\int_{0}^\infty {\cos(xt)\over 1+t^2}dt I'm supposed to solve this using Laplace Transformations.

Let f(x) denote the integral, and assume temporarily that x>0. This makes no difference since f(x) is even by definition. Then its Laplace transform Lf(s) defines a continuous function on ${\displaystyle s\in (0,\mathrm{\infty })}$ (in fact, it defines an analytic function on R(s)>0). Thus we may assume further that ${\displaystyle s\ne 1}$ and then rely on the continuity argument. Then
${\displaystyle \mathcal{L}f\left(s\right)={\int }_0^{\mathrm{\infty }}f\left(x\right)e^{-sx}dx}$
${\displaystyle ={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{\cos \left(xt\right)}{1+t^2}{e}^{-sx}dtdx}$
${\displaystyle \stackrel{\cdot }{=}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{\cos \left(xt\right)}{1+t^2}{e}^{-sx}dxdt}$
${\displaystyle ={\int }_0^{\mathrm{\infty }}\frac{1}{1+t^2}\frac{s}{s^2+t^2}dt}$
${\displaystyle =\frac{s}{1-s^2}{\int }_0^{\mathrm{\infty }}(\frac{1}{s^2+t^2}-\frac{1}{1+t^2})dt}$
${\displaystyle =\frac{s}{1-s^2}\frac{\pi }{2}(\frac{1}{s}-1)=\frac{\pi }{2}\frac{1}{1+s}}$
Here, the change of order of integration (∗) is justified by the dominated convergence theorem. Though we proved this for ${\displaystyle s\ne 1}$, it remains valid by continuity argument as mentioned above. Then by the uniqueness of the Laplace transform, we find that
${\displaystyle f\left(x\right)=\frac{\pi }{2}{e}^{-x}}$