I am trying to find a good upper bound on \begin{align*} f(x)={\rm erf}\left(\frac{x+d}{b}\right)-{\rm erf} \left(\frac{x-d}{b}\right) \end{align*} here $d>0$
I know that $f(x)$ is symmetric around origin so one bound would be $f(x) \le f(0)$.
But after plotting the function I realized there should be abound of the form \begin{align*} f(x) \le a e^{-|x|^n/c} \end{align*} where $a,n,c>0$.
It's easy to verify that I should be $a=2$ but what about $n$ and $c$?
Note: I added a more accurate upper bound at the end, but it may not be computational useful.
Since $erf(x) =\frac{2}{\sqrt{\pi}}\int_0^x e^{-x^2}dx $,
$\begin{array}\\ f(x) &={\rm erf}\left(\frac{x+d}{b}\right)-{\rm erf} \left(\frac{x-d}{b}\right)\\ &=\frac{2}{\sqrt{\pi}}\int_{(x-d)/b}^{(x+d)/b} e^{-t^2}dt\\ &=\frac{2}{\sqrt{\pi}}\int_{-d/b}^{d/b} e^{-(t+x/b)^2}dt\\ &=\frac{2}{\sqrt{\pi}}\int_{-d/b}^{d/b} e^{-t^2-2tx/b-(x/b)^2}dt\\ &=\frac{2e^{-(x/b)^2}}{\sqrt{\pi}}\int_{-d/b}^{d/b} e^{-t^2-2tx/b}dt\\ &<\frac{2e^{-(x/b)^2}}{\sqrt{\pi}}\int_{-d/b}^{d/b} e^{-2tx/b}dt \quad\text{(since } e^{-t^2} < 1)\\ &=\dfrac{2e^{-(x/b)^2}}{\sqrt{\pi}} \dfrac{e^{-2tx/b}}{-2tx/b}\big|_{-d/b}^{d/b}\\ &=\dfrac{2e^{-(x/b)^2}}{\sqrt{\pi}} \dfrac{e^{2xd/b^2}-e^{-2xd/b^2}}{2tx/b}\\ &=\dfrac{2e^{-(x/b)^2}\sinh(e^{2xd/b^2})}{\sqrt{\pi}tx/b}\\ \end{array} $
This might be reasonable for small $d/b$.
Here is a more accurate upper bound:
$f(x) =\frac{2e^{-(x/b)^2}}{\sqrt{\pi}}\int_{-d/b}^{d/b} e^{-t^2-2tx/b}dt $. Since $e^t \ge 1+t$, $e^{t^2} \ge 1+t^2$, so $e^{-t^2} \le \frac1{1+t^2}$.
Therefore
$$f(x) \le\frac{2e^{-(x/b)^2}}{\sqrt{\pi}}\int_{-d/b}^{d/b} \dfrac{e^{-2tx/b}}{1+t^2}dt $$
However, according to Wolfram Alpha, $$\int \frac{e^{a x}}{1+x^2} dx = \frac{-1}{2 i e^{-i a}} (e^{2 i a} Ei(a (x-i))-Ei(a (x+i)))+constant $$
(where $Ei$ is the exponential integral) so I don't know how useful this is.