We want to show that $$F(x)=\sqrt{\pi } \sum _{n=0}^{\infty } \frac{(-1)^n H_{2 n+1}(x)}{2^{3 n+3} \Gamma \left(n+\frac{3}{2}\right)}$$ where $F(x)$ is the Dawson Integral ($F(x)=\exp \left(-x^2\right) \int_0^x \exp \left(u^2\right) \, du$) and $H_n(x)$ is the $n$ th Hermite polynomial.
Since $F(x)$ is odd, we can write $F(x)=\sum _{n=0}^{\infty } c_{2 n+1} H_{2 n+1}(x).$ where $$c_{2 n+1}=\frac{\int_{-\infty }^{\infty } \exp \left(-x^2\right) F(x) H_{2 n+1}(x) \, dx}{2^{2 n+1} (2 n+1)! \sqrt{\pi }}$$ Employing the generating function for the Hermite polynomials we can write $$\int_{-\infty }^{\infty } \exp \left(-(x-t)^2\right) F(x) \, dx=\sum _{n=0}^{\infty } \frac{t^n \int_{-\infty }^{\infty } F(x) \exp \left(-x^2\right) H_n(x) \, dx}{n!}$$ How do we proceed to get the coefficient of $t^n$ ($n$ odd) on the left hand side ?
We will proceed to evaluate the integral appearing above in $c_{2 n+1}$. The Dawson Integral can be expressed in the form $$F(x)=\int_0^{\infty } \exp \left(-y^2\right) \sin (2 x y) \, dy$$ Hence, $$\int_{-\infty }^{\infty } \exp \left(-x^2\right) F(x) H_{2 n+1}(x) \, dx=\int_0^{\infty } \exp \left(-y^2\right) \int_{-\infty }^{\infty } \exp \left(-x^2\right) H_{2 n+1}(x) \sin (2 x y) \, dx \, dy$$ To evaluate the $x$-integral on the right hand side we invoke the generating function for the Hermite polynomials. $$\exp \left(-x^2\right) \sin (2 y x) \exp \left(2 x t-t^2\right)=\sum _{n=0}^{\infty } \frac{H_n(x) t^n \exp \left(-x^2\right) \sin (2 y x)}{n!}$$ Integrating both sides with respect to $x$ from $-\infty$ to $\infty$ we obtain $$\int_{-\infty }^{\infty } \exp \left(-(x-t)^2\right) \sin (2 y x) \, dx=\sum _{n=0}^{\infty } \frac{t^n \int_{-\infty }^{\infty } H_n(x) \exp \left(-x^2\right) \sin (2 y x) \, dx}{n!}$$ The integral on the left is readily evaluated as $\sqrt{\pi } \exp \left(-y^2\right) \sin (2 t x y)$. Expanding this expression in a power series in $t$ we obtain the realtion $$\sqrt{\pi } \exp \left(-y^2\right) \sum _{n=0}^{\infty } \frac{(-1)^n (2 y t)^{2 n+1}}{(2 n+1)!}=\sum _{n=0}^{\infty } \frac{t^{2 n+1} \int_{-\infty }^{\infty } H_{2 n+1}(x) \exp \left(-x^2\right) \sin (2 y x) \, dx}{(2 n+1)!}$$ where we have recognized that the integral vanishes for even values of $n$. Equating coefficient of like powers of $t$ we have $$\int_{-\infty }^{\infty } H_{2 n+1}(x) \exp \left(-x^2\right) \sin (2 y x) \, dx=(-1)^n \sqrt{\pi } \exp \left(-y^2\right) (2 y)^{2 n+1}$$ We then obtain $$\int_{-\infty }^{\infty } \exp \left(-x^2\right) F(x) H_{2 n+1}(x) \, dx=(-1)^n \sqrt{\pi } 2^{2 n+1} \int_0^{\infty } \exp \left(-2 y^2\right) y^{2 n+1} \, dy$$ The integral on the right hand side evaluates to $(-1)^n \sqrt{\pi } 2^{n-1} n!$ Finally, we find the desired coefficient $$c_{2 n+1}=\frac{(-1)^n n!}{2^{n+2} (2 n+1)!}$$ which can be shown to be equivalent to that given in the original problem statement