Upper bound on Gaussian Integral from 0 to x, divided by x

I am looking for upper bounds on $$F(x) = \frac{1}{x} \int_{0}^x e^{-t^2/2} dt$$ in terms of $$x$$. Obviously, $$F(x) \le \frac{\sqrt{\pi}}{2x}$$, but can we get faster decay in terms of $$x$$?

• For $x>1$, $$\int_0^x\mathrm e^{-t^2/2}\,\mathrm dt\ge\int_0^1\mathrm e^{-t^2/2}\,\mathrm dt\ge\int_0^1\mathrm e^{-1/2}\,\mathrm dt=\mathrm e^{-1/2}>0,$$ so $F(x)$ is really of order $\frac1x$ when $x$ is large. Commented Aug 5 at 7:42
• if $x>2$, you can bound $\exp(-t^2/2)$ with $\exp(-t/2)$ Commented Aug 5 at 7:56

$$F(x) = \frac{1}{x} \int_{0}^x \text{e}^{-\frac{t^2}{2}} \text{d}t = \sqrt{\frac{\pi}{2}}\cdot\frac{\text{erf}\left(\frac{x}{\sqrt{2}}\right)}{x}$$

You can use the following known inequality (see this post):

$$\text{erf}(x) \leq \sqrt{1-\text{e}^{-\frac{4}{\pi}x^2}}$$

such that

$$\text{erf}\left(\frac{x}{\sqrt{2}}\right) \leq \sqrt{1-\text{e}^{-\frac{2}{\pi}x^2}}$$

and you end up with:

$$F(x)\leq \sqrt{\frac{\pi}{2}}\cdot\frac{\sqrt{1-\text{e}^{-\frac{2}{\pi}x^2}}}{x}$$

for $$x>0$$.