= Solution
The <convolution> is $(u*v)(x)=\int_{-\infty}^\infty u(t)v(x-t)dt$. In this transform convention, the <convolution theorem> says $\widehat{u*v}=\widehat u\,\widehat v$, and <Parseval's identity> says
$$
\int_{-\infty}^\infty|u(x)|^2dx=\frac1{2\pi}\int_{-\infty}^\infty|\widehat u(k)|^2dk.
$$
To match the integral's scale exactly, set $u=f/2$ and $h(x)=\tfrac12\mathbf1_{[-1,1]}(x)=g(2\pi x)/2$. Then $\widehat u=1/(1+k^2)$ and $\widehat h=\sin k/k$, using the value one at zero. Their convolution is the <box smoothing of a two-sided exponential> with both scales equal to one:
$$
w(x)=(u*h)(x)=\begin{cases}\tfrac12(1-e^{-1}\cosh x),&|x|\leq1,\\\tfrac12\sinh1\,e^{-|x|},&|x|\geq1.\end{cases}
$$
Its transform is precisely $\sin k/[k(1+k^2)]$. Since $w$ is even, <Parseval's identity> reduces the requested integral to
$$
J=\pi\left[\int_0^1(1-e^{-1}\cosh x)^2dx+\sinh^2(1)\int_1^\infty e^{-2x}dx\right].
$$
Use $\int_0^1\cosh x\,dx=\sinh1$, $\int_0^1\cosh^2x\,dx=1/2+\sinh2/4$ and $\int_1^\infty e^{-2x}dx=e^{-2}/2$. Simplifying gives
$$
\boxed{J=\frac\pi4(1+5e^{-2}).}
$$
The apparent singularity at $k=0$ is removable, and the integrand decays as $O(|k|^{-6})$, so the integral converges.
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