For which $\alpha$ does the integral $\int\limits_2^{+\infty}\frac{e^{\alpha x}}{(x-1)^{\alpha}\ln x}\mathrm{d}x$ converge? 
For which values of $\alpha$ does the integral
  $$\int\limits_2^{+\infty}\frac{e^{\alpha x}}{(x-1)^{\alpha}\ln x}\mathrm{d}x$$ converge?

I'm lost here.
For simplicity, let us denote the above integral as $I_1$
I was able to prove (I noticed that it's kind of standard exercise though) that: $$I_2 = \int\limits_2^{+\infty}\frac{1}{(x-1)^{\alpha}\ln^{\beta} x}$$


*

*$I_2$ converges for $\alpha > 1$ and $\forall\ \beta$.

*converges for $\alpha = 1$ and $\beta > 1$. For $\beta \leq 1$
it diverges.

*diverges for $\alpha < 1$ and $\forall\ \beta$.
So my strategy was to use $I_2$ to prove the converges/divergence of $I_1$, by means of inequalities, but I get nothing from this.
For instance I did the following:
For $\alpha > 1$ we have that
$$0 < \int\limits_2^{+\infty} \frac{\mathrm{d}x}{x^{\alpha}\ln x} \leq  \int\limits_2^{+\infty}\frac{e^{\alpha x}}{(x-1)^{\alpha}\ln x} \mathrm{d}x$$
That is, for $\beta = 1$
$$0 < I_2 \leq I_1$$
From 1. we know that $I_2$ converges, but this doesn't tell us anything about the convergence of $I_1$. 
The answer given by my textbook is that $I_1$ converges for $\alpha < 0$, but I don't know how to arrive at that conclusion.
 A: Hint
What you've shown about the integral w/o the exponential factor is actually somewhat harder conceptually than the integral at hand. If $\alpha< 0$ then the exponential factor is an exponential decay which overpowers any of the other behavior and will make the integral converge. If $\alpha > 0,$ it's an exponential growth that will likewise make it diverge. $\alpha = 0$ you've already handled.
A: The main point is that exponentials are always going to dominate polynomials Specifically, for any $\epsilon>0$ there exists a constant $C_\epsilon$ such that $e^x\ge C_\epsilon(x-1)^\epsilon$ for all $x\ge2$. This implies that for $\alpha>0$, $e^{\alpha x}\ge C_\epsilon^\alpha(1-x)^{\alpha\epsilon}$, so choosing $\epsilon<1$ such that $0<\alpha(1-\epsilon)<1$, we find
$$\frac{e^{\alpha x}}{(x-1)^\alpha\log x}\ge C_\epsilon^\alpha\frac1{(x-1)^{-\alpha(1-\epsilon)}\log x}$$
and so by comparison with $I_2$ you know $I_1$ diverges. Similarly, if $\alpha<0$ then $e^{\alpha x/2}\le C_2^{\alpha}(x-1)^{2\alpha}$ and hence
$$\frac{e^{\alpha x}}{(x-1)^\alpha\log x}\le C_2^{\alpha/2}\frac{1}{(x-1)^{-\alpha}\log x},$$
so by comparison with $I_2$ you know $I_1$ converges.
