Expanding Dawson's integral in a series of Hermite functions 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 ?
 A: 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
