Asymptotic expansion of $\operatorname{arsinh}(c/t)$ as $t\rightarrow 0$

Question

let $c>0$ be a constant and let $f:(0,\infty)\rightarrow\mathbb{R}$ be defined as $f(t):=\operatorname{arsinh}(\frac{c}{t})$, where $\operatorname{arsinh}$ denotes the inverse of sinus hyperbolicus. That function has a singularity at $t=0$, and I am interested in an asymptotic expansion in powers of $t$ as $t\rightarrow 0$.

By just guessing and using the rule of l'Hospital, I proved that $f(t)\sim \frac{c}{t}$ as $t\rightarrow 0$. Is it possible to obtain a full asymptotic expansion by a systematic approach?

Best wishes

Answer 1

Reference Gradshtyn & Ryshik

Gradshteyn, I. S.; Ryzhik, I. M.; Zwillinger, Daniel (ed.); Moll, Victor (ed.), Table of integrals, series, and products. Translated from the Russian. Translation edited and with a preface by Victor Moll and Daniel Zwillinger, Amsterdam: Elsevier/Academic Press (ISBN 978-0-12-384933-5/hbk; 978-0-12-384934-2/ebook). xlv, 1133 p. (2015). ZBL1300.65001.

Formula 1.641.2
As $z \to 0$, which corresponds to $t \to \infty$:

\begin{align} \operatorname{asinh} z &= z - \frac{1}{2\cdot3}\;z^3 + \frac{1\cdot3}{2\cdot4\cdot5}\;z^5 -\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot7}\;z^7+\dots \\ &=\sum_{k=0}^\infty \frac{(-1)^k(2k)!z^{2k+1}}{2^{2k}(k!)^2(2k+1)} = z\;{}_2F_1\left(\frac{1}{2},\frac{1}{2};\frac{3}{2};-z^2\right) \end{align}

Formula 1.642.1
As $z \to \infty$, which corresponds to $t \to 0$: \begin{align} \operatorname{asinh} z &= \log(2z) + \frac{1}{2}\;\frac{1}{2z^2} -\frac{1\cdot3}{2\cdot4}\;\frac{1}{4z^4} +\dots \\ &= \log(2z) + \sum_{k=0}^\infty \frac{(-1)^{k+1}(2k)!z^{-2k}}{2^{2k}(k!)^2 2k} \end{align}

Answer 2

You have the following expansion of $\operatorname{argsinh}x$, based on the Taylor expansion of its derivative $\frac1{\sqrt{1+x^2}}$: \begin{align} &\operatorname{argsinh}x=\\ &\phantom{{}={}}x-\frac12\frac{x^3}3+\frac{1\cdot 3}{2\cdot 4}\frac{x^5}5-\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{x^7}7 +\dots+(-1)^{2n+1}\frac{(2n-1)!!}{(2n)!!}x^{2n+1}+\dotsm \end{align}

Answer 3

A little correction

We use argsinh instead of arcsinh, because we are talking about an argument in an exponential terms $\dfrac{e^x-e^{-x}}{2}$ and not an arc as in trigonometric function.

Then we have that formula (proved by inversion of $\sinh(y)+\left(1+\sinh^2(y)\right)^{\frac{1}{2}}=e^y$

$$ \operatorname{argsinh}(x)=\ln(x+\left(1+x^2\right)^{\frac{1}{2}}) $$

So we have

$$\operatorname{argsinh}(x)=\ln(x) + \ln\left(1+\left(1+\frac{1}{x^2}\right)^{\frac{1}{2}}\right)$$

Which gives :

$$\operatorname{argsinh}(x)\sim_\infty \ln(x) $$

For further development use asymptotic development inside the right side term.