How can we show that $\int_{0}^{\infty}x\sin^2(ax) e^{-x^2}\mathrm dx={aF(a)\over 2}?$ Consider the integral $(1)$, which has a closed form in term of Dawson's integral.

$$\int_{0}^{\infty}x\sin^2(ax) e^{-x^2}\mathrm dx={aF(a)\over 2}\tag1$$
  Where $F(a)$ is the Dawson's Integral

How can we prove $(1)?$
I accidentally found it, while trying to search for some other functions.
 A: We have that $$\begin{align} I\left(a\right)  = & \int_{0}^{x}x\sin^{2}\left(ax\right)e^{-x^{2}}dx \\ = &-\frac{1}{4}\int_{0}^{\infty}xe^{-x^{2}-2iax}dx-\frac{1}{4}\int_{0}^{\infty}xe^{-x^{2}+2iax}dx+\frac{1}{2}\int_{0}^{\infty}xe^{-x^{2}}dx \end{align}$$ so let us analyze $$ I_{1}\left(a\right)= -\frac{1}{4}\int_{0}^{\infty}xe^{-x^{2}-2iax}dx.$$It is not so hard to see that $$ \begin{align} I_{1}\left(a\right)= & -\frac{e^{-a^{2}}}{4}\int_{0}^{\infty}xe^{-\left(x+ia\right)^{2}}dx \\ = & -\frac{e^{-a^{2}}}{4}\int_{0}^{\infty}\left(x+ia\right)e^{-\left(x+ia\right)^{2}}dx+\frac{e^{-a^{2}}ia}{4}\int_{0}^{\infty}e^{-\left(x+ia\right)^{2}}dx \\ = & -\frac{1}{8}+\frac{e^{-a^{2}}ai\sqrt{\pi}\left(1-i\textrm{erfi}\left(a\right)\right)}{8} \end{align}$$ where $\textrm{erfi}\left(z\right)$ is the imaginary error function. In a similar manner $$\begin{align} I_{2}\left(a\right)= & -\frac{1}{4}\int_{0}^{\infty}xe^{-x^{2}+2iax}dx \\ = & -\frac{e^{-a^{2}}}{4}\int_{0}^{\infty}xe^{-\left(x-ia\right)^{2}}dx \\= & -\frac{e^{-a^{2}}}{4}\int_{0}^{\infty}\left(x-ia\right)e^{-\left(x-ia\right)^{2}}dx-\frac{e^{-a^{2}}ia}{4}\int_{0}^{\infty}e^{-\left(x-ia\right)^{2}}dx \\ = & -\frac{1}{8}-\frac{e^{-a^{2}}ai\sqrt{\pi}\left(1+i\textrm{erfi}\left(a\right)\right)}{8} \end{align}$$ and obiously $$I_{3}=\frac{1}{2}\int_{0}^{\infty}xe^{-x^{2}}dx=\frac{1}{4}.$$ So finally $$I\left(a\right)=I_{1}\left(a\right)+I_{2}\left(a\right)+I_{3}=\frac{e^{-a^{2}}a\sqrt{\pi}\textrm{erfi}\left(a\right)}{4}=\color{red}{\frac{aF\left(a\right)}{2}}$$ as wanted.
A: For sure, Marco Cantarini provided the good and formal solution of the problem.
I have a modest solution based on series. Starting using $$\sin^2(ax)=\frac{1-\cos(2ax)}2=\sum_{n=1}^\infty \frac{(-1)^{n-1} 2^{2 n-1} a^{2 n} }{(2 n)!}x^{2 n}$$ we then have $$x \sin^2(ax) e^{-x^2}=\sum_{n=1}^\infty \frac{(-1)^{n-1} 2^{2 n-1} a^{2 n} }{(2 n)!}x^{2 n+1}e^{-x^2}$$ $$I=\int_0^\infty x \sin^2(ax) e^{-x^2}\,dx=\sum_{n=1}^\infty\frac{(-1)^n 2^{2 n-2} a^{2 n} \Gamma (n+1)}{(2 n)!}=-\frac{\sqrt{\pi }} 4\sum_{n=1}^\infty \frac{ (-1)^n a^{2 n}}{ \Gamma \left(n+\frac{1}{2}\right)}$$ which simplifies to $$I=\frac{\sqrt{\pi }}{4}  a\, e^{-a^2} \text{erfi}(a)=\frac{a F(a)}{2}$$
