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Let $a \in \mathbb{R}$ and $r>0$. Could we find a constant $C>0$ and $0<n<1$ such that \begin{equation}\int_{a-r}^{a+r} e^{-x^2} dx \leq Cr^n.\end{equation} for all $r>0$.

My effort: I try for $r>1$. Since $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$, we have \begin{equation} \int_{a-r}^{a+r} e^{-x^2} dx \leq \sqrt{\pi} \leq \sqrt{\pi}r^{1/2}. \end{equation} How to find $C$ and $n$ for $r>0$?

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For starters, you can consider $a=0$ as that will make the integral as large as possible. If you handle that, any other $a$ will work. – Ross Millikan Feb 10 '13 at 1:49
up vote 1 down vote accepted

Since we were not asked for the best $C,n$ that works, we can be sloppy. Note that if $n=0$ were allowed, we could just take $n=0, C=\sqrt \pi$ and be done. We can also note that the integral is less than $2r$ as the integrand is less than $1$ everywhere but $0$. Also, for $r \lt 1, \sqrt r \gt r$. So let us choose $C=2, n=\frac 12$. For $r \lt 1,\int_{-r}^{r} e^{-x^2} dx\lt 2r \lt 2\sqrt r =Cr^{\frac 12}$. For $r \ge 1,\int_{-r}^{r} e^{-x^2} dx\lt \sqrt \pi \lt Cr^{\frac 12}$. I'm sure there are tighter limits.

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