2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.