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How to prove that, for any integer $d \geq 1$ and any $\alpha > 1/2$,

$$\int_{n^{\alpha}}^{\infty} r^{d-1} e^{-\frac{r^2 d}{n}} dr$$ goes to $0$ with $n$?

This can be interpreted (up to a multiplicative constant) as the integral of a multivariate gaussian distribution over points having distance $> \sqrt{n}$ from the origin.

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    $\begingroup$ Realise that $\exp(x)=e^x$ $\endgroup$ – Jan Jan 25 '16 at 17:42
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    $\begingroup$ Change variables so that you have $\exp(-r^2/2)$ instead. Then integrate by parts until the exponent on the $r$ is either $0$ or $1$. Then study the result. $\endgroup$ – Ian Jan 25 '16 at 17:48
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Let $u = r^2d/n$ so that

$$ \int_{n^{\alpha}}^{\infty} r^{d-1} e^{-\frac{r^2 d}{n}} \, \mathrm{d}r = \tfrac{1}{2} (n/d)^{d/2} \int_{n^{2\alpha-1}d}^{\infty} u^{(d-2)/2} e^{-u} \, \mathrm{d}u. $$

From the L'hospital's rule, we know that

$$ \frac{\int_{x}^{\infty} u^{(d-2)/2} e^{-u} \, \mathrm{d}u}{x^{(d-2)/2} e^{-x}} \sim 1 \quad \text{as } x \to \infty. $$

Therefore we have

\begin{align*} \int_{n^{\alpha}}^{\infty} r^{d-1} e^{-\frac{r^2 d}{n}} \, \mathrm{d}r &\sim \tfrac{1}{2} (n/d)^{d/2} \cdot (n^{2\alpha-1}d)^{(d-2)/2} e^{-n^{2\alpha-1}d} \\ &= c n^{\beta} e^{-n^{2\alpha-1}d} \end{align*}

for $c = 1/(2d)$ and $\beta = \alpha(d-2)+1$. Taking limit as $n \to \infty$ gives the desired result.

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  • $\begingroup$ Interesting approach, significantly different from mine (in that I left the square in the exponent). +1. $\endgroup$ – Ian Jan 25 '16 at 20:58

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