= Solution
Use <differentiation under the integral sign>, justified by the integrable bound $|x\sin(tx)e^{-x^2/2}|\leq xe^{-x^2/2}$. This gives
$$
F'(t)=-\int_0^\infty x\sin(tx)e^{-x^2/2}\,dx
=\int_0^\infty\sin(tx)\,d(e^{-x^2/2}).
$$
<Integration by parts> makes the <boundary> term vanish: <sine> vanishes at zero and the Gaussian tends to zero at infinity. Therefore
$$
\boxed{F'(t)=-t\int_0^\infty\cos(tx)e^{-x^2/2}\,dx=-tF(t)}.
$$
The <integrating factor> $e^{t^2/2}$ makes $(e^{t^2/2}F)'=0$, so $F(t)=Ce^{-t^2/2}$. At $t=0$, the given half-line <Gaussian integral> fixes $C=\sqrt{\pi/2}$. Thus
$$
\boxed{F(t)=\sqrt{\frac\pi2}\,e^{-t^2/2}}.
$$
This is the even cosine-transform form of the <Fourier transform of a Gaussian>.
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