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Relationship between arctanh and arctan or there is a mistake

  1. May 14, 2013 #1
    $$
    \int_0^{\nu}\frac{d\nu'}{(1 + e\cos(\nu'))^2}
    $$
    Consider
    \begin{align}
    \frac{d}{d\nu'}\frac{\sin(\nu')}{1 + e\cos(\nu')} = \frac{\cos(\nu') + e}{(1 + e\cos(\nu'))^2}\\
    \frac{d}{d\nu'}\frac{e\sin(\nu')}{1 + e\cos(\nu')} = \frac{1}{1 + e\cos(\nu')} + \frac{e^2 - 1}{(1 + e\cos(\nu'))^2}
    \end{align}
    We can now isolate a little bit easier integral.
    $$
    \int_0^{\nu}\frac{d\nu'}{(1 + e\cos(\nu'))^2}
    = \frac{e}{e^2 - 1}\frac{\sin(\nu)}{1 + e\cos(\nu)}
    - \frac{1}{e^2 - 1}\int_0^{\nu}\frac{d\nu'}{1 + e\cos(\nu')}
    $$
    After integrating, we end up with
    $$
    \int_0^{\nu}\frac{d\nu'}{(1 + e\cos(\nu'))^2}
    = \frac{e}{e^2 - 1}\frac{\sin(\nu)}{1 + e\cos(\nu)}
    - \frac{2}{(e^2 - 1)^{3/2}}\tanh^{-1}\left[\sqrt{\frac{1 - e}{1 + e}}\tan\left(\frac{\nu}{2}\right)\right] = \frac{\mu^2}{h^3}t.
    $$


    The solution is
    $$
    \frac{\mu^2}{h^3}t = \frac{1}{(1 - e^2)^{3/2}}\left[2\arctan\left[\sqrt{\frac{1 - e}{1 + e}}\tan\frac{\nu}{2}\right] - \frac{e\sqrt{1 - e^2}\sin\nu}{1 + e\cos\nu}\right]
    $$
    The second term will work when simplified but I have a arctanh.

    What went wrong or is there a sligh trick?
     
    Last edited: May 14, 2013
  2. jcsd
  3. May 15, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    arctan(iz)=i*arctanh(z). The difference between the two ways of writing it is probably whether you take e>1 or e<1, since you have things like sqrt(1-e) floating around, which could be imaginary or not.
     
    Last edited: May 15, 2013
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