Can you give me some hint on how to show that $$\lim_{y\to0^+}\frac{\int_0^\infty \exp(-y\cosh (x))\text dx}{\log y}=-1?$$
I tried to delimit from above and from below the function $x\mapsto−y\cosh(x)$ with some simple functions (simple means for which I can explicitly evaluate the integral by anti-differentiation) whose integrals from 0 to ∞ are asymptotic to $-\log y$ as $y$ approaches $0+$. But I failed in finding them.
Tricky problem. By using the change of variable $x=\text{arccosh}(u)$ we are left with: $$ \int_{1}^{+\infty}\frac{e^{-yu}}{\sqrt{u^2-1}}\,du=e^{-y}\int_{0}^{+\infty}\frac{e^{-yu}}{\sqrt{u(u+2)}}\,du\tag{1} $$ hence we just need to study the asymptotic behaviour of the Laplace transform of $\frac{1}{\sqrt{u(u+2)}}$ in a right neighbourhood of the origin. It is well-known that: $$ \mathcal{L}\left(\frac{1}{\sqrt{u(u+2)}}\right) = e^y\,K_0(y), \tag{2}$$ where $K_0$ is a modified Bessel function of the second kind. You can recover the asymptotic form by applying a modified form of Laplace method: as stated by Wikipedia, too (last line before the "Properties" paragraph) if $y$ is in a right neighbourhood of the origin we have: $$ K_0(y)\approx-\log\left(\frac{y}{2}\right)-\gamma \tag{3}$$ and the result follows.