Cfrgtkky

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 expansion about $r_0$. I appreciate any hint.

Thanks!

If you are interested, here is the general form of the integral:
$$I=int_{r_1}^{r_2}frac{dr}{rsqrt{Phi^2(r,r_1)-1}}$$
where
$$Phi(u,v)=frac{uexp{(k(u))}}{vexp{(k(v))}}$$
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
$$lim_{r_1rightarrow r_2} I =frac{pi}{sqrt{1+r_0 k''(r_0)/k'(r_0)}}$$

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

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

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 $rexp(-r)$ you see that it is monotonous on the interval $rin[0,1]$ and $rin[1,infty]$. 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
$$I(1)- I(r_1) = int_0^1left[frac{left(-30 x^2+34 x-5right) epsilon ^2}{9 sqrt{(1-x) x}}+frac{2 left(472
x^3-858 x^2+422 x-33right) epsilon ^3}{135 sqrt{(1-x) x}}+frac{2 (2 x-1) epsilon
}{3 sqrt{(1-x) x}}right]dx= frac{pi}{12} epsilon^2 + frac{pi}{18} epsilon^3+O(epsilon^4),.$$

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

Firstly, the integral
$$I = intlimits_{r_1}^{r_2}dfrac{e^r,mathrm dr}{r^2sqrt{left(dfrac{e^{r_1}}{r_1}right)^2-left(dfrac{e^r}rright)^2}}tag1$$
exists iff $r_2le 1,$ because the function $dfrac1r e^r$ has minimum at $r=1.$

Taking in account that
$$mathrm dleft(dfrac{e^r}rright)=left(dfrac1r - dfrac1{r^2}right)e^r,mathrm dr,$$
one can get
$$I = intlimits_{r_1}^{r_2}dfrac{e^r,mathrm dr}{r^2sqrt{left(dfrac{e^{r_1}}{r_1}right)^2-left(dfrac{e^r}rright)^2}} = intlimits_{r_1}^{r_2}dfrac{e^r,mathrm dr}{rsqrt{left(dfrac{e^{r_1}}{r_1}right)^2-left(dfrac{e^r}rright)^2}} - intlimits_{r_1}^{r_2}dfrac1{sqrt{left(dfrac{e^{r_1}}{r_1}right)^2-left(dfrac{e^r}rright)^2}}dleft(dfrac {e^r}rright),mathrm dr$$
$$=intlimits_{r_1}^{r_2}dfrac{e^r,mathrm dr}{rsqrt{left(dfrac{e^{r_1}}{r_1}right)^2-left(dfrac{e^r}rright)^2}} - mathrm{arcsin}left(dfrac {r_1}{r}e^{r-r_1}right)Big|_{r_1}^{r_2}= I_1 + mathrm{arccos}left(dfrac{r_1}{r_2}e^{r_2-r_1}right),tag1$$
where
$$I_1 = intlimits_{r_1}^{r_2}dfrac{e^r,mathrm dr}{rsqrt{left(dfrac{e^{r_1}}{r_1}right)^2-left(dfrac{e^r}rright)^2}}.tag2$$
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
$$dfrac {e^x} {xsqrt{dfrac{e^{2a}}{a^2}-dfrac{e^{2x}}{x^2}}}
= dfrac e{sqrt{dfrac{e^{2a}}{a^2}-e^2}}
- dfrac{e^{2a+1}(x-1)^2}{2left(e^2 a^2-e^{2a}right) sqrt{dfrac{e^{2a}}{a^2}-e^2}}
+ dfrac{e^{2a+1}(x-1)^3}{3left(e^2 a^2-e^{2a}right) sqrt{dfrac{e^{2a}}{a^2}-e^2}}$$
$$ + dfrac{3e^{4a+1}(x-1)^4}{8left(e^{2a}-e^2 a^2right)^2 sqrt{dfrac{e^{2a}}{a^2}-e^2}}
+ dfrac{left(-4e^{2a+3}a^2-11e^{4a+1}right)(x-1)^5}{30left(e^{2a}-e^2a^2right)^2sqrt{dfrac{e^{2a}}{a^2}-e^2}} + dots$$
(see also Wolfram Alpha), one can get the estimation

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

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