Speed of convergence of an integral (whose complete version gives the Mascheroni constant)

Question

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.

Answer 1

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 $\Theta\left(\frac{\log k}{k}\right)$.

Answer 2

Hint: For $x>0$, put $F(x)=\int_0^x\exp(-t)\log tdt$, integrate by parts using that $1-\exp(-t)$ is a primitive of $\exp(-t)$, and show that $$F(x)=(x\log x)(\frac{1-\exp(-x)}{x}-\frac{1}{x\log x}\int_0^x \frac{1-\exp(-t)}{t}dt)$$