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: $$I_2 = \int\limits_2^{+\infty}\frac{1}{(x-1)^{\alpha}\ln^{\beta} x}$$

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

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

  3. 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.

  • $\begingroup$ Try to estimate what $\frac{e^{\alpha x}}{(x - 1)^\alpha}$ will be, then apply your previous conclusions. $\endgroup$ – Chris Apr 16 '17 at 19:46
  • $\begingroup$ @Chris Something like this: for $x\to +\infty: \frac{e^{\alpha x}}{(x - 1)^\alpha} \sim \left(\frac{e^x}{x}\right)^{\alpha} = \alpha e^{\alpha x}$ (after applying L'hopital). So for $x\to +\infty$, $\frac{e^{\alpha x}}{(x-1)^{\alpha}\ln x} \sim \frac{\alpha e^{\alpha x}}{\ln x}$. And from here we get that the integral converges for $\alpha < 0$. $\endgroup$ – Jazz Apr 16 '17 at 19:55


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.


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.


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