I'm trying to resolve the elliptic integral that appears in the Kepler problem, i.e., \begin{align} I=\int \frac{dx}{\sqrt{\alpha +\beta x+\gamma x^2}}. \end{align} Goldstein says the the integral is \begin{align} \int \frac{dx}{\sqrt{\alpha +\beta x+\gamma x^2}}=\frac{1}{\sqrt{-\gamma}}\arccos \left(-\frac{\beta+2\gamma x}{\sqrt{\beta^2-4\alpha\gamma}}\right). \end{align} I don't know how to resolve the integral.

My attempt: \begin{align} \int \frac{dx}{\sqrt{\alpha +\beta x+\gamma x^2}}=\int \frac{dx}{\sqrt{\alpha-\frac{\beta^2}{4\gamma}+\gamma\left(x+\frac{\beta}{2\gamma}\right)^2}} \end{align} Changing variables like $x=\frac{\beta}{2\gamma}\tan^2 \theta\rightarrow dx=\frac{\beta}{2\gamma}2\tan\theta \sec\theta\; d\theta$, \begin{align} \therefore I=\int\frac{\frac{\beta}{2\gamma}2\tan\theta\sec\theta\;d\theta}{\sqrt{\alpha-\frac{\beta^2}{4\gamma^2}+\gamma\left[\frac{\beta}{2\gamma}(\tan^2\theta+1)\right]^2}}=\int\frac{\frac{\beta}{\gamma}\tan\theta\sec\theta}{\sqrt{\alpha+\sec^4\theta}}\;d\theta=\frac{\beta}{\gamma}\int\frac{\sin\theta}{\sqrt{\alpha \cos^4\theta+1}}\;d\theta. \end{align}



\begin{align} I&=\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=\int\frac{dx}{\sqrt{\alpha+\gamma\left(\frac{\beta}{\gamma}x+x^2\right)}}=\int\frac{dx}{\sqrt{\alpha+\gamma\left(x+\frac{\beta}{2\gamma}\right)^2-\frac{\beta^2}{4\gamma}}} \end{align} If $y=x+\frac{\beta}{2\gamma}$: \begin{align} I&=\int\frac{dy}{\sqrt{\alpha+\gamma y^2-\frac{\beta^2}{4\gamma}}}=\int\frac{dy}{\sqrt{\frac{4\alpha\gamma-\beta^2}{4\gamma}+\gamma y^2}} \end{align} If $q=\beta^2-4\alpha\gamma$, \begin{align} I=\int\frac{dy}{\sqrt{\gamma y^2-\frac{q}{4\gamma}}} \end{align} Then $y=-\frac{\sqrt{q}}{2\gamma}\cos\theta$ and $dy=\frac{\sqrt{q}}{2\gamma}\sin\theta\;d\theta$, \begin{align} \therefore I=\frac{\sqrt{q}}{2(-\gamma)}\int\frac{-\sin\theta\;d\theta}{\sqrt{\frac{q}{4\gamma}(\cos^2\theta-1)}}=\frac{\sqrt{q}}{2(-\gamma)}\int\frac{-\sin\theta\;d\theta}{\sqrt{\frac{q}{4(-\gamma)}(1-\cos^2\theta)}}=-\frac{1}{\sqrt{-\gamma}}\;\theta\\ \boxed{\therefore \int \frac{dx}{\sqrt{\alpha +\beta x+\gamma x^2}}=-\frac{1}{\sqrt{-\gamma}}\arccos\left(-\frac{\beta+2\gamma x}{\sqrt{\beta^2-4\alpha\gamma}}\right)} \end{align}


migrated from physics.stackexchange.com Mar 7 '18 at 4:26

This question came from our site for active researchers, academics and students of physics.

  • 1
    $\begingroup$ After completing the square, first do an affine change of variables $y=ax+b$ so that the thing in the square root becomes $c(1+y^2)$ for some constant $c$. Then set $y=\cos(\theta)$. You will find that the resulting integrand is just a constant. $\endgroup$ – Yly Mar 6 '18 at 22:50
  • 1
    $\begingroup$ I think this is an issue of pure mathematics. Methods of solution are found in standard calculus books. The context is irrelevant : replacing Kepler with Nash or Darwin makes absolutely no difference to the question. No physics is applied in the method of solution. I think this belongs on Mathematics SE. $\endgroup$ – sammy gerbil Mar 7 '18 at 1:36
  • $\begingroup$ @Yly, I obtained the result, but I'm missing something in the denominator: $\frac{1}{\sqrt{-\gamma}}$. After I completed squares, I think there's something about the discrminant $\beta^2-4\alpha\gamma$ that I cannot see that avoids me obtaining the full result. $\endgroup$ – hyriusen Mar 8 '18 at 22:36
  • $\begingroup$ @hyriusen I suggest you edit your work into the question. The $1/\sqrt{-\gamma}$ should come from the first linear change of variables. $\endgroup$ – Yly Mar 8 '18 at 22:41
  • $\begingroup$ @hyriusen Also, I made a slight typo in the first comment: you want the thing in the square root to be $c(1-y^2)$. $\endgroup$ – Yly Mar 8 '18 at 22:42

Your Answer

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

Browse other questions tagged or ask your own question.