Dawson function

Remarks

Example: $F(0) = 0$, and $F$ is odd because $e^{t^2}$ is even. The product rule gives the differential equation $F'(x) = 1 - 2x F(x)$, with $F(0) = 0$.

Non-example: $x \mapsto e^{x^2} \int_0^x e^{-t^2}\, dt$ is not the Dawson function; the signs of the exponents are swapped, and this function grows without bound.

With the imaginary error function $\operatorname{erfi} x = \frac{2}{\sqrt{\pi}} \int_0^x e^{t^2}\, dt$ one has $F(x) = \frac{\sqrt{\pi}}{2} e^{-x^2} \operatorname{erfi} x$. For large $x$, $F(x) \sim \frac{1}{2x}$. Some texts call this $D_+(x)$ and also use $D_-(x) = e^{x^2} \int_0^x e^{-t^2}\, dt$, which is the non-example above.