I'm stuck on the following question:
Prove that: $\mathcal{F_x}[e^{-2\pi K_0 |x|}](K)=\frac {1}{\pi}\frac{K_0}{K^2+K_0^2}$
Using: $\int_0^{\infty}e^{-\omega T}\cos(\omega t)d\omega=\frac{T}{t^2+T^2}$
My attempt:
$F(K)=\mathcal{F_x}(f(x))(K)$
$ F(K)=\int_{-\infty}^{\infty}f(x)e^{-2\pi iKx }dx$
$ =\int_{-\infty}^{\infty}e^{-2\pi K_0|x|}e^{-2\pi iKx }dx$
(Edit, worked on it some more): now I split the integrals and write them in the sine cosine form:
$ F(K)=\int_{-\infty}^{0}e^{2\pi K_0x}e^{-2\pi iKx }dx+\int_{0}^{\infty}e^{-2\pi K_0x}e^{-2\pi iKx}$
$F(K)=\int_{-\infty}^{0}(\cos(2\pi Kx)-i\sin(2\pi K x))e^{2\pi K_0x}dx+\int_{0}^{\infty}(\cos(2\pi Kx)-i\sin(2\pi K x))e^{-2\pi K_0x}dx$
From here on I don't know how to continue, any help is appreciated $:)$