I'm studying Landau, Lifshitz - Mechanics. Could someone help me with this problem ? =)

Problem $\S12$ $2(b)$ (Page 27 3rd Edition) Determine the period of oscillation, as a function of the energy, when a particle of mass $m$ moves in fields for which the potential energy is

$(b) \quad U=-U_{0}/\cosh^{2}\alpha x \quad -U_0<E<0.$

Attempt at solution:

The period is given by

$$ T=4 \sqrt{\frac{m}{2}}\int\frac{dx}{\sqrt{E+\frac{U_{0}}{\cosh^{2}\alpha x}}}.$$

How can I evaluate this integral? I know that the answer is $$T=(\pi/\alpha)\sqrt{2m/|E|}.$$

  • $\begingroup$ Welcome to Physics! Please see our homework policy. We expect homework problems to have some effort put into them, and deal with conceptual issues. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with @Manishearth to notify me) $\endgroup$ Commented Jan 26, 2013 at 8:26
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    $\begingroup$ @Manishearth your motivation to close this question it is not fully true 1)"This question is unlikely to help any future visitors" people have favorite the question and, for example, I'm a future visitor and this is a very important question given that I'm studying fully the Landau course as research 2)"it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation (...)" Landau course is widely known and recommended with very interesting homework-questions for every physicist, and geographically I'm in brazil enjoying the answer below $\endgroup$
    – user78217
    Commented Feb 15, 2017 at 15:44
  • $\begingroup$ @Manishearth Possibly, after trying many substitutions and still not getting the right integral (my own experience), do you really want the OP to dump the attempts here? ;) $\endgroup$
    – Cheng
    Commented Sep 11, 2022 at 5:37

2 Answers 2



  1. From energy conservation $$\frac{1}{2}m\dot{x}^2 ~=~ E-U,$$ with even/symmetric potential $$U ~=~ - \frac{U_0}{c^2},\qquad c~:=~\cosh(\alpha x), \qquad 0<-E<U_0,$$ one gets the quarter period $$\frac{T}{4}~=~ \sqrt{\frac{m}{2}} \int_0^{x_1}\frac{dx}{\sqrt{E-U}},$$ where $x_1>0$ is the upper turning point determined by the condition $E=U(x_1)$.

  2. Prove that $$\int_0^{x_1}\frac{dx}{\sqrt{E-U}}~=~\frac{1}{\alpha\sqrt{|E|}}\int_0^{\sqrt{a}}\frac{ds}{\sqrt{(a-s^2)}}, $$ where $$ s~:=~\sinh(\alpha x),\qquad ~a:=~ -U_0/E-1~>~0.$$

  3. Show that the last integral does not depend on $a>0$. Choose $a=1$, and perform a substitution $s=\sin(t)$.

  • $\begingroup$ For 3. Use Leibniz integral rule $\endgroup$
    – Cheng
    Commented Sep 11, 2022 at 6:41

Some idea to evaluate the integral: $$\int\frac{dx}{\sqrt{E+\frac{U_{0}}{\cosh^{2}\alpha x}}}=\int \cosh\alpha x\frac{dx}{\sqrt{E \cosh^{2}\alpha x+U_{0}}} = \int \frac{d(\sinh\alpha x)/\alpha}{\sqrt{(E+U_0) + E \sinh^{2}\alpha x }}$$ The last equality use the identity $\cosh^2 x - \sinh^2 x = 1$.

Using $y=\sinh \alpha x$ and factoring $|E|$ out of the square root (because $E<0$), you should then be able see it as a standard integral of the form $\int dy/\sqrt{a^2-y^2} = \sin^{-1}(y/a)$


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