Let $a>0$ be a small parameter and consider a fixed integer $n\geq 0$. Is it true that $$ \int_0^a \frac{(a-x)^n e^{-1/x}}{x^2}\ dx=n! a^{2n}e^{-1/a}(1+O(a)) $$ as $a\to 0$?
I have verified this for $n\leq 4$, essentially by applying integration by parts with the asymptotic $$ \int_0^a \frac{e^{-1/x}}{x}\ dx=ae^{-1/a}\left[\sum_{k=0}^{2n}(-1)^kk! a^k+O(a^{2n+1})\right] $$
Substitute $\alpha = \frac{1}{a}$ ($\alpha\rightarrow \infty$ as $a\rightarrow 0$)and $t=\frac{1}{x}$ so that the integral transforms as $$I(\alpha)=\frac{1}{\alpha ^n}\int_{\alpha}^{\infty}\big(t-\alpha\big)^nt^{-n}e^{-t}dt.$$ Expanding the first term in the integrand $$I(\alpha)=\frac{1}{\alpha ^n}\Big[\sum_{r=0}^{n}{n\choose r}(-\alpha)^r\int_{\alpha}^{\infty}t^{-r}e^{-t}dt\Big].$$The integral in above is the incmplete gamma function $\Gamma(-r+1, \alpha)$. Rewriting, one has $$I(\alpha)=\frac{1}{\alpha^n}\sum_{r=0}^{n}{n\choose r}(-\alpha)^r \Gamma(-r+1, \alpha).$$ The asymptotics for the incomplete gamma function is $$\Gamma(n, \alpha)\sim \alpha^{n-1}e^{-\alpha}\Big[1+\frac{n-1}{\alpha}+\frac{(n-1)(n-2)}{\alpha^2}+\cdot\cdot\cdot\Big]$$ which in the current case, can written as $$\Gamma(-r+1,\alpha)\sim\alpha^{-r}e^{-\alpha}\sum_{s=0}^{\infty}(-1)^s\frac{(r+s-1)!}{(r-1)!}\frac{1}{\alpha^s}.$$ Using the above expressions, it can be easily verified that $$I(\alpha)\sim\frac{e^{-\alpha}}{\alpha ^n}\Big[\sum_{r=0}^{n}{n\choose r}(-1)^r\sum_{s=0}^{\infty}(-1)^s\frac{(r+s-1)!}{(r-1)!}\frac{1}{\alpha^s}\Big].$$ Upon interchanging the summations, one has to look at sums of the form $$\frac{1}{\alpha^s}\sum_{r=0}^{n}(-1)^r{n\choose r}\frac{(r+s-1)!}{(r-1)!}.$$Now, we may note that $$-\frac{d^s}{dx^s}x^{s-1}(1+x)^n=-\sum_{r=0}^{n}{n\choose r}\frac{(r+s-1)!}{(r-1)!}x^{r-1}$$ The first non vanishing derivative at $x=-1$ occurs for $s=n$. The required estimate follows from here.