 Solution

The convolution is $(u\ast v)(x)={\int }_{-\infty }^{\infty }u(t)v(x-t)dt$. In this transform convention, the convolution theorem says $\widehat{u\ast v}=\hat{u}\hat{v}$, and Parseval's identity saysTo match the integral's scale exactly, set $u=f/2$ and $h(x)=\frac{1}{2}{1}_{[-1,1]}(x)=g(2\pi x)/2$. Then $\hat{u}=1/(1+k^2)$ and $\hat{h}=\sin k/k$, using the value one at zero. Their convolution is the box smoothing of a two-sided exponential with both scales equal to one:Its transform is precisely $\sin k/[k(1+k^2)]$. Since $w$ is even, Parseval's identity reduces the requested integral toUse ${\int }_0^1\cosh xdx=\sinh 1$, ${\int }_0^1{\cosh }^{2}xdx=1/2+\sinh 2/4$ and ${\int }_1^{\infty }{e}^{-2x}dx=e^{-2}/2$. Simplifying givesThe apparent singularity at $k=0$ is removable, and the integrand decays as $O(|k{|}^{-6})$, so the integral converges.