As the title states.
I've determined that for $a>0$ and $m\in\mathbb N$,
$$\int_1^\infty dy \,e^{-ay}\frac{y^m}{\sqrt{y^2-1}} = (-1)^mK^{(m)}_0(a),$$
where $K^{(m)}_0(a)$ is the $m^\mathrm{th}$ derivative of the usual modified Bessel function of zeroth order evaluate at $a$.
By integrating under the integral sign, I find that for $a>0$ $$\int_1^\infty dy \,e^{-ay}\frac{y^{-1}}{\sqrt{y^2-1}} = \int_a^\infty da'K_0(a') = \frac{\pi}{2}-\frac{\pi a}{2}\big( K_0(a)L_{-1}(a) + K_1(a)L_0(a) \big),$$ where $L_n(z)$ is the modified Struve function.
If I integrate under the integral sign again I'm stuck with an expression for which I'm unaware of a closed form solution, $$\int_1^\infty dy \,e^{-ay}\frac{y^{-2}}{\sqrt{y^2-1}} = \int_a^\infty da' \Big[\frac{\pi}{2}-\frac{\pi a'}{2}\big( K_0(a')L_{-1}(a') + K_1(a')L_0(a') \big)\Big].$$
Does anyone know of a closed form solution to the title integral, the above integral, or an equivalent? Thank you very much!
EDIT: following from Paul's answer below, one can find a somewhat nice closed form solution for the result of integration, $\operatorname{Ki}_2(z)$. Using Eq. 10.43.17 from Olver, https://dlmf.nist.gov/10.43#E17, with $\operatorname{Ki}_0(z)=K_0(z)$ and $\operatorname{Ki}_1(z) = \frac{\pi}{2}-\frac{\pi a}{2}\big( K_0(a)L_{-1}(a) + K_1(a)L_0(a) \big)$ one has that $$\int_1^\infty dy \,e^{-ay}\frac{y^{-2}}{\sqrt{y^2-1}} = \operatorname{Ki}_2(a) = a K_1(a) + \frac\pi2a\big( aK_0(a)L_{-1}(a)+aK_1(a)L_0(a)-1 \big).$$
The proposed integral is a Bickley-Naylor function. It can be defined as \begin{equation} \operatorname {Ki} _{n}(x)=\int _{1}^{\infty }{\frac {e^{-xt}\,dt}{t^{n}{\sqrt {t^2-1}}}} \end{equation} For $n=2$, which is the case of interest, \begin{align} \operatorname {Ki} _{2}(x)=1-{\frac {\pi }{2}}x-&{\frac {x^{2}}{2}}\left(\gamma +\ln \left({\frac {x}{2}}\right)\right)\sum _{k=0}^{\infty }{\frac {(x^{2}/4)^{k}}{k!(k+1)!(2k+1)}}+\\ &+{\frac {x^{2}}{4}}\sum _{k=0}^{\infty }{\frac {(4k+3)(x^{2}/4)^{k}}{k!(k+1)!(2k+1)^{2}}}+{\frac {x^{2}}{2}}\sum _{k=1}^{\infty }{\frac {(x^{2}/4)^{k}\Phi (k+1)}{k!(k+1)!(2k+1)}} \end{align} where $\gamma$ is the Euler constant and $\phi(k+1)=1+1/2+1/3+\cdots+1/k$.
and for $x\to\infty$, $\operatorname {Ki} _{2}(x)\simeq e^{-x}(\pi/\sqrt{2x})$.
The Wikipedia link provides several properties and references.