This question is related to this previous question of mine.
Suppose I define the function $G : [1,\infty) \times \mathbb{R} \to \mathbb{R}$ as the following integral: $$ G(x,y) = \int_x^\infty \frac{\ln(t)^2}{(t^2 -1) ( \ln(t)^2 + y^2 )^2} \,dt $$
Again by goofing around on Wolfram Alpha, I know the following facts about this function:
- $G(x,y) \to 0$ as $x \to \infty$, for all $y \in \mathbb{R}$ (even $y=0$).
- $G(1,y)$ is of finite value for all $y \neq 0$. In fact, I think that this function has a singularity (over the domain $[1,\infty) \times \mathbb{R}$) $only$ at the point $(x,y) = (1,0)$
- $G(x,y) \to 0$ as $y \to \infty$ for all $x \in [1, \infty)$
My question: what are the asymptotics of this function for $x \to \infty$?
I know that $$ \frac{\ln(t)^{2}}{(t^{2} -1) ( \ln(t)^2 + y^{2} )^2} \sim \frac{\ln(t)^{2}}{t^{2} ( \ln(t)^2 + y^{2} )^2} $$ as $t \to \infty$. I am unsure how I can use this to give the asymptotics of the function...is there a way to write the asymptotics as some function $f(x,y)$, which is valid for $x \to \infty$ and $all$ $y \in \mathbb{R}$? I am confused how I can understand asymptotics of a such a function when there are two variables involved.
Accurate approximates of the integral can be obtained from asymptotic series (below). One have to chose the number of terms $K$ and $H$ : $$ G(x,y) = \int_x^\infty \frac{\ln(t)^2}{(t^2 -1) ( \ln(t)^2 + y^2 )^2} \,dt $$ $$G(x,y)\simeq \sum_{k=0}^K\sum_{h=0}^H\frac{(-1)^h(h+1)}{(2k+1)^{2h-1}}y^{2h}\Gamma\left(-2h-1\:,\: (2k+1)\ln(x) \right)$$ where $\Gamma$ is the Incomplete Gamma function. The first terms are : $$G(x,y)\simeq \Gamma\left(-1\:,\:\ln(x)\right) -2y^2\Gamma\left(-3\:,\:\ln(x)\right) +3y^4\Gamma\left(-5\:,\:\ln(x)\right)+...\\+\Gamma\left(-1\:,\:3\ln(x)\right)-\frac{2}{3}y^2\Gamma\left(-3\:,\:3\ln(x)\right)+...$$ For $x$ large, all the Gamma terms are small.
Note: For a good approximate it is better to chose large $H$ and small $K$ than large $K$ and small $H$.
Asymptotic series expansion: