Name of special function used by Wolfram integrator Integrating $e^{-r}/\sqrt{2t-r}$ with respect to $r$ between $r=t$ and $r=2t$ using this widget gives the answer $2e^{-t}F(\sqrt{t})$. However the widget doesn't say what $F$ is. 
I have looked on Wolfram's documentation on common special functions
but there's nothing there resembling a plain $F$.
What is this function $F$?
Many thanks for your help.
 A: The function $F$ is the Dawson integral or Dawson function (references: reference.wolfram.com, MathWorld, Wikipedia), given by
$$
F(x) = D_+(x) = e^{-x^2} \int_0^x e^{t^2}dt.
$$
You can find this out by using the main Wolfram Alpha interface rather than the widget you linked to (the circled text below will be interactive):


Remarks on a few properties of $F$, since you expressed an interest in the comments: 
Clearly $F(x)>0$ for all $x>0$. So to prove that $\lim_{x\to\infty}F(x)=0$, it suffices to show that given a fixed $\epsilon > 0$, we have $F(x) < 2\epsilon$ for all sufficiently large $x$. Since $t \mapsto e^{t^2-x^2}$ is an increasing function $[0, x] \to (0,1]$,
$$
F(x)
= \int_0^{x-\epsilon} e^{t^2 - x^2} dt + \int_{x-\epsilon}^x e^{t^2 - x^2} dt
< x e^{(x-\epsilon)^2-x^2} + \epsilon.
$$
Now
$x e^{(x-\epsilon)^2-x^2} = x e^{- 2x\epsilon + \epsilon^2} = \frac{\exp(\epsilon^2)}{2\epsilon} y e^{-y}$ where we have set $y = 2\epsilon x$. Thus we can use the fact that $\lim_{y \to \infty} y e^{-y} = 0$ to obtain the desired bound.
As a consequence, $\lim_{t \to \infty} t^3 e^{-t} F(\sqrt t) = 0$ because both $t^3 e^{-t}$ and $F(\sqrt t)$ tend to $0$ as $t \to \infty$.
