I am interested in the function $F : (1, \infty) \to \mathbb{R}$ defined as: $$ F(x) \ = \ \int_{x}^{\infty} \frac{dt}{\mathrm{ln}(t)( t^2 - 1 )} $$
There is of course, no closed-form expression for the above integral and I am only interested in the behaviour of this function for $x \in (1,\infty)$
By goofing around on wolfram alpha, I've found that:
$f(x)$ seems to have a finite value for all $x \in (1,\infty)$
$f(x) \to 0$ as $x \to \infty$
$f(x) = + \infty$ as $x \to 1^{+}$
Is there a way to understand the asymptotics of this function as $x \to 1^{+}$? Or understanding it's expansion as $x \to \infty$?
I am trying to approximate this function at these two limits, and have never encountered how to deal with approximating functions that are defined as integrals....
For $x\to\infty$, it will behave as $\int_x^\infty\frac{dt}{t^2 \ln t}$, i.e. asymptotically equivalent to $\frac{1}{x\ln x}$. See e.g. this recent question.
For $x\to1^+$, observe that $$ \frac{1}{(t^2-1)\ln t} = \frac{1}{(t+1)(t-1)\ln(1+ t-1)} \operatorname*{\sim}_{t\to 1^+}\frac{1}{2(t-1)^2} $$ using the Taylor expansion to first order of $\ln(1+u)$ at $0$. By the same kind of comparison theorems,$^{(\dagger)}$ we thus have $$ \int_x^\infty \frac{dt}{(t^2-1)\ln t} \operatorname*{\sim}_{x\to 1^+}\int_x^\infty \frac{dt}{2(t-1)^2} = \frac{1}{2(x-1)} $$
$(\dagger)$ Specifically, here: if $f,g\geq 0$ are locally integrable with $f\sim_0 g$ and the integral of $f$ diverges at $a$, then so does that of $g$, and $\int_x^\infty f \sim_{x\to a}\int_x^\infty g$.