I need your help evaluating this integral: $$I=\int_0^\infty F(x)\,F\left(x\,\sqrt2\right)\frac{e^{-x^2}}{x^2} \, dx,\tag1$$ where $F(x)$ represents Dawson's function/integral: $$F(x)=e^{-x^2}\int_0^x e^{y^2} \, dy = \frac{\sqrt{\pi}}{2} e^{-x^{2}} \operatorname{erfi}(x).\tag2$$
Dawson's function can also be represented by the infinite integral $$F(x) = \frac{1}{2} \int_{0}^{\infty} e^{-t^{2}/4} \sin(xt) \, dt.$$
Since $F(x)$ behaves like $x$ near $x=0$ and like $\frac{1}{2x}$ for large values of $x$, we know that integral $(1)$ converges.
$$I=\frac{\pi^{3/2}}8\left(\sqrt2-4\right)+\frac{3\,\pi^{1/2}}2\arctan\sqrt2$$