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For a given $\alpha \in (0,2)$ How fast does

\begin{equation} \int_{\pi/h}^\infty{\exp(-p^\alpha)}\,\mathrm{d}p \end{equation}

go to zero as $h$ goes to zero? Any upper bound on the speed of convergence would be great!

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1 Answer

Hint: use integration by parts

Hint2: to simplify the calculation. Substitute first $x=p^\alpha$

Hint3: you should get the result $\sim (\pi/h)^{1-\alpha} e^{-(\pi/h)^\alpha} \alpha^{-1}$

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Thanks Fabian, after performing the integration by part and the substitution, I get to: $\int_{(\pi/h)^\alpha}^\infty x^{1/\alpha} e^{-x}\, dx$. How do I proceed from there? –  angry_pacifist Aug 6 '12 at 17:12
    
You should IBP the other way round (integral $\exp(-x)$). The resulting integral $\int_{(\pi/h)^\alpha}^\infty\!dx\, x^{1/\alpha-2} e^{-x}$ you can bound using $|x^{1/\alpha -2}| \leq (\pi/h)^{1-2\alpha}$. –  Fabian Aug 6 '12 at 19:19
    
Thanks but could you explain a bit more? For instance, take $\alpha = \frac{1}{3}$, you get $x^{\frac{1}{\alpha}-2}= x$ which is unbounded on the interval under consideration so there has to be something wrong? –  angry_pacifist Aug 7 '12 at 13:41
    
Thanks for that but I get to $\int_{(\frac{\pi}{h})^h}^\infty x^{\frac{1}{\alpha}-2}e^{-x}dx$ However, $x^{\frac{1}{\alpha}-2}$ is an increasing function for $\alpha \in (0,.5)$; take $\alpha = \frac{1}{3}$ and you obtain $x^{\frac{1}{\alpha}-2} = x$ which is clearly unbounded so there has to be something wrong with your bound, right? –  angry_pacifist Aug 7 '12 at 13:55
    
@angry_pacifist: oh, I was under the impression that $\alpha >1$. For $\alpha\leq 1/2$ you indeed have to be a bit more clever. Just integrate once more by parts and then use the estimate... (then you will manage to show it up to $\alpha > 1/3$) and so on. –  Fabian Aug 7 '12 at 14:57
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