Background
Consider the following divergent integral:
$$ \int_{-\infty}^\infty e^{-e^x} e^{ax} x^b dx $$
Where $a$ and $b$ are positive integers such that $b > a$. Multipling $e^{-x} \cdot e^x$ inside the integral.
$$ \int_{-\infty}^\infty e^{-e^x} e^x e^{(a-1)x} x^b dx $$
Using integration by parts and noticing $(e^{-e^x})' = -e^{-e^x}e^x $:
$$ \implies - \int_{-\infty}^\infty (e^{-e^x})' e^{(a-1)x} x^b dx = - \int_{-\infty}^\infty e^{-e^x} (e^{(a-1)x} x^b)' dx $$
Again multiplying $e^{-x} \cdot e^x$ inside the integral and repeating the process:
$$ = - \int_{-\infty}^\infty e^{-e^x} e^x \cdot e^{-x}(e^{(a-1)x} x^b)' dx $$
$$ = \int_{-\infty}^\infty (e^{-e^x})'\cdot e^{-x}(e^{(a-1)x} x^b)' dx $$
$$ = \int_{-\infty}^\infty (e^{-e^x})\cdot (e^{-x}(e^{(a-1)x} x^b)')' dx $$
This trick will only work $a$ times. The leading order term in $x$ will be something like: $ a!\int_{-\infty}^\infty e^{-e^x} x^b dx $
My idea was to take:
$$ \lim_{n \to \infty} \frac{\int_{-n}^n e^{-e^x} e^{ax} x^b dx + (-1)^a (a-1)! \int_{-n}^n e^{-e^x} x^b dx}{\int_{-n}^n e^{-e^x} x^{b-1} dx} $$
Now one can use two ways two solve this:
The first by using L'hopital rule and differentiating. The other is by using integration by parts. I've tried to do this myself but I've got disagreement between my answers (I think I'm messing up the calculation).
Question
Is my idea correct? What is the identity gained after these manipulations? Furthermore, what is the identity of gained after the $x^{k}$ coefficient using limits and integration by parts? for:
$$ \lim_{n \to \infty} \frac{\int_{-n}^n e^{-e^x} e^{ax} x^b dx + (-1)^a (a-1)! \int_{-n}^n e^{-e^x} x^b dx}{\int_{-n}^n e^{-e^x} x^{b-1} dx} = ?$$