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How does $$\beta=\theta-\int\frac{\alpha }{r^2\sqrt {2mE+\frac{2me^2z}{r}-\frac{\alpha^2}{r^2}}}dr\\\implies \frac{1}{r}=\frac{me^2z}{\alpha^2}-\sqrt{\frac{m^2e^4z^2}{\alpha^2}+\frac{2mE}{\alpha^2}}\cos(\theta-\beta-\pi/2)$$ ??

This implication occured in Kepler's problem solution by Hamilton-Jacobi method, i've tried hard to prove this implication but unable to reach to correct conclusion. Here's the original problem-enter image description here

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  • This is indeed difficult. I was looking at some notes my professor wrote at the time but unfortunately he did not calculate this integral... It is written in portuguese but perhaps you could get some ideas. link:https://www.dropbox.com/s/4tff34i03rh3ag6/integrais%20complexos.PDF?dl=0 – aadcg Dec 06 '17 at 12:18
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    If I'm not mistaken, your goal is to calculate $\int\frac{dx}{\sqrt{-C+2Bx-Ax^{2}}}$ where I made the following change of variable: $x=1/r$ and called $A=\alpha ^2$, $B=me^2z$ and $C=-2mE$. Perhaps make a new question on how to calculate this integral. – aadcg Dec 06 '17 at 12:23
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    From what I understand from my notes the strategy is as follows: Let $r_1$ and $r_2$ be the roots of the polynomial $-Ax^2+2Bx-C$. We want to express the change in $x$ around the average $(r_1 + r_2 )/2$ with maximum "deviation" given by $(r_1 - r_2 )/2$. Hence $x=(r_1 + r_2) /2 +( r_1 - r_2 )/2 \sin \phi$. I don't understand why this is clever and I can't develop from here... – aadcg Dec 06 '17 at 12:33
  • Connected : (https://math.stackexchange.com/q/182229) – Jean Marie Dec 12 '17 at 06:38

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