Prove a integral equation if $x > 0$, show that
$$f(x)=\int_0^\infty \frac{\cos(xt)}{1+t^2} \, dt= \frac{\pi}{2}e^{-x}$$
The problem appears in the book Differential Equations: With Applications and Historical Notes, under the chapter derivative and integral of Laplace transformation. So I tried to apply the equation 
$$\int_0^\infty \frac {f(t)} t \, dt = \int_0^\infty F(p)\,dp,$$
where $F(p) = L[f(x)](p)$, the Laplace transformation of $f(t)$. However, I failed to manipulate the expression to apply this equality.
I need your help. Please not be misled by my approach, it might be totally wrong.
 A: You can find a lot of ways to prove your equality here. Another way (which I love) is the following. We have that $$I=\int_{0}^{\infty}\frac{\cos\left(xt\right)}{1+t^{2}}dt=\frac{1}{2}\int_{0}^{\infty}\frac{e^{ixt}+e^{-ixt}}{1+t^{2}}dt
 $$ $$=\frac{1}{4}\int_{0}^{\infty}\frac{e^{ixt}+e^{-ixt}}{1-it}dt+\frac{1}{4}\int_{0}^{\infty}\frac{e^{ixt}+e^{-ixt}}{1+it}dt
 $$ $$=\frac{e^{-x}}{4i}\int_{0}^{\infty}\frac{ixt\left(e^{x+ixt}+e^{x-ixt}\right)}{t\left(x-ixt\right)}dt+\frac{e^{-x}}{4i}\int_{0}^{\infty}\frac{ixt\left(e^{x+ixt}+e^{x-ixt}\right)}{t\left(x+ixt\right)}dt$$ $$=\frac{e^{-x}}{4i}\int_{0}^{\infty}\frac{ixte^{x+ixt}}{t\left(x+ixt\right)}+\frac{ixte^{x-ixt}}{t\left(x-ixt\right)}dt+\frac{e^{-x}}{4i}\int_{0}^{\infty}\frac{ixte^{x-ixt}}{t\left(x+ixt\right)}+\frac{ixte^{x+ixt}}{t\left(x-ixt\right)}dt
 $$ so if we apply the complex version of the Frullani's theorem to the functions $$f\left(t\right)=\frac{te^{x-t}}{x-t}
 $$ and $$g\left(t\right)=\frac{te^{x-t}}{x+t}
 $$ we get $$I=\frac{e^{-x}}{2i}\log(-1)=\color{red}{\frac{\pi}{2}e^{-x}}
 $$ as wanted.
