The integral

$$I = \int_{-\infty}^\infty \frac{e^{-\varepsilon x^2}} { \sqrt{1+x^2} } dx$$

is convergent for $\varepsilon > 0$ and can even be given in terms of the Bessel function $K_0$. As $\varepsilon \to 0$ it is divergent and $I \sim -\log \varepsilon$. What would be the simplest way to derive the above leading term in an asymptotic $\varepsilon \to 0$ expansion directly in terms of the above integral?

Clearly, if one uses the exact result in terms of $K_0$ and then for instance uses the second order differential equation satisfied by $K_0$ it is quite simple to derive the $\log\varepsilon$ form of the divergence. But I'm looking for a way to derive this directly from the above integral. The reason is that I have a much more complicated integral to analyze which can not be given in a closed form but the above simple integral captures its difficulty so would like to understand this one first.

Another related question: the integral

$$J = \int_0^{2\pi} e^{-\sin(x)^2/\varepsilon^2}dx $$

can also be evaluated exactly in terms of the Bessel function $I_0$ which result will imply $J \sim \varepsilon$ as $\epsilon\to 0$. But again, directly from the integral what is the simplest way to see this leading term in the $\varepsilon\to 0$ expansion?