= Solution
The interval indicator has transform
$$
\boxed{\widetilde g(k)=\int_{-b}^be^{-ikx}\,dx=\frac{2\sin(bk)}k,}
$$
with <continuous> value $2b$ at $k=0$. By the <convolution theorem>, $h=f*g$ has transform $4a\sin(bk)/[k(a^2+k^2)]$. Fourier inversion therefore expresses the requested integral as
$$
I(x):=\int_{\mathbb R}\frac{\sin(bk)e^{ikx}}{k(a^2+k^2)}\,dk
=\frac\pi{2a}h(x),\qquad h(x)=\int_{-b}^be^{-a|x-y|}\,dy.
$$
For $x>b$, $|x-y|=x-y$, so $h(x)=2e^{-ax}\sinh(ab)/a$. Symmetry gives the same expression with $|x|$ for $x<-b$. For $|x|\leq b$, splitting the integral at $y=x$ gives
$$
h(x)=\frac{1-e^{-a(x+b)}}a+\frac{1-e^{-a(b-x)}}a
=\frac2a\bigl(1-e^{-ab}\cosh(ax)\bigr).
$$
Consequently \b[the value for every real $x$ is]
$$
\boxed{I(x)=\begin{cases}
\displaystyle\frac\pi{a^2}\sinh(ab)e^{-a|x|},&|x|\geq b,\\[2mm]
\displaystyle\frac\pi{a^2}\left(1-e^{-ab}\cosh(ax)\right),&|x|\leq b.
\end{cases}}
$$
The 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 $O(|k|^{-3})$.
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