I am trying to compute the following integral during my research, I have some intuitive belief that this limit will goes to 0 as x increases. However, I failed to prove it, any one has some ideas?
$$\lim_{x\rightarrow \infty} \int_0^\infty \frac{\sqrt{x} \exp(-0.5 x w^2)}{w^{p} K_p(w)} d w = 0$$
Notice here:
$K_p$ is the modified Bessel function with $p \in (0, 0.5)$. $w^p K_p(w)$ is part of Matern covariance function.
$\int_0^\infty \sqrt{x} \exp(-0.5 x w^2) d w $ is some constant does not depend on $x$ (Gaussian density).
The limit is not zero. Assume that $p$, with $\operatorname{Re}(p)>0$, is fixed. First we note that $$ \mathop {\lim }\limits_{w \to 0^ + } w^p K_p (w) = 2^{p - 1} \Gamma (p), $$ and $$ K_p (w) \sim \sqrt {\frac{\pi }{{2w}}} {\rm e}^{ - w} ,\quad w \to + \infty $$ (see, for instance, $\S10.30$ of the DLMF). Now consider $$ \sqrt x \int_0^{ + \infty } {{\rm e}^{ - \frac{1}{2}xw^2 } \left| {\frac{1}{{w^p K_p (w)}} - \frac{{2^{1 - p} }}{{\Gamma (p)}}} \right|{\rm d}w} = \int_0^{ + \infty } {{\rm e}^{ - \frac{1}{2}t^2 } \left| {\frac{1}{{(t/\sqrt x )^p K_p (t/\sqrt x )}} - \frac{{2^{1 - p} }}{{\Gamma (p)}}} \right|{\rm d}t}. $$ From the asymptotic properties of the modified Bessel function, $$ \left| {\frac{1}{{(t/\sqrt x )^p K_p (t/\sqrt x )}} - \frac{{2^{1 - p} }}{{\Gamma (p)}}} \right| \ll (1 + t)^{1/2 - p} {\rm e}^t $$ for all $t>0$, uniformly in $x\ge 1$, say. Consequently, by the Lebesgue dominated convergence theorem, $$ \mathop {\lim }\limits_{x \to + \infty } \sqrt x \int_0^{ + \infty } {{\rm e}^{ - \frac{1}{2}xw^2 } \left| {\frac{1}{{w^p K_p (w)}} - \frac{{2^{1 - p} }}{{\Gamma (p)}}} \right|{\rm d}w} = 0. $$ Therefore, $$ \mathop {\lim }\limits_{x \to + \infty } \sqrt x \int_0^{ + \infty } {{\rm e}^{ - \frac{1}{2}xw^2 } \frac{{{\rm d}w}}{{w^p K_p (w)}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt x \int_0^{ + \infty } {{\rm e}^{ - \frac{1}{2}xw^2 } \frac{{2^{1 - p} }}{{\Gamma (p)}}{\rm d}w} = \frac{{2^{1/2 - p} \sqrt \pi }}{{\Gamma (p)}}. $$