TITLE: Integrating using Laplace Transforms
QUESTION [12 upvotes]: $$\int_{0}^\infty {\cos(xt)\over 1+t^2}dt $$
I'm supposed to solve this using Laplace Transformations.
I've been trying this since this morning but I haven't figured it out. Any pointers to push me in the right direction?

REPLY [4 votes]: Laplace approach :
$$  I= \int\limits_{0}^\infty {\cos(xt)\over 1+t^2}\,\mathrm dt=\left[ \int\limits_{0}^\infty {\cos(xt)\over 1+t^2} e^{-st}\,\mathrm dt \right]_{s=0} =\mathcal{L}\left[ {\cos(xt)\over 1+t^2} \right]_{s=0}$$
$$ f(t)=\frac{\cos(xt)}{1+t^2}  $$
$$ (1+t^2)f(t)=\cos(xt) $$
$$ \downarrow\mathcal{L}$$
$$  \frac{\mathrm d^2}{\mathrm ds^2}F(s)+F(s)=\frac{s}{s^2+x^2} $$
We have second order differential equation, after we find $ F(s) $
we put zero instead of $ s $.
Where we have used :
$$  \boxed{\mathcal{L} \big\{ t^nf(t)\big\}=(-1)^n \dfrac{\mathrm d^n}{\mathrm ds^n}F(s)} $$
$$  \boxed{\mathcal{L}\big\{\cos(wt) \big\} = \dfrac{s}{s^2+w^2}} $$