Taking limit with hyperbolic functions

Question

I have a problem with evaluating $$\sinh^{-1}(C \sinh (ax))\bigg|_{-\infty}^{+\infty}$$ where $C$ and $a$ are real positive constants.

Answer 1

You can solve for $$ y= \sinh^{-1}(C \sinh (ax))$$ explicitly: Let $z = e^y$, then $$ \frac{1}{2}\left( z + \frac{1}{z} \right) = \frac{C}{2} \left( e^{ax} - e^{-ax} \right) $$ which yields a quadratic equation the positive solution of which (after all, $z$ is $e^y$ for $y$ real) is $$z = C\frac{e^{ax} - e^{-ax} + \sqrt{(e^{ax}-e^{-ax})^2+4}}{2} $$

For large positive $x$ this is asymptotic to $Ce^{ax}$ while for large negative $x$ you can do an expansion of the square root and find that this is asymptotic to $1$.

So $y$, which will be $\ln z$, goes to zero at negative infinity, and to $$ ax \ln C $$ as $x$ goes to positive infinity, and $ax \ln C$ is the answer.

Answer 2

For positive $x$, $$ \begin{align} \sinh^{-1}(u) &=\log\left(u+\sqrt{u^2+1}\right)\\ &=\log\left(C\sinh(ax)+\sqrt{C^2\sinh^2(ax)+1}\right)\\ &=\log(C)+\log(\sinh(ax))+\log\left(1+\sqrt{1+\frac1{C^2\sinh^2(ax)}}\right)\\ &=\log(C)+ax+\log\left(1+e^{-2ax}\right)+\log\left(\frac12\left(1+\sqrt{1+\frac1{C^2\sinh^2(ax)}}\right)\right)\\ &=ax+\log(C)+O\left(\tfrac{C^2+1}{C^2}e^{-2ax}\right) \end{align} $$ Since $\sinh$ is an odd function, $$ \left.\sinh^{-1}(C\sinh(ax))\right|_{-\infty}^{+\infty}=\infty $$