Fourier-Transformation of $\exp(-|t|)$

Question

Using the Fourier-inversion-theorem I have to show that

$$\frac{\pi}{2}\exp(-|t|)=\int_0^\infty\frac{\cos(\omega t)}{1+\omega^2}\mathrm d\omega$$

Can anyone give me a hint on how to show it?

Thank you very much!

Answer 1

Note first that$$\int_0^\infty\frac{\cos(\omega t)d\omega}{1+\omega^2}=\frac12\int_{\Bbb R}\frac{\cos(\omega t)d\omega}{1+\omega^2}=\frac12\int_{\Bbb R}\frac{\exp(i\omega t)d\omega}{1+\omega^2},$$where the second $=$ uses the fact that the final integrand's imaginary part is odd. We're trying to evaluate the Fourier transform of $\frac12\frac{1}{1+\omega^2}$, if we define the Fourier transform of $f(\omega)$ as $\int_{\Bbb R}f(\omega)\exp(i\omega t)d\omega$. With this definition, Fourier inversion is $\frac{1}{2\pi}\int_{\Bbb R}\tilde{f}(t)\exp(-i\omega t)dt$. But$$\frac{1}{2\pi}\int_{\Bbb R}\frac{\pi}{2}\exp(-i\omega t-|t|)dt=\frac14\int_{\Bbb R}\exp(-i\omega t-|t|)dt=\frac12\Re\int_0^\infty\exp(-(i\omega+1)t)dt,$$using$$\int_{-\infty}^0\exp(-i\omega t-|t|)dt=\int_{-\infty}^0\exp(-(i\omega-1)t)dt=\int_0^\infty\exp(-(1-i\omega)u)du=\overline{\int_0^\infty\exp(-(i\omega+1)t)dt}.$$Hence$$\frac{1}{2\pi}\int_{\Bbb R}\frac{\pi}{2}\exp(-i\omega t-|t|)dt=\frac12\Re\frac{1}{1-i\omega}=\frac12\frac{1}{1+\omega^2},$$so we're done.