Quasiconcavity of incomplete gaussian integral From visual experiments , it appears that the set
$$
S_r = \left\{ (x,y) \text{ s.t. } \int_x^y e^{-t^2} dt \geq r \right\}
$$
is convex for $r \geq 0$. Or equivalently, the function
$$
f(x,y) = \int_x^y e^{-t^2} dt = \frac{\sqrt{\pi}}{2}(\operatorname{erf}(y) - \operatorname{erf}(x))
$$
is quasiconcave over the domain $y \geq x$. Is this true? If so, how might one prove this? 
 A: The unimaginative approach is to compute the curvature of level sets of $f$. The formula looks a bit complicated, but since we only care about the sign, the positive denominator can be ignored. Also, $f_{xy}=0$ for this function, leaving us with 
$$
-(f_{xx}f_y^2 + f_{yy}f_x^2) = 2e^{-2x^2-2y^2}(ye^{y^2}-xe^{x^2}) > 0 
$$
(The positive sign means the center of curvature is in the direction of the gradient of $f$.) 
One can modify the above argument to avoid talking about curvature. The  vector $\vec v=(-f_y,f_x)$ determines the direction of tangent line to level curves $f=c$. We want the function to be less than $c$ on this line, at least near the point of tangency -- this will mean that the upper level set is convex. To this end, we need the directional second-order derivative, which is obtained by plugging $\vec v$ into the Hessian of $f$. The Hessian being diagonal, we get
$$\begin{pmatrix} -f_{y} & f_{x}\end{pmatrix}
 \begin{pmatrix} f_{xx} & 0 \\ 0 & f_{yy}\end{pmatrix}\begin{pmatrix} -f_{y} \\ f_{x}\end{pmatrix} = f_{xx}f_y^2 + f_{yy}f_x^2<0
$$
confirming that $f<c$ on the tangent line near the point of tangency.
A: To answer my own question, it looks like another approach is to look at
$$
\log f(x,y)
$$
which it doesn't seem too difficult to prove is concave. This implies that the upper level sets of $ f(x,y) $ are convex.
