Can you help me an approach to solve the integrand given constants $a, b> 0$, and $m, n\in\mathbb{N}$?$$\int_{1}^{\infty}\frac{\left ( a\left ( x^{2}- 1 \right ) \right )^{m}}{1+ x}\frac{\exp\left ( -a\left ( x^{2}- 1 \right ) \right )}{\left ( 1+ b\left ( x^{2}- 1 \right ) \right )^{n}}{\it d}x$$The challenge: I tried to approximate, or to use Mellin transform to deal with $\exp\left ( -a\left ( x^{2}- 1 \right ) \right )$ (if successfully, then come to Gradshteyn and Ryzhik's Table of Integrals, Series, and Products) using Taylor series reasoning, but it turned out to be too simplistic.
Attempt: I gave a try with $m= 0, n= 1$, we need to evaluate the improper integral:$$\int_{1}^{\infty}\frac{1}{1+ x}\frac{\exp\left ( -a\left ( x^{2}- 1 \right ) \right )}{1+ b\left ( x^{2}- 1 \right )}{\it d}x$$Since $x\geq 1$, let $u= \sqrt{x+ 1}$, and ${\it d}x= \frac{1}{2\sqrt{u+ 1}}{\it d}u$. Substitute into the integrand, and it becomes:$$\frac{1}{2}\int_{0}^{\infty}\frac{e^{-au}}{\left ( 1+ bu \right )\sqrt{u+ 1}\left ( 1+ \sqrt{u+ 1} \right )}{\it d}u= \frac{1}{2}\int_{0}^{\infty}\frac{\sqrt{u+ 1}- 1}{u}\frac{e^{-au}}{1+ bu}{\it d}u$$Unfortunately, I cannot proceed because I have not found an effective method to approximate $\frac{\sqrt{u+ 1}- 1}{u}$ over the interval $\left ( 0, \infty \right )$.
I need to gather more ideas for solving this from you all. Thank you for taking the time to consider this problem.