Speed of convergence of an integral (whose complete version gives the Mascheroni constant) Consider the following integral: $$ I(k):= \int_k^{+\infty} \frac{e^{-1/x} \log(x)}{x^2} dx\\\ =-\int_0^{1/k} e^{-t} \log t \, dt, $$ where $k>0$. It is known that $\lim_{k \to 0} I(k)=\gamma$, where $\gamma$ denotes the Euler-Mascheroni constant. Henceforth, it must be that $\lim_{k\to +\infty}I(k)=0$. At which rate does $I(k)$ decays to $0$ as $k \to +\infty$? I guess there's some known result about that.

If $k$ is large, $t$ is near zero, so $I(k)$ is roughly (to $0$th order) equal to $$-\int_0^{1/k} \log t \,dt = -\frac{1}{k}\left(\log\left(\frac{1}{k}\right) - 1\right)=\frac{1}{k}(\log k + 1)$$

That is, $I(k)$ is $\Theta\left(\frac{\log k}{k}\right)$.