I have to calculate the Fourier transform of $f(x) = e^{-a|x|}$ with the definition $\widehat{f}(t) = \int_{\mathbb{R}} f(x) e^{-itx} dx$.
Then, I have to deduce that for $a>0$, $\dfrac{a}{\pi} \int_{\mathbb{R}} \dfrac{\cos(2\pi xt)}{a^2+t^2} dt = e^{-2\pi a|x|}$.
I calculated $\widehat{f}(t) = \dfrac{2a}{t^2+a^2}$.
I know the Fourier inversion theorem, but I'm unable to deduce the desired resultat. Could someone help me ?
Using residues you get:
For $x > 0$: $$I = \frac{a}{\pi} \int_{\mathbb{R}} \frac{\cos(2\pi xt)}{a^2+t^2} dt = \frac{a}{\pi}Re\left( \int_{-\infty}^{\infty}\frac{1}{a^2+t^2}e^{2\pi ixt}dt \right)$$
$$\int_{-\infty}^{\infty}\frac{1}{a^2+t^2}e^{2\pi ixt}dt = 2\pi i Res_{ia}\left( \frac{1}{a^2+z^2}e^{2\pi ixz} \right)= 2\pi i \lim_{z\rightarrow ia}\left( \frac{(z-ia)}{(z+ia)(z-ia)}e^{2\pi ixz}\right)= 2\pi i\frac{1}{2ia}e^{-2\pi xa} = \frac{\pi}{a} e^{-2\pi xa}$$ For $x<0$ note that $\cos(2\pi xt) = \cos(- 2\pi |x|t) = \cos(2\pi |x|t)$. So, it follows
$$I = \frac{a}{\pi} \int_{\mathbb{R}} \frac{\cos(2\pi xt)}{a^2+t^2} dt = \frac{a}{\pi}\frac{\pi}{a} e^{-2\pi |x|a} = e^{-2\pi |x|a}$$