inverse fourier transform of w*e^w

Question

I have the function

\begin{align} F^{-1}\{{λe^{-|λ|}}\} \end{align}

How can we find the inverse Fourier transform?

The correct answer is:

\begin{align} \frac{-2ix}{π(1+x^2)^2} \end{align}

Can somebody explain to me what happened? Thanks in advance.

Answer 1

$$ \mathcal F\{f(x)\}=\hat{f}(\lambda)= \int_{-\infty}^\infty f(x) e^{-i\lambda x}\, dx $$

The Fourier transform of $f(x)=e^{-a|x|}$ is $\hat{f}(\lambda)=\frac{2a}{a^2 + \lambda^2} $ and then for the duality property $\mathcal F\{\hat{f}(x)\}= 2\pi f(-\lambda)$ we have $$\mathcal F\left\{\frac{2a}{a^2 + x^2} \right\}=2\pi e^{-a|-\lambda|}$$ and for the differentiation property $\mathcal F\{{f'}(x)\}=i\lambda \hat{f}(\lambda)$, and then $$ \mathcal F\left\{\left(\frac{2a}{a^2 + x^2}\right)'\right\}=\mathcal F\left\{-\frac{4ax}{(a^2 + x^2)^2}\right\}=i\lambda \left(2\pi e^{-a|-\lambda|}\right) $$ Thus we have $$ \mathcal F^{-1}\left\{\lambda e^{-a|\lambda|}\right\}=\frac{4ax}{2\pi i(a^2 + x^2)^2}=\frac{-2iax}{\pi (a^2 + x^2)^2} $$ and for $a=1$ $$ \mathcal F^{-1}\left\{\lambda e^{-|\lambda|}\right\}=-\frac{2ix}{\pi (1 + x^2)^2} $$