An integral related to the theory of isotropic harmonic oscillator

Question

The classical Hamiltonian for an isotropic harmonic oscillator in polar coordinates is \begin{equation}\tag{1} H = \frac{1}{2m}\left(p_\rho^2 + \frac{p_\varphi^2}{\rho^2}\right) + \frac{1}{2}m\omega^2\rho^2, \end{equation} where $m$ is the mass and $\omega$ is the angular frequency. The action variable is \begin{equation}\tag{2} J_\rho = \frac{1}{\pi}\int_{\rho_1}^{\rho_2}d\rho\sqrt{2mE - \frac{\beta_\varphi^2}{\rho^2} - (m\omega)^2\rho^2}, \end{equation} where $\rho_{1,2}$ are the turning points of the oscillator and $\beta_\varphi$ is the constant action $J_\varphi$. They are the positive roots of the equation \begin{equation}\tag{3} 2mE - \frac{\beta_\varphi^2}{\rho^2} - (m\omega^2)^2\rho^2 = 0. \end{equation} How to show that the integral evaluates to \begin{equation}\tag{4} \int_{\rho_1}^{\rho_2}d\rho\sqrt{a - \frac{b^2}{\rho^2} - c^2\rho^2} = \frac{\pi}{2}\left(\frac{a}{2c} - |b|\right)? \end{equation}

This equation is reported in H. Iro's book A Modern Approach to Classical Mechanics, (2016).

Answer 1

I obtain this result ?

I )

$$J_r=\frac 1\pi\int \sqrt{F}\,dr\quad, F=-{\frac {{P_{{\varphi }}}^{2}}{{r}^{2}}}-{r}^{2}{c}^{2}+a$$

II)

$$ \text{result}=\left(J\right)_r\bigg|_{r_l}^{r_h}={\rm signum} \left( {\frac {\sqrt {2\,a-2\,\sqrt {{a}^{2 }-4\,{c}^{2}{P_{{\varphi }}}^{2}}}}{c}} \right) \,P_{{\varphi }}=\pm P_{{\varphi }} =\text{constant}$$

where $~r_l~,~r_h~$ are the negative and positive solutions of $~F(r)=0$

$$r_l = \sqrt{\frac{a - \sqrt{a^2 - 4P_\varphi^2c^2}}{2c^2}} \\ r_h = \sqrt{\frac{a + \sqrt{a^2 - 4P_\varphi^2c^2}}{2c^2}} $$

Remark

It is not Elliptic Integral. I obtained the result with MAPLE program

Maple result

$$\begin{eqnarray}\pi\,J_r &=&\frac{1}{2}\,\sqrt {-{P_{{\varphi }}}^{2}-{r}^{4}{c}^{2}+a{r}^{2}} - \\ & & \frac{1}{4c}\,a\tan^{-1} \left( -{\frac {{r}^{2}c}{\sqrt {-{P_{{\varphi }}}^{2}-{r}^{4} {c}^{2}+a{r}^{2}}}}+\frac{1}{2}\,{\frac {a}{c\sqrt {-{P_{{\varphi }}}^{2}-{r}^ {4}{c}^{2}+a{r}^{2}}}} \right) + \\ & & \frac{1}{4}\,P_{{\varphi }}\pi + iP_{{\varphi }}\ln(r) - \\ & & \frac{1}{2}\,i P_{{\varphi }}\ln \left( 2\,i{P_{{\varphi }}}^{2}-ia{r}^{2}+2\,P_{{ \varphi }}\sqrt {-{P_{{\varphi }}}^{2}-{r}^{4}{c}^{2}+a{r}^{2}} \right) & & \end{eqnarray} $$

hence $$\pi\,\frac {\partial\,J_r}{\partial r}= {\frac {\sqrt {-{P_{{\varphi }}}^{2}-{r}^{4}{c}^{2}+a{r}^{2}}}{r}}=\sqrt{F}$$

Answer 2

This is not yet and answer but a continuation of @Eli's analysis.

At the turning points, $-P_\varphi^2 - r^4c^2 + ar^2 = 0$. The turning points themselves are \begin{equation} r^2 = \frac{a \pm \sqrt{a^2 - 4P_\varphi^2c^2}}{2c^2} \end{equation}
so that \begin{equation} 2r^2c^2 - a = \pm\sqrt{a^2 - 4P_\varphi^2c^2}. \end{equation} Therefore, at the turning point $r_h$, the argument of $\tan^{-1}$ is $-\infty$ while that of $r_l$ is $+\infty$. Thus, \begin{eqnarray*} \pi J_r(r_h) &=& \frac{a}{4}\frac{\pi}{2} + \frac{P_\varphi\pi}{4} + iP_\varphi\ln r_h - \frac{iP_\varphi}{2}\ln(i(2P_\varphi^2 - ar_h^2)) \\ &=& \frac{a\pi}{8c} + \frac{P_\varphi\pi}{4} + iP_\varphi\ln r_h - \frac{iP_\varphi}{2}\ln(e^{i\pi/2}(2P_\varphi^2 - ar_h^2)) \\ &=& \frac{a\pi}{8c} + \frac{P_\varphi\pi}{4} + iP_\varphi\ln r_h + \frac{P_\varphi\pi}{4} - \frac{iP_\varphi}{2}\ln(2P_\varphi^2 - ar_h^2) \\ &=& \frac{a\pi}{8c} + \frac{P_\varphi\pi}{2} + \frac{iP_\varphi}{2}\ln r_h^2 - \frac{iP_\varphi}{2}\ln(2P_\varphi^2 - ar_h^2) \\ &=& \frac{a\pi}{8c} + \frac{P_\varphi\pi}{2} + \frac{iP_\varphi}{2}\ln\frac{r_h^2}{2P_\varphi^2 - ar_h^2} \end{eqnarray*} Similarly, \begin{equation} \pi J_r(r_l) = -\frac{a\pi}{8c} + \frac{P_\varphi\pi}{2} + \frac{iP_\varphi}{2}\ln\frac{r_l^2}{2P_\varphi^2 - ar_l^2} \end{equation}
Therefore, \begin{equation} \pi J_r(r_h) - \pi J_r(r_l) = \frac{a\pi}{4c} + \frac{iP_\varphi}{2}\ln\frac{r_h^2(2P_\varphi^2 - ar_l^2)} {r_l^2(2P_\varphi^2 - ar_h^2)} \end{equation}

We are done if we show that the second expression on RHS is equal to \begin{equation} -\frac{\pi}{2}|P_\varphi|. \end{equation}