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:
$$J(0,\beta)=\operatorname{Ci}(b)\sin(b)+\frac{\pi-2\operatorname{Si}(b)}{2}\cos(b)$$
 A: You can perform a further reduction:
$$ J(\alpha,\beta)=\int_{0}^{+\infty}\frac{\cos(\alpha u)}{u^2+1}e^{-\beta u}\,du =\text{Re}\int_{0}^{+\infty}\frac{1}{u^2+1}e^{-(\beta-\alpha i)u}\,du$$
thus all you need is the Laplace transform of $\frac{1}{u^2+1}$:
$$ K(c) = \int_{0}^{+\infty}\frac{e^{-cu}}{u^2+1}=\int_{0}^{\pi/2}\exp\left(-c\tan\theta\right)\,d\theta,\qquad c\in\mathbb{C},\text{Re}(c)\geq 0.$$
By the self-adjointness of the Laplace transform and the fact that $\mathcal{L}^{-1}\left(\frac{1}{u^2+1}\right)=\sin(s), \mathcal{L}(e^{-cu})=\frac{1}{c+s} $ we have
$$ K(c) = \int_{0}^{+\infty}\frac{\sin(s)}{s+c} \,ds $$
and the relation with the sine and cosine integrals is now obvious.
A: In the same lines as Jack D´Aurizio did, start with your integral
$$J\left(a,b,c\right)=\int_{0}^{\infty} \frac{\cos(ax)e^{-cx}}{b^{2}+x^{2}}dx$$
which can be rewriten as
$$J\left(a,b,c\right)=\text{Re}\left\{\frac{1}{b}\int_{0}^{\infty} \frac{e^{-x(b(c-ia))}}{1+x^{2}}dx\right\}$$
set $(b(c-ia))=s$ to get
$$\text{I}\left(s,b\right)=\frac{1}{b}\int_{0}^{\infty} \frac{e^{-sx}}{1+x^{2}}dx$$
to simplify, consider the version
$$\text{I}\left(s\right)=\int_{0}^{\infty} \frac{e^{-sx}}{1+x^{2}}dx$$
Now differentiate $\text{I}\left(s\right)$ with respect to $s$ twice to get
$$I''\left(s\right)=\int_{0}^{\infty} \frac{x^{2}e^{-sx}}{1+x^{2}}dx$$
Adding $I''\left(s\right)$ and $I\left(s\right)$
$$I''\left(s\right)+I\left(s\right)=\int_{0}^{\infty} e^{-sx}dx=\frac{1}{s}$$
This non homogeneous second order ODE can be solved by the method of  variation of parameters. The two linear independent solutions of the homogeneous equations are given by
$$u_{1}(s)=\cos(s)$$
$$u_{2}(s)=\sin(s)$$
The the general solution is given by
$$I_{g}\left(s\right)=A(s)\cos(s)+B(s)\sin(s)$$
where
$$A(s)=-\int_{}^{}\frac{1}{W}u_{2}(s)f(s)ds$$
and
$$B(s)=\int_{}^{}\frac{1}{W}u_{1}(s)f(s)ds $$
$W=u_{1}u_{2}'-u_{2}u_{1}'$ is the Wronskian which is $1$ here, and $f(s)=\frac{1}{s}$
putting all together
$$I_{g}\left(s\right)=-\cos(s)\int_{}^{s}\frac{\sin(t)}{t}dt +\sin(s)\int_{}^{s}\frac{\cos(t)}{t}dt$$
But $I(s)$ and all its derivatives vanish at $s=\infty$, and therefore
$$I_{g}\left(s\right)=\cos(s)\int_{s}^{\infty}\frac{\sin(t)}{t}dt -\sin(s)\int_{s}^{\infty}\frac{\cos(t)}{t}dt$$
$$\boxed{I\left(s\right)=\sin(s)Ci(s)+\cos(s)\left(\frac{\pi}{2}-Si(s)\right)}$$
A: 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:
$$I(s,t,k)=\int_0^\infty\frac{\cos(tx)}{x^2+k^2}e^{-sx}dx=\frac{\pi e^{-tk}}{2k}\left[\operatorname{Ci}(sk)\sin(sk)+\frac{\pi-2\operatorname{Si}(sk)}{2}\cos(sk)\right]$$
