Find h(x) when $\hat{h}(y)=\frac{1}{(1+y^2)^2}$.
$$\hat{h}(y)=\frac{1}{(1+y^2)^2} =\frac{1}{1+y^2} \times \frac{1}{1+y^2} =\hat{\frac{1}{2}e^{-|x|}} \times \hat{\frac{1}{2}e^{-|x|}}$$
Let $f(x)=\frac{1}{2}e^{-|x|}$
$$\hat{h}(y)=\frac{1}{(1+y^2)^2}=\hat{f} \times \hat{f} = \hat{f*f}$$
So $h(x)=f*f=\int_R f(y)f(x-y)dy=\frac{1}{4} \int_R e^{-|y|}e^{-|x-y|}dy$
How do I go from here?