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 $x$?
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$\begingroup$ 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. $\endgroup$– nejimbanCommented Aug 5 at 7:42
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$\begingroup$ if $x>2$, you can bound $\exp(-t^2/2)$ with $\exp(-t/2)$ $\endgroup$– Sine of the TimeCommented Aug 5 at 7:56
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1 Answer
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Your function is:
$$F(x) = \frac{1}{x} \int_{0}^x \text{e}^{-\frac{t^2}{2}} \text{d}t = \sqrt{\frac{\pi}{2}}\cdot\frac{\text{erf}\left(\frac{x}{\sqrt{2}}\right)}{x}$$
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 $x>0$.
