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I'm trying to calculate the asymptotics of the following integral: For $\alpha \in (0,1/2)$,

$$I(\epsilon) = \int^1_\epsilon s^{\alpha -3/2} \exp \left\{ -\frac{s^{2\alpha -1}}{2} \right\} \exp \left\{ \frac{\alpha s^{2 \alpha - 2}}{8} \right\} ds \qquad \text{as }\epsilon \searrow 0.$$

How quickly does it go to $\infty$ as $\epsilon \searrow 0$?

Mathematica fails to calculate the integral analytically, but numerical calculations give $$ \frac{\log \log I(\epsilon)}{\log \epsilon} \sim 2 \alpha - 1.8 \quad (\text{1.8 is not exact}) $$ or what is the same, $$ I(\epsilon) \sim \exp\{ \epsilon^{2\alpha - 1.8} \} \qquad \text{as }\epsilon \searrow 0.$$

Any ideas on how to show this analytically? Many thanks!

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Non-formal approach: the second factor is a continuous function, so we can discard it. The third one diverges much faster than the first one, so we can ignore the effect from the first one, too. Thus, we need to estimate the divergence rate of the integral of the third factor, which should be much simplier. –  TZakrevskiy Jul 1 '13 at 11:56
    
Substitute $u=s^{2\alpha-2}$. This turns $s^{2\alpha-1}$ into $u^\beta$. Expand the first exponential into a Taylor series in $u^\beta$, and you are left with a bunch of integrals of $\exp(\alpha u/8) u^{\theta}$, integrated from 1 to $\epsilon^{2\alpha-2}$. Each term looks a bit like a gamma function times the $\exp(\alpha u/8)$. –  Michael Jul 1 '13 at 12:22
    
I fail to understand your result, which appears to imply that the integral approaches $1$ as $\epsilon \to 0$. or am I missing something? –  Ron Gordon Jul 1 '13 at 15:04

2 Answers 2

I would start by substituting $t \mapsto 1/s$ to get

$$I(\epsilon) = \int_1^{1/\epsilon} \frac{dt}{t^{\alpha+1/2}} \, e^{-t^{1-2 \alpha}/2} e^{\alpha t^{2-2 \alpha}/8}$$

Now recognize that the rightmost term in the integrand dominates as $\epsilon \to 0$, so ignore the subdominant exponential to recognize that

$$I(\epsilon) \sim \int_1^{1/\epsilon} \frac{dt}{t^{\alpha+1/2}} e^{\alpha t^{2-2 \alpha}/8} \quad (\epsilon \to 0)$$

Rescale by letting $u \mapsto t^{2-2 \alpha}$; we now have

$$I(\epsilon) \sim \int_1^{1/\epsilon^{2-2 \alpha}} du \, u^{-(3/2 + \alpha)/(2-2 \alpha)} \, e^{\alpha u/8} $$

Upon an integration by parts, we see that

$$I(\epsilon) \sim \frac{8}{\alpha} \epsilon^{\frac{3}{2}+\alpha} \exp{\left(\frac{\alpha}{8 \epsilon^{2-2 \alpha}} \right)} + \frac{8}{\alpha} \frac{\frac{3}{2}+\alpha}{2-2 \alpha}\int_1^{1/\epsilon^{2-2 \alpha}} du \, e^{\alpha u/8}\, u^{-(3/2 + \alpha)/(2-2 \alpha) - 1} $$

The term in the right is subdominant to that on the left; thus we may say that, to leading order,

$$I(\epsilon) \sim \frac{8}{\alpha} \epsilon^{\frac{3}{2}+\alpha} \exp{\left(\frac{\alpha}{8 \epsilon^{2-2 \alpha}} \right)} \quad (\epsilon \to 0)$$

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Thanks Ron. This is the best I have been able to do analytically too. But numerics suggest the correct asymptotics are $I(\epsilon) = o( \exp(\kappa \epsilon^{2 \alpha - 1.8}))$. Any ideas? –  zab Jul 1 '13 at 17:24
1  
@zab: The numerics simply make no sense in that case. I do not have access to Mathematica right now where I could back this up, but try a semilog plot of the data, i.e. $$\log[\epsilon^{-3/2-\alpha} I(\epsilon)] = \log{(8/\alpha)} + \frac{\alpha}{8 \epsilon^{2-2 \alpha}}$$ –  Ron Gordon Jul 1 '13 at 17:40
    
Thanks Ron. Trying it now. –  zab Jul 1 '13 at 18:55

Using that procedure, with $\int_1^N u^m e^u du$ approximated by $N^m e^N$, I got this answer:

$$\left(\frac{8}{\alpha}\right)^{\gamma}\frac{1}{2-2\alpha} \exp\left(\frac{3-2\alpha}{4\alpha-4}\log\epsilon+\frac{\alpha \epsilon^{2\alpha-2}}{8} -\frac{1}{2} \left(\frac{8}{\alpha}\right)^{\beta}\epsilon^{2\alpha-1}\right)\\ \beta = \frac{2\alpha-1}{2\alpha-2}\\ \gamma = \frac{2\alpha-1}{4\alpha-4} $$ But I don't know whether my approximations are bigger or smaller than the terms I kept, except the biggest term.

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