Let the Fourier transform of a function and its inverse be given as:
$\mathcal{F}[f](\lambda) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t)e^{-i\lambda t} dt$
$f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \mathcal{F}[f](\lambda)e^{i\lambda t} d\lambda$
For $a>0$, let $\psi_a:\mathbb{R}\rightarrow \mathbb{R}$ be given by:
$\psi_a(t) = \frac{1}{2\pi}\frac{2a}{a^2+t^2}$
I want to show that the fourier transform of $\psi_a$ is $\mathcal{F}[\psi_a](\lambda) = \frac{1}{\sqrt{2\pi}}e^{-a|\lambda|}$. I know that the fourier transform of $\frac{1}{1+t^2}$ is $\mathcal{F}[\frac{1}{1+t^2}](\lambda) = \sqrt{\frac{\pi}{2}}e^{-|\lambda|}$. I figured I could use this fact to somehow find the transform of $\psi_a$ using properties regarding the fourier transform of a translation or a rescaling, but it seems to me that $\psi_a$ is neither a rescaling or a translation of $\frac{1}{1+t^2}$. Am I missing something?
Let \begin{align} \psi_a(t)&=\frac{a}{a^2+t^2} \\ \phi(t)&=\frac{1}{1+t^2} \end{align} then \begin{equation} \psi_a(t)=\frac{1}{a}\phi(t/a). \end{equation} Assume knowledge of the Fourier transform of a function $f(t)$, then a the function $g(t)=C f(t/a)$ has Fourier transform \begin{equation} \mathcal{F}[g](\lambda)=C\int_{\mathbb{R}}dt e^{-i\lambda t} f(t/a)=C a\int_{\mathbb{R}}dt' e^{-i a \lambda t'} f(t')=C a \, \mathcal{F}[f](a\lambda), \end{equation} Applying this relation to our case \begin{equation} \mathcal{F}[\psi_a](\lambda)=\mathcal{F}[\phi](a\lambda) \end{equation}