The context of the question is that I was trying to derive the asymptotics of the modified Bessel function of the second kind: $$ K_\alpha(z) = \int_0^\infty e^{-z\cosh(t)}\cosh(\alpha t)dt $$ for $z$ small, which Wikipedia gives as $$ K_\alpha(z) \sim \begin{cases} -\log(z/2) - \gamma, &\alpha = 0, \\ \frac{\Gamma(\alpha)}{2}(2/z)^\alpha, &\alpha > 0. \end{cases} $$
My attempt is to make a change of variable $x = \cosh(t)$ so that $t = \text{arccosh}(x)$ and $dx = \sinh(t)dt$ so that $$ K_\alpha(z) = \int_1^\infty e^{-zx}\cosh(\alpha\,\text{arccosh}(x))\frac{dx}{\sinh(\text{arccosh}(x))}. $$ If $\alpha = 0$ for simplicity say then $\sinh(\text{arccosh}(x)) = \sqrt{x^2-1}$ so making another substitution $x = y + 1$ we obtain $$ K_0(z) = e^{-z} \int_0^\infty e^{-zy} \frac{dy}{\sqrt{y(y+2)}} $$ which I recognise as a Laplace transform. How does one proceed from here to obtain (either the $\alpha = 0$ case or the general case) asymptotics for $K_0(z)$ as $z \rightarrow 0^+$? (Is this even the right approach?)
In general is there a way to study the asymptotics of the Laplace transform $\int_0^\infty e^{-zt}f(t)dt$ for $z \rightarrow 0^+$? I'm asking this question since most answers I've found seem to relate to asymptotics as $z \rightarrow \infty$ and I'm also interested in a more general discussion of techniques to study asymptotics of integrals of this form.
Let's consider the case $\alpha=0$ first. $$K_0(z) = \int_0^\infty e^{-z\cosh t}dt$$ Making the substitution $x=\cosh t$ $$K_0=\int_1^\infty e^{-zx}\frac{dx}{\sqrt{x^2-1}}=e^{-z}\int_0^\infty e^{-zt}\frac{dt}{\sqrt{t+2}\sqrt t}=2e^{-z}\int_0^\infty e^{-2zx^2}\frac{dx}{\sqrt{x^2+1}}$$ Using $d\big(\ln(t+\sqrt{t^2+1})\big)=\frac{dt}{\sqrt{1+t^2}}$ and integrating by part (also making several substitutions) $$K_0=2e^{-z}e^{-2zt^2}\ln(t+\sqrt{t^2+1})\Big|_{t=0}^\infty+8ze^{-z}\int_0^\infty e^{-2zt^2}\ln(t+\sqrt{t^2+1})tdt$$ $$=2e^{-z}\int_0^\infty e^{-x}\ln\frac{\sqrt x+\sqrt{x+2z}}{\sqrt{2z}}dx=-e^{-z}\ln(2z)+2e^{-z}\int_0^\infty e^{-x}\ln(\sqrt x+\sqrt{x+2z})dx$$ In the second integral the main contribution comes from the region $x>z$. Indeed, $$|\int_0^{2z} e^{-x}\ln(\sqrt x+\sqrt{x+2z})dx|<|\int_0^{2z}\ln(\sqrt {2z})dx|<|\int_0^{2z}\ln(\sqrt {2x})dx|=O(z\ln z)\ll1$$ Therefore, to find the main asymptotics term, we can drop $2z$ in the second integral and (with the required accuracy) integrate from zero to $\infty$. Also, $e^{-z}=1+O(z)\sim 1$. Taking all together $$K_0\sim-\ln(2z)+2\int_0^\infty e^{-x}\ln(2\sqrt x)dx=-\ln(2z)+2\ln2+\int_0^\infty e^{-x}\ln xdx$$ $$K_0\sim \ln\frac{2}{z}-\gamma$$ For $\alpha> 0$ we are acting in the same way. $$K_\alpha(z) = \int_0^\infty e^{-z\cosh(t)}\cosh(\alpha t)dt=\int_0^\infty\frac{e^{-z-x}\cosh\Big(\alpha\cosh^{-1}(1+\frac{x}{z})\Big)}{\sqrt x\sqrt{x+2z}}dx$$ Using $\cosh^{-1}(a)=\ln(a+\sqrt{a^2-1})$ (we consider $x>0$), after easy transformations we get $$K_\alpha(z)=\frac{e^{-z}}{2}\int_0^\infty\frac{dx\,e^{-x}}{\sqrt x\sqrt{x+2z}}\bigg(\Big(1+\frac{x}{z}+\sqrt{\frac{x^2}{z^2}+\frac{2x}{z}}\,\,\Big)^\alpha+\Big(1+\frac{x}{z}+\sqrt{\frac{x^2}{z^2}+\frac{2x}{z}}\,\,\Big)^{-\alpha}\bigg)$$ Again, the main contribution to the integral comes from the region $x>z$ $$\int_0^{2z}\frac{dx\,e^{-x}}{\sqrt x\sqrt{x+2z}}\bigg(\Big(1+\frac{x}{z}+\sqrt{\frac{x^2}{z^2}+\frac{2x}{z}}\,\,\Big)^\alpha+\Big(1+\frac{x}{z}+\sqrt{\frac{x^2}{z^2}+\frac{2x}{z}}\,\,\Big)^{-\alpha}\bigg)$$ $$<\frac{1}{2}\int_0^{2z}\frac{dx}{\sqrt x\sqrt{x+2z}}\Big((2+\sqrt3)^\alpha+1\Big)<\frac{(2+\sqrt3)^\alpha+1}{2\sqrt{2z}}\int_0^{2z}\frac{dx}{\sqrt x}=O(1)$$ Therefore, keeping the main terms ($\frac{x}{z}>1$) in the integrand, we get the main asymptotics term: $$K_\alpha(z)\sim\frac{1}{2}\int_0^\infty\frac{e^{-x}\big(2\frac{x}{z}\big)^\alpha}{x}dx=\frac{1}{2}\Big(\frac{2}{z}\Big)^\alpha\int_0^\infty e^{-x}x^{\alpha-1}dx=\frac{\Gamma(\alpha)}{2}\Big(\frac{2}{z}\Big)^\alpha\gg1$$