Compute $\int_0^\infty\frac{\cos(xt)}{1+t^2}dt$ Let $x \in \mathbb R$
Find a closed form of $\int_0^\infty\dfrac{\cos(xt)}{1+t^2}dt$ . 
Let me give some context: this is an exercise from an improper integrals course for undergraduates. My teacher was not able to give a valid solution without  resorting to Fourier series. 
He actually wanted to prove first that the function is a solution of the ODE $$ y-y'' =0$$ without succeeding in proving it. 
Anyway the result sould be $\dfrac{\pi}{2}e^{-x}$
Thanks for any help on this subject. 

We can also find the differential equation by just differentiating under the integral sign, after rewriting a little:
$$F(x) = \int_0^\infty \frac{\cos (xt)}{1+t^2}\,dt$$
is continuous by the dominated convergence theorem, with $F(0) = \frac{\pi}{2}$ known and $\lvert F(x)\rvert \leqslant \frac{\pi}{2}$. For $x \neq 0$, we can integrate by parts,
$$\begin{align}
F(x) &= \left[\frac{1}{x}\cdot \frac{\sin (xt)}{1+t^2}\right]_0^\infty + \frac{2}{x} \underbrace{\int_0^\infty \frac{t\sin (xt)}{(1+t^2)^2}\,dt}_{G(x)}\\
&= \frac{2}{x}G(x).
\end{align}$$
$G(x)$ can be differentiated under the integral sign by the dominated convergence theorem, and
$$G'(x) = \int_0^\infty \frac{t^2\cos (xt)}{(1+t^2)^2}\,dt = F(x) - \underbrace{\int_0^\infty \frac{\cos (xt)}{(1+t^2)^2}\,dt}_{H(x)}.$$
$H(x)$ can also be differentiated under the integral sign by the dominated convergence theorem, and
$$H'(x) = -G(x).$$
Thus
$$\begin{align}
F'(x) &= -\frac{2}{x^2}G(x) + \frac{2}{x}G'(x)\\
&= -\frac{1}{x}F(x) + \frac{2}{x}\left(F(x) - H(x)\right)\\
&= \frac{1}{x}F(x) - \frac{2}{x}H(x),\quad\text{and}\\
F''(x) &= -\frac{1}{x^2}F(x) + \frac{1}{x} F'(x) + \frac{2}{x^2}H(x) - \frac{2}{x}H'(x)\\
&=  -\frac{1}{x^2}F(x) + \frac{1}{x^2}F(x) - \frac{2}{x^2}H(x) + \frac{2}{x^2}H(x) + \frac{2}{x}G(x)\\
&= \frac{2}{x}G(x)\\
&= F(x).
\end{align}$$
From the boundedness of $F$ and the differential equation together with the initial value $F(0)$ we then obtain
$$F(x) = \frac{\pi}{2}e^{-\lvert x\rvert}.$$
