 Solution

For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence