Asymptotics of a divergent integral

Question

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!

Answer 1

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.

Answer 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,

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