Answer
$$e^{-1 / \ln 2} $$
Convergent
Work Step by Step
Let $\frac{1}{x}=y,\ \ \ \frac{-1}{x^2}dx=dy$ , then
\begin{align*}
\int_{0}^{\ln 2} x^{-2} e^{-1 / x} d x &= \int_{\infty}^{1 / \ln 2} \frac{y^{2} e^{-y} d y}{-y^{3}}\\
&=\int_{1 / \ln 2}^{\infty} e^{-y} d y\\
&=\lim _{b \rightarrow \infty}\int_{1 / \ln 2}^{b} e^{-y} d y\\
&=\lim _{b \rightarrow \infty}\left[-e^{-y}\right]\bigg|_{1 / \ln 2}^{b}\\
&=\lim _{b \rightarrow \infty}\left[-e^{-b}-\left(-e^{-1 / \ln 2}\right)\right]\\
&=0+e^{-1 / \ln 2}\\
&=e^{-1 / \ln 2}
\end{align*}
So the integral converges.
