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I am working through the textbook "Mechanics", from the series "Course of Theoretical Physics " by Landau and Lifshitz. In Chapter 3, where the authors talk about integrating the equation of motion $E=\frac{1}{2}m\dot{x}^2+U(x)$, yielding the period T: $$T=2\sqrt{\frac{m}{2}}\int_{x_1}^{x_2}\frac{dx}{\sqrt{E-U(x)}}$$ Where $x_1$ and $x_2$ are the points at which $E=U(x)$, or the endpoints of the motion, and we are integrating only over a half period, so we need a factor of 2 in front of the integral. Problem 2, part (b) asks us to calculate the period $T$ for a particle subject to a potential $U=-U_0/\cosh^2(\alpha x)$ and $E\in(-U_0, 0)$. I then obtained the following integral $$T=2\sqrt{\frac{2m}{U_0}}\int_0^{x_0} \frac{\cosh(\alpha x)}{\sqrt{1-\frac{\cosh^2(\alpha x)}{\cosh^2(\alpha x_0)}}}\,dx$$ Where $x_0$ denotes the point of maximum amplitude of the motion, where $E=-U_0/\cosh^2(\alpha x_0)$, and instead of integrating over a half period, I integrated over a quarter period and multiplied the result by 2 again.

  1. Are my bounds correct? I am not sure how to make use of the condition $E\in(-U_0, 0)$.
  2. If so, how you explicitly calculate this integral? I have tried using the Binomial series expansion, only to end up with a Hypergeometric series, as well as different $u$-substitution.
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    $\begingroup$ The integral is calculated exactly, make a replacement $y=\sinh(\alpha x), dy=\alpha \cosh(\alpha x)dx$ and use identity $\cosh^2(\alpha x)=1+\sinh^2(\alpha x)=1+y^2$. $\endgroup$ Jun 5, 2019 at 16:52
  • $\begingroup$ Thank you for the comment. I've managed to solve the integral. $\endgroup$
    – Zachary
    Jun 5, 2019 at 18:14
  • $\begingroup$ Did you manage to calculate the period? $\endgroup$ Jun 5, 2019 at 18:17
  • $\begingroup$ Yes, I ended up getting $T=\frac{\pi}{\alpha}\sqrt{\frac{2m}{-E}}$ where $E=-U_0/\cosh^2(\alpha x_0)<0$ and $x_0$ is the amplitude of the motion. I verified my solution with the textbook by Landau and Lifshitz. $\endgroup$
    – Zachary
    Jun 5, 2019 at 18:29

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