= Solution
On any bounded parameter interval $|t|\leq T$, the <derivative> of the integrand has the integrable majorant
$$
\left|\frac\partial{\partial t}\bigl[e^{-x^2}\cos(2xt^2)\bigr]\right|
=|4xt e^{-x^2}\sin(2xt^2)|\leq4Tx e^{-x^2}.
$$
The <dominated convergence theorem> therefore justifies <differentiation under the integral sign> at every real $t$, giving
$$
f'(t)=-4t\int_0^\infty xe^{-x^2}\sin(2xt^2)\,dx.
$$
Use <integration by parts> in this absolutely convergent integral, with $xe^{-x^2}\,dx=-\tfrac12d(e^{-x^2})$. Both boundary terms vanish, since the sine is zero at $x=0$ and the Gaussian tends to zero at infinity. Hence
$$
\int_0^\infty xe^{-x^2}\sin(2xt^2)\,dx
=t^2\int_0^\infty e^{-x^2}\cos(2xt^2)\,dx=t^2f(t).
$$
It follows that $f'=-4t^3f$, or equivalently $(e^{t^4}f(t))'=0$. This identity holds on all of $\mathbb R$, including zero. With the supplied Gaussian integral value,
$$
\boxed{f(t)=\frac{\sqrt\pi}{2}e^{-t^4}\qquad(t\in\mathbb R).}
$$
The fourth power comes from the printed phase $2xt^2$; the parameter has been differentiated with that phase throughout.
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