unique root problem of a special function involving the error function I was looking for the root of the following function on the domain $x\geq 0$:
$$F(x)=(x+a)e^{x^2}(1−erf(x))−\frac{b}{\sqrt π}$$
where
$$erf(x)=\frac{2}{\sqrt \pi}\int_0^x e^{-t^2}dt$$
is the familiar error function. Also $a>0$, $0<b<1$.
I tried several numerical solutions for different values of $a$ and $b$. It seems that there is at most one root on $[0,\infty)$. However I am not able to prove it since $F(x)$ is not monotone in $x$. Is there any hints about the proof?
What I know about this function is
$$F(0)=a-\frac{b}{\sqrt\pi},F(\infty)=\frac{1-b}{\sqrt\pi}>0.$$
Thank you!
 A: Since the standard asymptotic expansion
as $x \to \infty$ is
$1-erf(x)
=\dfrac{e^{-x^2}}{x\sqrt{\pi}}(1-\dfrac1{2x^2}+O(\dfrac1{x^4}))
$
(see
https://en.wikipedia.org/wiki/Error_function#Asymptotic_expansion
),
$\begin{array}\\
F(x)
&=(x+a)e^{x^2}(1−erf(x))−\frac{b}{\sqrt π}\\
&=(x+a)e^{x^2}(\dfrac{e^{-x^2}}{x\sqrt{\pi}}(1-\dfrac1{2x^2}+O(\dfrac1{x^4})))−\frac{b}{\sqrt π}\\
&=\dfrac{(x+a)}{x\sqrt{\pi}}((1-\dfrac1{2x^2}+O(\dfrac1{x^4})))−\frac{b}{\sqrt π}\\
&=\dfrac{1+a/x}{\sqrt{\pi}}((1-\dfrac1{2x^2}+O(\dfrac1{x^4})))−\frac{b}{\sqrt π}\\
&=\dfrac{1-b}{\sqrt{\pi}}+\dfrac{a}{x\sqrt{\pi}}-\dfrac{1}{2\sqrt{\pi} x^2}+O(\dfrac1{x^3})\\
\end{array}
$
For a root,
approximately
$0
=(1-b)+a/x
$
or
$x = -a/(1-b)
$.
This seems to show that
there is no root for large $x$.
If another term is taken,
$0
=\dfrac{1-b}{\sqrt{\pi}}+\dfrac{a}{x\sqrt{\pi}}-\dfrac{1}{2\sqrt{\pi} x^2}
$
or
$0
=x^2(1-b)+ax-\frac12
$.
The root of this is
$x
=\dfrac{-a+\sqrt{a^2+2(1-b)}}{2(1-b)}
$.
You might try the
power series expansions
for small $x$
of
$erf(x)
= \dfrac{2}{\sqrt{\pi}}(x-\dfrac{x^3}{3}+O(x^5))
$
and
$e^{-x^2}
=1-x^2+O(x^4)
$.
