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 $\Phi(r_1,r_2)=1$, $r_1<r_2$.

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: The interval of integral

$\uparrow$ The interval of the integral and the integrand

integral as a function of r2

$\uparrow$ 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.


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 $\Phi(r_1,r_2)=1$, $r_1<r_2$.

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 $I=\pi-\frac{\pi}{12}\epsilon^2+O(\epsilon^3)$: secondOrder

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

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

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.