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I don't understand why the following is true, explanation would be greatly appreciated! Suppose $$E(x,y)=\gamma x^{2n}+{y^2\over a}$$ where $\gamma >0, a>0,n\in \mathbb Z^+$.

And we define $$\alpha = {1\over 2\pi}\oint y\,\,\,dx$$ where the path is where $E, \gamma$ are constants.

Why then does it follow that $$a^{n\over n+1}E=\gamma^{1\over n+1}\left({n\pi\alpha\over f}\right)^{2n\over n+1}$$ where $f=\int_0^1(1-u)^{1\over 2}u^{1-2n\over 2n}\,\,du$?


I first tried making $y$ the subject then substitute its expression in terms of $E$ and $x$ into the integral of $\alpha$. Here I can take $E, \gamma$ to be constants, since they are so on the path. This gives $$\alpha = {\sqrt a\over 2\pi}\oint (E-\gamma x^{2n})^{1\over 2}dx$$. But what next? Or perhaps there is a different approach to begin with?

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Welcome to Math.SE! This is a strange calculations indeed; it makes one wonder in what context you encountered it. Can you provide a reference? –  user53153 Jan 4 '13 at 3:14
    
@PavelM : Thank you for the welcome! It is from applied mathematics where $\alpha$ is usually called the action variable and $E$ is the Energy/Hamiltonian. $y$ is the conjugate momentum of $x$. –  user55116 Jan 4 '13 at 3:18

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One possible approach is to use Green's formula, but I think you are already on the right track. In the integral $\oint (E-\gamma x^{2n})^{1\over 2}dx$ the square root should be treated with care, since it's negative half of the way. (The path is oval-shaped, symmetric with respect to both axes). Let's focus on the upper half of the path: the lower half will contribute exactly the same amount, because both $y$ and $dx$ will have opposite sign there. Thus, the upper half of integral is $\pi \alpha/\sqrt{a}$ in your notation

The variable $x$ ranges between $\pm (E/\gamma)^{1/(2n)}$, but by symmetry we can consider $$2\int_0^{(E/\gamma)^{1/(2n)}} (E-\gamma x^{2n})^{1/2}\,dx$$ instead. Now introduce the substitution $u=(\gamma/E)x^{2n}$. Since $x=(E/\gamma)^{1/(2n)}u^{1/(2n)}$, we have $dx=(E/\gamma)^{1/(2n)}\frac{1}{2n}u^{\frac{1-2n}{2n}}du$. Thus, the integral becomes $$2\int_0^1 E^{1/2}(1-u)^{1/2}(E/\gamma)^{1/(2n)}\frac{1}{2n}u^{\frac{1-2n}{2n}}\,du = \frac{E^{\frac{n+1}{2n}}}{n\,\gamma^{\frac{1}{2n}}}f $$ We have arrived at $$\frac{\pi\alpha}{\sqrt{a}} = \frac{E^{\frac{n+1}{2n}}}{n\,\gamma^{\frac{1}{2n}}}f$$ which, if there is any justice in the world, should be equivalent to your equation.

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Indeed! Thank you so much, Pavel! –  user55116 Jan 4 '13 at 4:06

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