solution to a general integral $\int_0^\infty \frac{\cos(tx)}{x^2+k^2}e^{-sx}dx$ I would like to find a general solution to the integral:
$$I(s,t,k)=\int_0^\infty \frac{\cos(tx)}{x^2+k^2}e^{-sx}dx$$
so far using the substitution $u=\frac xk$ I have managed to reduce this to:
$$I(s,t,k)=\frac 1k\int_0^\infty\frac{\cos(tku)}{u^2+1}e^{-sku}du$$
and then by defining $\alpha=tk,\beta=sk$ we can come up with a simpler integral:
$$J(\alpha,\beta)=\int_0^\infty\frac{\cos(\alpha u)}{u^2+1}e^{-\beta u}du$$

We can calculate that:
$$J_{\beta\beta}=\int_0^\infty\frac{u^2\cos(\alpha u)}{u^2+1}e^{-\beta u}du$$
$$=\int_0^\infty\cos(\alpha u)e^{-\beta u}du-J$$
$$=\frac{\beta}{\beta^2+\alpha^2}-J$$
$$J_{\alpha\alpha}=-J_{\beta\beta}$$

We now know that: $\nabla^2J=0$
Now to form a system of equations I found that:
$$J(0,0)=\frac \pi2$$
$$J(\alpha,0)=\frac{\pi}{2}e^{-\alpha}$$
However I am struggling to find a solution to $J(0,\beta)$ although I know that it satisfies the equation:
$$K''(\beta)+K(\beta)=\frac 1\beta,K(0)=\frac \pi2$$
It seems clear to me that $\lim_{\beta\to\infty}J(\alpha,\beta)=0$
so if I could solve for $K$ I should have everything I need to try and solve this problem.

I think its obvious but I should add that:
$$I(s,t,k)=\frac 1kJ(tk,sk)$$
Basically, could anyone help me find $J(0,\beta)$ or proceed with solving the pde I stated. Thanks!

EDIT
wolfram alpha gives:
$$J(0,\beta)=\operatorname{Ci}(b)\sin(b)+\frac{\pi-2\operatorname{Si}(b)}{2}\cos(b)$$

I have managed to solve this in the end by assuming that the equation $J$ is separable then solving for this, and the answers lining up with the boundary conditions that be know for $J$, back substituting then gives us:
$$I(s,t,k)=\int_0^\infty\frac{\cos(tx)}{x^2+k^2}e^{-sx}dx=\frac{\pi e^{-tk}}{2k}\left[\operatorname{Ci}(sk)\sin(sk)+\frac{\pi-2\operatorname{Si}(sk)}{2}\cos(sk)\right]$$
