Evaluate $\int_a^1 \theta^{-n} e^{-\frac{1}{(\theta-1)^2}} d\theta, 0In short, I'm interested in the rate of this thing goes to 0 as $a \to 1$. I'm able to get an approximation by $\int_a^1 \theta^{-n} e^{-\frac{1}{(\theta-1)^2}} d\theta \le e^{-\frac{1}{(a-1)^2}}\int_a^1 \theta^{-n}  d\theta$. But I'm just curious if it's possible to have an exact expression or simply a tighter bound. It looks a bit like the Laplace function but not quite. I'm not very good with integral. Thank you very much for the help.
 A: Yes, this bound is not very good since the exponential part in the integrand makes the decay much faster than the $\sim (1-a)$ you get from the regular part left inside. What will get you closer is Taylor expanding the exponent in the integral about $a$ to get $$\frac{1}{(a-1)^2} - \frac{2}{(a-1)^3}(\theta-a).$$ The $\theta^{-n}$ in the integral can be omitted in the leading order, as can the effect of the limit of the integral so doing the integral from $-\infty$ to $0$ gives $$ \frac{(1-a)^3}{2}e^{-1/(1-a)^2},$$ which is two powers of $(1-a)$ faster convergence to zero than the upper bound you got.
It isn't an upper bound (if anything a lower bound, actually), but I believe it is the correct leading order in an expansion like $$ \frac{(1-a)^3}{2}e^{-1/(1-a)^2}(1+O(1-a)).$$ Note the $n$-dependence will only appear at higher order.

After a few more minutes of thinking, it occurs to me that you can rewrite the integral as $$ \int_{1/(1-a)^2}^\infty \frac{e^{-x}}{2x^{3/2}(1-1/\sqrt{x})^n}dx$$ which admits a nice orderly expansion in $n$ under the integral sign to give an expansion in incomplete Gamma functions $\int_z^{\infty} e^{-x}/x^{3/2+n/2}dx,$ whose asymptotics as $z\to\infty$ you could calculate or look up.
