Bound on Bessel potential

Question

Let $s\in\mathbb{C}$. For a complex number $z$, $Re(z)>0$, consider the Bessel potential $$K_s(z)=\int_0^{+\infty}e^{-z\cosh t}\cosh(st)dt$$ I need to prove that, if $|z|\leq 1$, then $$|K_s(z)|\leq C_s|z|^{-|Re(s)|}$$ where $C_s$ is a constant depending only on $s$.

Answer 1

$$\cosh(st) = \cosh(\Re(s)t+i\Im(s)t) = \cos(\Im(s)t)\cosh(\Re(s)t),$$ hence: $$\|\cosh(st)\|\leq \exp(|\Re(s)|t),$$ while $$\|\exp(-z\,\cosh t)\|=\exp(-\Re(z)\cosh t)\leq\exp\left(-\frac{1}{2}\Re(z)\,e^t\right),$$ so: $$\begin{eqnarray*}\|K_s(z)\|&\leq&\int_{0}^{+\infty}\exp\left(|\Re(s)|t-\frac{1}{2}\Re(z)\,e^t\right)\,dt\\&=&\left(\frac{1}{2}\Re(z)\right)^{-|\Re(s)|}\Gamma\!\left(|\Re(s)|,\Re(z)\right)\end{eqnarray*}$$ and you can take $C_s$ as:

$$ C_s =2^{|\Re(s)|}\cdot \Gamma(|\Re(s)|).$$