Use the representation as Fourier integral to prove that $$\int_0^\infty \displaystyle\frac{cos(xw)}{1+x^2}dw=\frac{\pi}{2}e^{-x}$$
I am really confuse with this kind of problem, I do not know well how to use the Fourier Transform.
Somebody can give some advice to solve it?
the integral on left side does not coverge, i think you mistyped something there is something similar $\int_0^{\infty}\frac{cos(ωx)}{(1 + x²)} dx= \frac{1}{2}\int_{-\infty}^{\infty}\frac{cos(ωx)}{(1 + x²)} dx = \frac{1}{2}Re(\int_{-\infty}^{\infty}\frac{e^{i\omega x}}{(1 + x²)} dx ) $
use residue theory from Complex Analysis.
Consider closed curve C (counterclockwise ) [-R, R] U C', where C' is the top halfcircle of R > 1 from R to -R.
Then, $\int_C \frac{e^{i\omega z}}{1 + z^2} dz = \int_{C'} \frac{e^{i\omega z}}{1 + z^2} dz + \int_{-R}^{R} \frac{e^{i\omega z}}{1 + z^2} dz $
The singularity inside C iz z = i. $\int_C \frac{e^{i\omega z}}{1 + z^2} dz = 2\pi i\lim_{z\rightarrow i}\frac{(z-i)e^{i\omega z}}{1 + z^2} = \pi e^{-\omega}$ by residue theorem
$\int_{C'} \frac{e^{i\omega z}}{1 + z^2} dz = \int_{0}^\pi \frac{e^{i\omega Re^{it}}ie^{it}}{1 + R^2e^{2it}} dt $
lets show that this converge to 0 when $R \rightarrow \infty$
$|\int_{0}^\pi \frac{e^{i\omega Re^{it}}ie^{it}}{1 + R^2e^{2it}} dt |\le \int_{0}^\pi |\frac{e^{i\omega Re^{it}}iRe^{it}}{1 + R^2e^{2it}}| dt \le$
$\int_{0}^\pi \frac{|e^{i\omega R(cos t +i sin t)}|R}{|1 + R^2e^{2it}|} dt =\int_{0}^\pi \frac{|e^{-\omega R sin t)}|R}{|1 + R^2e^{2it}|} dt \le \int_{0}^\pi \frac{|e^{-\omega R sin t)}|R}{R^2-1} dt \le $
$\int_{C'} \frac{e^{i\omega z}}{1 + z^2} dz = \int_{0}^\pi \frac{e^{i\omega Re^{it}}ie^{it}}{1 + R^2e^{2it}} dt = |\int_{0}^\pi \frac{e^{i\omega Re^{it}}ie^{it}}{1 + R^2e^{2it}} dt |\le \int_{0}^\pi |\frac{e^{i\omega Re^{it}}iRe^{it}}{1 + R^2e^{2it}}| dt \int_{0}^\pi \frac{|e^{i\omega R(cos t +i sin t)}|R}{|1 + R^2e^{2it}|} dt =\int_{0}^\pi \frac{|e^{-\omega R sin t)}|R}{|1 + R^2e^{2it}|} dt \le \int_{0}^\pi \frac{|e^{-\omega R sin t)}|R}{R^2-1} dt \le \int_{0}^\pi \frac{R}{R^2-1} dt = \frac{\pi R}{R^2-1} \rightarrow0 $
when $ r\rightarrow \infty$ keep in mind sin is positive on $(0, 2\pi)$ and $R>1$ since $ R\rightarrow \infty$
so when $ R\rightarrow \infty$
$\pi e^{-\omega}=0 + \int_{-R}^{R} \frac{e^{i\omega z}}{1 + z^2} dz $
$\int_0^{\infty}\frac{cos(ωx)}{(1 + x²)} dx= \frac{1}{2}Re(\pi e^{-\omega} )=\frac{\pi e^{-\omega}}{2} $