@robjohn Is there any reason why we couldn't just do it in two steps and apply Plancherel's theorem twice?
$$ \begin{align} \int_{0}^{\infty} \frac{\sin(2x)\operatorname{Ci}^{2}(x)}{x} \, \mathrm dx &= -\int_{0}^{\infty}\frac{\sin(2x)\operatorname{Ci}(x)}{x} \int_{1}^{\infty} \frac{\cos(xt)}{t} \, \mathrm dt \, \mathrm dx \\ &= - \int_{1}^{\infty} \frac{1}{t} \int_{0}^{\infty} \frac{\sin(2x) \cos(tx)\operatorname{Ci}(x)}{x} \, \mathrm dx \, \mathrm dt \\ &= \int_{1}^{\infty} \frac{1}{t} \left( \int_{0}^{\infty} \frac{\sin(2x) \cos(xt)}{x} \int_{1}^{\infty} \frac{\cos(xu)}{u}\, \mathrm du …