Relationship between arctanh and arctan or there is a mistake

1. May 14, 2013

Dustinsfl

$$\int_0^{\nu}\frac{d\nu'}{(1 + e\cos(\nu'))^2}$$
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. May 15, 2013

Dick

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