To simplify the logarithmic expression and possibly find its inverse.

Question

While solving an equation, I came across an unexpectedly symmetric solution.

$$\frac{\log \left(\frac{a r+h}{\sqrt{(a r+h)^2-1}}+1\right)}{2 a}-\frac{\log \left(1-\frac{a r+h}{\sqrt{(a r+h)^2-1}}\right)}{2 a}$$

Now, I know that the 1st term of the expression corresponds to the hyperbolic cosine through the logarithmic transformation but the first term isn't so. Now, I want to ask if you know whether the 2nd term also transforms into something clever like that. Since I want to find the inverse of this solution, I think a short expression would make it beautiful to do so(?)

Thanks.

Answer 1

If we denote the numerator by $A$ and the denominator by $B > 0$, we have:
$\frac{1}{2a}(\log(\frac{A}{B} + 1) - \log(1 - \frac{A}{B})) = \frac{1}{2a}(\log(\frac{A + B}{B}) - \log(\frac{B - A}{B})) = \frac{1}{2a}(\log(A + B) - \log(B) - \log(B - A) + \log(B)) = \frac{1}{2a}\log(\frac{A + B}{B - A})$.
Now, you can simplify a bit :)

Edit: removed the conjugate quantity

Answer 2

Using $x=ar+h$ $$y=\log \left(1+\frac{x}{\sqrt{x^2-1}}\right)-\log\left(1-\frac{x}{\sqrt{x^2-1}}\right)$$ Make $x=\cosh(t)$ $$y=2 \tanh ^{-1}\left(\frac{\cosh (t)}{\sqrt{\sinh ^2(t)}}\right)=2 \tanh ^{-1}\left(\frac{x}{\sqrt{x^2-1}}\right)$$

To check it, using Taylor series, the first and last expressions give $$y=-2i x\left(1+\frac{x^2}{6}+\frac{3 x^4}{40}+O\left(x^6\right) \right)$$