For which values of $\alpha $ does the integral

$$\underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{{e}^{\alpha x}}{(x-1{)}^{\alpha}\mathrm{ln}x}\mathrm{d}x$$

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:

$${I}_{2}=\underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{1}{(x-1{)}^{\alpha}{\mathrm{ln}}^{\beta}x}$$

I2 converges for $\alpha >1$ and $\mathrm{\forall}\text{}\beta $.

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

diverges for $\alpha <1$ and $\mathrm{\forall}\text{}\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

$$0<\underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{\mathrm{d}x}{{x}^{\alpha}\mathrm{ln}x}\le \underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{{e}^{\alpha x}}{(x-1{)}^{\alpha}\mathrm{ln}x}\mathrm{d}x$$

That is, for $\beta =1$

$$0<{I}_{2}\le {I}_{1}$$

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.