Inverse Fourier transformation of $\frac{1}{(1+w^2)^2}$?

Question

I have tried to crack this via basic definition, but I am unable to solve the integral: $$\int_{-\infty} ^{\infty} \frac{1}{(1+w^2)^2} e^{-iwx} dw $$ Kindly guide me a bit. Thanks in anticipation.

Answer 1

You can avoid doing the integral if you know the Fourier transform of $1/(1+w^2)$ (which is $\pi e^{-|x|}$) and recall that the Fourier transform converts multiplication into convolution.

Therefore, for $g(x)=e^{-|x|}$, $$\int_{-\infty}^\infty\frac{e^{-iwx}\,dw}{(1+w^2)^2} =g*g(x)=\int_{-\infty}^\infty g(y)g(x-y)\,dy =\pi^2\int_{-\infty}^\infty\exp(-|y|-|x-y|)\,dy.$$ One can split this into three integrals over intervals where $-|y|-|x-y|$ is linear on each.