Taking limit with hyperbolic functions I have a problem with evaluating 
$$\sinh^{-1}(C \sinh (ax))\bigg|_{-\infty}^{+\infty}$$ where $C$ and $a$ are real positive constants.
 A: 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.
A: 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,
$$
\left.\sinh^{-1}(C\sinh(ax))\right|_{-\infty}^{+\infty}=\infty
$$
