 Solution

The interval indicator has transformwith continuous value $2b$ at $k=0$. By the convolution theorem, $h=f\ast g$ has transform $4a\sin (bk)/[k(a^2+k^2)]$. Fourier inversion therefore expresses the requested integral asFor $x>b$, $|x-y|=x-y$, so $h(x)=2e^{-ax}\sinh (ab)/a$. Symmetry gives the same expression with $|x|$ for $x<-b$. For $|x|\le b$, splitting the integral at $y=x$ givesConsequently the value for every real $x$ isThe two expressions agree at $x=\pm b$. This is the exponential kernel convolved with an interval indicator. The spectral integral is absolutely convergent: its apparent singularity at zero is removable, and the tail decays as $O(|k{|}^{-3})$.