= Solution
\b[Transform pair.] In the paper's symmetric normalization, the inverse <Fourier transform> is
$$
\boxed{g(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}\widetilde g(k)e^{ikx}\,dk.}
$$
Split the transform of $e^{-c|x|}$ at zero, with $c>0$:
$$
\begin{aligned}
\mathcal F[e^{-c|x|}](k)&=\frac1{\sqrt{2\pi}}\left[\int_{-\infty}^0e^{(c-ik)x}\,dx+\int_0^\infty e^{-(c+ik)x}\,dx\right]\\
&=\frac1{\sqrt{2\pi}}\left(\frac1{c-ik}+\frac1{c+ik}\right)=\boxed{\sqrt{\frac2\pi}\frac c{c^2+k^2}}.
\end{aligned}
$$
The inverse formula applied to this result gives
$$
e^{-c|x|}=\frac c\pi\int_{-\infty}^{\infty}\frac{e^{ikx}}{c^2+k^2}\,dk.
$$
Because the kernel is even, replacing the transform variable by $x$ and the argument by $k$ yields
$$
\boxed{\mathcal F\left[\frac1{x^2+c^2}\right](k)=\frac{\sqrt{2\pi}}{2c}e^{-c|k|}.}
$$
The factors of $\sqrt{2\pi}$ depend on the normalization and will also enter the <convolution theorem>.
\b[Convolution theorem.] For integrable <functions> define the <convolution>
$$
(f*g)(x)=\int_{-\infty}^{\infty}f(x-y)g(y)\,dy.
$$
Substitute this into the <Fourier transform>, interchange the two absolutely integrable factors using <Fubini's theorem>, and set $z=x-y$:
$$
\begin{aligned}
\widetilde{f*g}(k)&=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x-y)g(y)e^{-ikx}\,dy\,dx\\
&=\frac1{\sqrt{2\pi}}\left(\int f(z)e^{-ikz}\,dz\right)\left(\int g(y)e^{-iky}\,dy\right).
\end{aligned}
$$
Therefore
$$
\boxed{\widetilde{f*g}(k)=\sqrt{2\pi}\,\widetilde f(k)\widetilde g(k).}
$$
\b[Integral equation.] Put $f_a(x)=1/(x^2+a^2)$. Since this is even, the left side is $f_a*g$. The transformed equation is
$$
\frac\pi a e^{-a|k|}\widetilde g(k)=\frac{\sqrt{2\pi}}{2b}e^{-b|k|},
$$
so
$$
\widetilde g(k)=\frac a{b\sqrt{2\pi}}e^{-(b-a)|k|}.
$$
The assumption $b>a$ makes the new scale $d=b-a$ positive. Inverting with the previously established transform pair gives
$$
\boxed{g(x)=\frac{a(b-a)}{\pi b\,[x^2+(b-a)^2]}.}
$$
The <Cauchy convolution semigroup> gives a quick verification. With $K_c(x)=c/[\pi(x^2+c^2)]$, the formula is $g=(a/b)K_{b-a}$ and $f_a=(\pi/a)K_a$. Hence $f_a*g=(\pi/b)K_b=1/(x^2+b^2)$, as required. The solution is unique among integrable <functions> because the multiplier $\widetilde f_a(k)$ never vanishes.
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