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I am reading right now a book about classical mechanics (Landau Lifschitz) which contains an integral which I do not understand

$\int_{0}^{\alpha}\left [ \frac{dx_{2}(U)}{dU}- \frac{dx_{1}(U)}{dU}\right]dU\int_{U}^{\alpha}\frac{dE}{\sqrt{(\alpha - E)(E-U)}}$

Then the book states the second integral is elementary and results in $\pi$. And the result is written as:

$\pi\left[ x_{2}(U) - x_{1}(U) \right ]$.

I am not able to solve the second part of the integral nor understand it properly. Can anybody give me a hint on how to start. I was googling already some integral tables but was not able to find it.

zodiac
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1 Answers1

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This is a $\int\frac{du}{\sqrt{a^2 - u^2}}$ form, which you can find in integral tables and which most relatively thorough elementary calculus texts discuss. For a very complete treatment of the kinds of classical integration methods that would be familiar to students of the Landau/Lifschitz era, see A Treatise on the Integral Calculus, Volume I, by Joseph Edwards.

To put your integral into this form, complete the square of $$(\alpha - E)(E-U) = -E^2 + (\alpha + U)E - \alpha U,$$ which gives you $$ -\left(E \; - \; \frac{\alpha + U}{2} \right)^2 \;\; + \;\; \frac{(\alpha + U)^2}{4} \; - \; \alpha U $$ Therefore, $$\;a \; = \; \sqrt{\frac{(\alpha + U)^2}{4} \; - \; \alpha U \;} \;\;\; \text{and} \;\;\; u = E \; - \; \frac{\alpha + U}{2}$$ or $$\;a \; = \; \sqrt{\frac{(\alpha - U)^2}{4}\;} \;\;\; \text{and} \;\;\; u = E \; - \; \frac{\alpha + U}{2}$$ or $$\;a \; = \; \frac{1}{2}\alpha \; - \; \frac{1}{2}U \;\;\; \text{and} \;\;\; u = E \; - \; \frac{1}{2}\alpha - \frac{1}{2}U$$ For the variable change $u = E - \frac{1}{2}\alpha - \frac{1}{2}U,$ we have $du = dE$ and the limits $E = U$ to $E = \alpha$ become $u = \frac{1}{2}U - \frac{1}{2}\alpha$ to $u = \frac{1}{2}\alpha - \frac{1}{2}U.$

Thus, we get $$ \int_{E=U}^{E=\alpha}\frac{dE}{\sqrt{(\alpha - E)(E-U)}} \;\; = \;\; \int_{E=U}^{E=\alpha}\frac{du}{\sqrt{a^2 - u^2}} \;\; = \;\; \int_{u = \frac{1}{2}U - \frac{1}{2}\alpha}^{u = \frac{1}{2}\alpha - \frac{1}{2}U}\frac{du}{\sqrt{a^2 - u^2}}$$ $$= \;\; \left[ \sin^{-1}\left(\frac{u}{a}\right) \right]_{\frac{1}{2}U - \frac{1}{2}\alpha}^{\frac{1}{2}\alpha - \frac{1}{2}U} \;\; = \;\; \sin^{-1}\left(\frac{\frac{1}{2}\alpha - \frac{1}{2}U}{a}\right) \; - \; \sin^{-1}\left(\frac{\frac{1}{2}U - \frac{1}{2}\alpha}{a}\right) $$ $$ = \;\; \sin^{-1}\left(\frac{\frac{1}{2}\alpha - \frac{1}{2}U}{\frac{1}{2}\alpha - \frac{1}{2}U}\right) \; - \; \sin^{-1}\left(\frac{\frac{1}{2}U - \frac{1}{2}\alpha}{\frac{1}{2}\alpha - \frac{1}{2}U}\right)$$ $$= \;\; \sin^{-1}(1) - \sin^{-1}(-1) \;\; = \;\; \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \;\; = \;\; \pi $$

Dave L. Renfro
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  • Thank you for the answer. So in your book which tells me how to solve the indefinite integral that gives the $sinh^{-1}(\frac{x}{a})$. But what still puzzles me is the lower boundary of the integral limit is depending on the first integral of the product. And $\alpha$ is an arbitrary real constant. How am I then able to give $\pi$ as the answer. What am I missing? – zodiac Nov 05 '17 at 22:04
  • I just worked out the details in calculating the 2nd integral by itself (i.e. without worrying about the 1st integral), and I got $\pi.$ The details will be added to my answer as soon as I get them "typed up". – Dave L. Renfro Nov 05 '17 at 22:31
  • Apparently, even through $U$ appears in the 2nd integral, the value of the 2nd integral is independent of $U,$ in a similar way (for example) that the value of $(U-2)^2 - U^2 + 4U$ is independent of $U.$ – Dave L. Renfro Nov 05 '17 at 23:05