= Solution
{c}
The theorem gives
$$
F'(t)=-2\int_0^\infty xe^{-x^2}\sin(2xt)dx
=\int_0^\infty (e^{-x^2})'\sin(2xt)dx.
$$
<Integration by parts> has zero boundary term: the sine vanishes at zero and the <Gaussian> tends to zero at infinity. Hence $F'(t)=-2t\int_0^\infty e^{-x^2}\cos(2xt)dx=-2tF(t)$.
The same <Gaussian> majorant and dominated convergence give <continuity> at zero, where the allowed <Gaussian integral> gives $F(0)=\sqrt\pi/2$. The differential equation implies $(e^{t^2}F(t))'=0$; <continuity> at zero fixes that constant. Thus
$$
\boxed{F(t)=\frac{\sqrt\pi}{2}e^{-t^2}\qquad(0\leq t<b).}
$$
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