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While solving a physical problem from Landau, Lifshitz "Mechanics" book, I came across an integral:

$$\int_0^\delta \frac{du}{\sqrt{\left(\frac{\cosh\delta}{\cosh u}\right)^2-1}}.$$

In the book only the final answer for the problem is given, from which I deduce that this integral must be $\frac\pi2$.

I've tried feeding it to Wolfram Mathematica, but it wasn't able to evaluate it, returning unevaluated result. Evaluating it numerically confirms that this is a likely answer, but I haven't been able to prove this.

I've tried making a substitution $v=\frac{\cosh\delta}{\cosh u}$ and got this integral instead:

$$\gamma \int_1^\gamma \frac{dv}{v\sqrt{(\gamma^2-v^2)(v^2-1)}},$$

where $\gamma=\cosh\delta$, but still this doesn't give me a clue how to proceed. Also, I can't seem to eliminate the parameter ($\delta$ or $\gamma$), which shouldn't affect the result at all.

So, the question is: how can one evaluate this integral or at least prove that it's equal $\frac\pi2$?

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  • $\begingroup$ by inspection.. $\endgroup$ – user111072 Dec 21 '13 at 16:49
  • $\begingroup$ Notice that $\delta$ is a constant. What does this tell you about the integral? HINT Trigonometric substitution with sech. $\endgroup$ – Don Larynx Dec 21 '13 at 16:52
  • $\begingroup$ @DonLarynx I don't think I follow. From constness of $\delta$ I can take it out of integral leaving $\left(\cosh^{-2} u-\cosh^{-2} \delta\right)^{-1/2}$ under integral. If I then substitute $u=\text{arcosh} v$, I get $\left(\sqrt{(v^{-2}-\text{sech}^2 \delta)(v^2-1)}\right)$ under integral. But I don't know how to move on from this. Do I do what you meant? $\endgroup$ – Ruslan Dec 21 '13 at 17:33
  • $\begingroup$ Try expanding $\cosh x = (e^x+e^{-x})/2$ in both occurrences, and then making the change of variables $u = \frac12\log t$. That should change the integrand into a rational function of $t$ divided by the square root of a quadratic in $t$, on which you can do a trig substitution. $\endgroup$ – Greg Martin Dec 21 '13 at 18:03
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It's actually a lot simpler than this. Rewrite the integral as

$$\int_0^{\delta} du \frac{\cosh{u}}{\sqrt{\cosh^2{\delta}-\cosh^2{u}}} = \int_0^{\delta} du \frac{\cosh{u}}{\sqrt{\sinh^2{\delta}-\sinh^2{u}}}$$

Sub $y=\sinh{u}$ and the integral becomes

$$\int_0^{\sinh{\delta}} \frac{dy}{\sqrt{\sinh^2{\delta}-y^2}}$$

Now sub $y=\sinh{\delta}\, \sin{t}$ and the integral is

$$\int_0^{\pi/2} dt = \frac{\pi}{2}$$

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  • $\begingroup$ Or that, of course ;) $\endgroup$ – Daniel Fischer Dec 21 '13 at 22:50
  • $\begingroup$ @DanielFischer: yeah, a bunch of ways works. I think the presence of the cosh's makes people think the integral is harder than it really is. $\endgroup$ – Ron Gordon Dec 21 '13 at 22:51
  • $\begingroup$ Yes, probably, and the square root. Still, one needs a trained eye to spot your elegant way. $\endgroup$ – Daniel Fischer Dec 21 '13 at 22:52
  • $\begingroup$ Very well done. $\endgroup$ – Raul C. de Assis Dec 22 '13 at 1:25
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Continuing from your substitution:

$$\begin{aligned}\mathcal{I} &= \int_1^{\gamma} \frac{\gamma\,dv}{v\sqrt{(\gamma^2-v^2)(v^2-1)}}\\&=\int_0^{\frac{\pi}{2}}\frac{\gamma\,\sec^2 t dt}{1 +\gamma^2\tan^2 t}\quad (v^2=\cos^2 t+\gamma^2\sin^2 t)\\&=\int_0^{\infty}\frac{dw}{1 +w^2}\quad (w=\gamma\tan t)\\&=\frac{\pi}{2}\end{aligned}$$

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