r_1$$ close to $$1$$. Using (1), we expand to third order in $$\epsilon$$. We obtain $$I(1)- I(r_1) = \int_0^1\left[\frac{\left(-30 x^2+34 x-5\right) \epsilon ^2}{9 \sqrt{(1-x) x}}+\frac{2 \left(472 x^3-858 x^2+422 x-33\right) \epsilon ^3}{135 \sqrt{(1-x) x}}+\frac{2 (2 x-1) \epsilon }{3 \sqrt{(1-x) x}}\right]dx= \frac{\pi}{12} \epsilon^2 + \frac{\pi}{18} \epsilon^3+O(\epsilon^4)\,.$$

To obtain a higher order approximation, we need more terms in (1). In particular, we have $$r_2 = 1+\epsilon +\frac{2 \epsilon ^2}{3}+\frac{4 \epsilon ^3}{9}+\frac{44 \epsilon ^4}{135}+\frac{104 \epsilon ^5}{405}+\frac{40 \epsilon ^6}{189}+\frac{7648 \epsilon ^7}{42525}+\frac{2848 \epsilon ^8}{18225}+\frac{31712 \epsilon ^9}{229635}+\frac{23429344 \epsilon ^{10}}{189448875} +O(\epsilon^{11})\,.$$

Now, the expansion of the integral in $$\epsilon$$ yields $$I(r_1) = \pi -\frac{\pi \epsilon ^2}{12}-\frac{\pi \epsilon ^3}{18}-\frac{23 \pi \epsilon ^4}{576}-\frac{67 \pi \epsilon ^5}{2160}-\frac{7613 \pi \epsilon ^6}{311040}-\frac{21419 \pi \epsilon ^7}{1088640}-\frac{320153 \pi \epsilon ^8}{19906560}-\frac{31342051 \pi \epsilon ^9}{2351462400} + O(\epsilon^{10})\,.$$

• Thanks, Fabian! I have verified your derivation in the problem description. I really appreciate your inspirational solution. – Shengkai Li Dec 23 '18 at 4:49
• All right. You have proved $\lim\limits_{r_1\to r_2} = \pi.$ $(+1),$ – Yuri Negometyanov Dec 23 '18 at 8:27

Firstly, the integral $$I = \int\limits_{r_1}^{r_2}\dfrac{e^r\,\mathrm dr}{r^2\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}}\tag1$$ exists iff $$r_2\le 1,$$ because the function $$\dfrac1r e^r$$ has minimum at $$r=1.$$

Taking in account that $$\mathrm d\left(\dfrac{e^r}r\right)=\left(\dfrac1r - \dfrac1{r^2}\right)e^r\,\mathrm dr,$$ one can get $$I = \int\limits_{r_1}^{r_2}\dfrac{e^r\,\mathrm dr}{r^2\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}} = \int\limits_{r_1}^{r_2}\dfrac{e^r\,\mathrm dr}{r\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}} - \int\limits_{r_1}^{r_2}\dfrac1{\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}}d\left(\dfrac {e^r}r\right)\,\mathrm dr$$ $$=\int\limits_{r_1}^{r_2}\dfrac{e^r\,\mathrm dr}{r\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}} - \mathrm{arcsin}\left(\dfrac {r_1}{r}e^{r-r_1}\right)\Big|_{r_1}^{r_2}= I_1 + \mathrm{arccos}\left(\dfrac{r_1}{r_2}e^{r_2-r_1}\right),\tag1$$ where $$I_1 = \int\limits_{r_1}^{r_2}\dfrac{e^r\,\mathrm dr}{r\sqrt{\left(\dfrac{e^{r_1}}{r_1}\right)^2-\left(\dfrac{e^r}r\right)^2}}.\tag2$$ Note that $$I_1 \le I,$$ because $$r_2 \le1.$$

I cannot obtain the closed form for $$(2).$$

On the other hand, using Taylor series at $$x=1$$ in the form of $$\dfrac {e^x} {x\sqrt{\dfrac{e^{2a}}{a^2}-\dfrac{e^{2x}}{x^2}}} = \dfrac e{\sqrt{\dfrac{e^{2a}}{a^2}-e^2}} - \dfrac{e^{2a+1}(x-1)^2}{2\left(e^2 a^2-e^{2a}\right) \sqrt{\dfrac{e^{2a}}{a^2}-e^2}} + \dfrac{e^{2a+1}(x-1)^3}{3\left(e^2 a^2-e^{2a}\right) \sqrt{\dfrac{e^{2a}}{a^2}-e^2}}$$ $$+ \dfrac{3e^{4a+1}(x-1)^4}{8\left(e^{2a}-e^2 a^2\right)^2 \sqrt{\dfrac{e^{2a}}{a^2}-e^2}} + \dfrac{\left(-4e^{2a+3}a^2-11e^{4a+1}\right)(x-1)^5}{30\left(e^{2a}-e^2a^2\right)^2\sqrt{\dfrac{e^{2a}}{a^2}-e^2}} + \dots$$ (see also Wolfram Alpha), one can get the estimation

$$I_1 = \dfrac1{\sqrt{\dfrac{e^{2r_1-2}}{r_1^2}-1}} \int_{r_1}^{r_2}\Bigg(1 - \dfrac{e^{2r_1}(r-1)^2}{2\left(e^2 r_1^2-e^{2r_1}\right)} + \dfrac{e^{2r_1}(r-1)^3}{3\left(e^2 r_1^2-e^{2r_1}\right)}$$ $$+ \dfrac{3e^{4r_1}(r-1)^4}{8\left(e^{2r_1}-e^2 r_1^2\right)^2} + \dfrac{\left(-4e^{2r_1+3}r_1^2-11e^{4r_1}\right)(r-1)^5}{30\left(e^{2r_1}-e^2r_1^2\right)^2} + \dots\Bigg)\,\mathrm dr,$$

$$I_1 = \dfrac {1}{\sqrt{\dfrac{e^{2r_1-2}}{r_1^2}-1}}\Bigg((r_2-r_1) - \dfrac{e^{2r_1}\left((r_2-1)^3-(r_1-1)^3\right)}{6\left(e^2 r_1^2-e^{2r_1}\right)} + \dfrac{e^{2r_1+1}\left((r_2-1)^4-(r_1-1)^4\right)}{12\left(e^2 r_1^2-e^{2r_1}\right) }$$ $$+ \dfrac{3e^{4r_1+1}\left((r_2-1)^5-(r_1-1)^5\right)}{40\left(e^{2r_1}-e^2 r_1^2\right)^2} + \dfrac{\left(-4e^{2r_1+3}r_1^2-11e^{4r_1+1}\right)\left((r_2-1)^6-(r_1-1)^6\right)}{180\left(e^{2r_1}-e^2r_1^2\right)^2} + \dots\Bigg)$$

• Thanks, Yuri. I appreciate your time on this problem! – Shengkai Li Dec 23 '18 at 4:51