$\newcommand{\om}{\omega}\newcommand{\al}{\alpha}\newcommand{\la}{\lambda}\newcommand{\Ga}{\Gamma}$Inequality (14) in the paper by Glushak and Pokruchin referred to in your comment is equivalent to
\begin{align*}
&\Big\|\int_0^\infty t^{\nu+n}K_{\nu-n+1}(t\sqrt\mu)Y_k(t)x\,dt\Big\| \\
&\le M\|x\|\int_0^\infty t^{\nu+n-1/2}\exp((\om-\sqrt\mu)t)\,dt \tag{14}
\end{align*}
given that, by formula (3) in the paper,
\begin{equation}
\|Y_k(t)\|\le M\exp(\om t) \tag{0}
\end{equation}
for some real $\om\ge0$ and all real $t>0$. Here, $\nu=(k-1)/2$ (by p. 40 of the paper), $k\ge0$, and $n$ is a natural number.
One can easily see that inequality (14) is false in general (or, maybe, always).
However, one can save the conclusion (at the bottom of p. 42 of the paper) that
\begin{equation*}
\|R^k(\mu)\|\le M(k)/(\mu-\om_1)^n \tag{1}
\end{equation*}
for some real $M(k)>0$ not depending on $\mu$ or $n$, some real $\om_1$, and all real $\mu>\om_1$, which makes the Hille--Yosida theorem applicable.
Indeed, for real $u>0$ and $\al$
\begin{equation*}
K_\al(u)=\int_0^\infty e^{-u\cosh z}\cosh\al z\,dz \tag{2}
\end{equation*}
by the second display under the picture in Section Modified Bessel functions: $I_\al$, $K_\al$. By rescaling, without loss of generality, $\|x\|=1$ and (0) holds with $M=1$. So, by (2), with
\begin{equation}
m:=n-\nu+1,\quad l:=2\nu+2=k+1\ge1,\quad\mu>4\om^2,\quad r:=\om/\sqrt\mu\in(0,1/2),
\end{equation}
the left-hand side of (14) is no greater than
\begin{align*}
&\int_0^\infty dt\, t^{\nu+n}e^{\om t}
\int_0^\infty dz\,e^{-t\sqrt\mu\,\cosh z}\cosh mz \\
&=\int_0^\infty dz\,\cosh mz
\int_0^\infty dt\, t^{\nu+n}e^{-t(\sqrt\mu\,\cosh z-\om)} \\
&=\Ga(\nu+n+1)\int_0^\infty \frac{dz\,\cosh mz}{(\sqrt\mu\,\cosh z-\om)^{m+l}} \\
&=\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{m+l}}\int_0^\infty \frac{dz\,\cosh mz}{(\cosh z-r)^{m+l}} \\
&\ll\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{m+l}}\int_0^\infty dz\,f(z)^m e^{-lz}, \tag{3}
\end{align*}
where $a\ll b$ means that $|a|\le Cb$ for some real $C>0$ not depending on $n$ or $\mu$ or $\om$ (as long as $r$ is small enough) and
\begin{equation*}
f(z):=\frac{e^z}{\cosh z-r}.
\end{equation*}
Note that $\max_{z>0}f(z)=(1-r^2)^{-1}$. So, the left-hand side of (14) is
\begin{align*}
&\ll\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{m+l}}\, (1-r^2)^{\nu-1-n}
\ll\frac{\Ga(\nu+n+1)}{(\sqrt\mu)^{n+\nu+3}}\, (1-r^2)^{-n}=:C_n(\mu).
\end{align*}
So, by the equality in formula (14) in the paper, for $\mu>\om_1:=1+4\om^2$,
\begin{align*}
\|R^k(\mu)\|&\ll\frac{C_n(\mu)}{2^{n+\nu-1}(n-1)!\Gamma(\nu+1)\mu^{n/2-k/4}} \\
&\ll\frac{\Ga(\nu+n+1)(1-r^2)^{-n}}{2^{n+\nu-1}(n-1)!\Gamma(\nu+1)\mu^n} \\
&\ll\frac{\Ga(\nu+n+1)(4/3)^n}{2^{n+\nu-1}(n-1)!\Gamma(\nu+1)\mu^n} \\
&\ll\frac{M(k)}{(\mu-\om_1)^n},
\end{align*}
which proves (1).
Edit: In fact, $\max_{z>0}f(z)=2(1-r^2)^{-1}=:c>2$ -- I previously missed the factor $2$ here. So, the last integral in (3) is $\asymp c^m/\sqrt m$. Since the bounding was lossless (up to inessential constant factors) everywhere in the above reasoning, this appears to invalidate the entire result in the paper by Glushak and Pokruchin.
