# "For which alpha does the integral int_2^+inf ( e^(alpha x)/((x−1)^alpha ln x) converge? For which values of α does the integral int_2^+inf ( e^(alpha x)/((x−1)^alpha ln x) dx converge?

For which α does the integral $\underset{2}{\overset{+\mathrm{\infty }}{\int }}\frac{{e}^{\alpha x}}{\left(x-1{\right)}^{\alpha }\mathrm{ln}x}\mathrm{d}x$ converge?
For which values of $\alpha$ does the integral
$\underset{2}{\overset{+\mathrm{\infty }}{\int }}\frac{{e}^{\alpha x}}{\left(x-1{\right)}^{\alpha }\mathrm{ln}x}\mathrm{d}x$
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:
${I}_{2}=\underset{2}{\overset{+\mathrm{\infty }}{\int }}\frac{1}{\left(x-1{\right)}^{\alpha }{\mathrm{ln}}^{\beta }x}$
I2 converges for $\alpha >1$ and .
converges for $\alpha =1$ and $\beta >1$. For $\beta \le 1$ it diverges.
diverges for $\alpha <1$ and .
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
$0<\underset{2}{\overset{+\mathrm{\infty }}{\int }}\frac{\mathrm{d}x}{{x}^{\alpha }\mathrm{ln}x}\le \underset{2}{\overset{+\mathrm{\infty }}{\int }}\frac{{e}^{\alpha x}}{\left(x-1{\right)}^{\alpha }\mathrm{ln}x}\mathrm{d}x$
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.
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Cremolinoer
The main point is that exponentials are always going to dominate polynomials Specifically, for any $ϵ>0$ there exists a constant ${C}_{ϵ}$ such that ${e}^{x}\ge {C}_{ϵ}\left(x-1{\right)}^{ϵ}$ for all $x\ge 2$. This implies that for $\alpha >0$, ${e}^{\alpha x}\ge {C}_{ϵ}^{\alpha }\left(1-x{\right)}^{\alpha ϵ}$, so choosing $ϵ<1$ such that $0<\alpha \left(1-ϵ\right)<1$, we find
$\frac{{e}^{\alpha x}}{\left(x-1{\right)}^{\alpha }\mathrm{log}x}\ge {C}_{ϵ}^{\alpha }\frac{1}{\left(x-1{\right)}^{-\alpha \left(1-ϵ\right)}\mathrm{log}x}$
and so by comparison with ${I}_{2}$ you know ${I}_{1}$ diverges. Similarly, if $\alpha <0$ then ${e}^{\alpha x/2}\le {C}_{2}^{\alpha }\left(x-1{\right)}^{2\alpha }$ and hence
$\frac{{e}^{\alpha x}}{\left(x-1{\right)}^{\alpha }\mathrm{log}x}\le {C}_{2}^{\alpha /2}\frac{1}{\left(x-1{\right)}^{-\alpha }\mathrm{log}x},$
so by comparison with ${I}_{2}$ you know ${I}_{1}$ converges.
###### Did you like this example?
hotonglamoz
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.