$\operatorname{Coth}(\alpha_n a) \to i$ when $n \to \infty $

Question

I have a question about the following statement from the article

• Alves, M., Rivera, J.M., Sepúlveda, M., Villagrán, O.V. and Garay, M.Z., The asymptotic behavior of the linear transmission problem in viscoelasticity, Math. Nachr. 287 (2014) pp 483-497, doi:10.1002/mana.201200319, (author pdf)

They state on page 8 that:

Let $\alpha = \dfrac{\rho_1\lambda^2}{\kappa_1 + i\kappa_2\lambda}$ where $\rho_1, \kappa_1, \kappa_2 > 0$. then $$ \operatorname{Coth}(i\alpha L) \to 1, \ \ \text{ when } \ \ n \to \infty $$ where $L > 0$.

Well, firstly, I think they are changing $\alpha$ to $\alpha_{n}$ and leaving $n$ hidden. And doing some mathematical manipulations, I'm getting to

Let $x_n = \Re(\alpha_n L)$ and $y_n = \Im(\alpha_n L)$. Using some properties, I can get $$ \operatorname{Coth}(i\alpha_n L) = \dfrac{\cos(x_n)\sin(x_n)}{\cosh^2 (y_n) - \cos^2(x_n)} - \dfrac{\tanh(y_n)} {\frac{\cos^2(x_n)}{\cosh^2 (y_n)} - 1} $$

From the equality above can I conclude what I want? Why?

Answer 1

$\newcommand\la\lambda\newcommand\al\alpha\newcommand\ka\kappa\newcommand\th\theta$First of all, you copied the expression for $\al$ incorrectly. On p. 489 of the paper linked by you (the fourth display from the bottom of the page), we find $$\al=r(\la)e^{i\th/2},$$ where $\mathbb R\ni\la=\la_n\to\infty$, $$r(\la):=\frac{\rho_1^{1/2}\la}{\sqrt[4]{\ka_1^2+\ka_2^2\la^2}}\to\infty, \quad e^{i\th/2}\to\frac{1-i}{\sqrt2}, $$ and $\rho_1,\ka_1,\ka_2$ are fixed positive real numbers. So, $$|e^{-2i\al L}|=|\exp(-2iLr(\la)e^{i\th/2})| =\exp(\Re(-2iLr(\la)e^{i\th/2})) \\ =\exp(-(\sqrt2+o(1))\,Lr(\la))\to0.$$ Thus, $$\coth(i\al L)=\frac{1+e^{-2i\al L}}{1-e^{-2i\al L}}\to\frac{1+0}{1-0}=1,$$ as desired.