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For what $\alpha\in\mathbb{R}$ does $$\int_{1}^{\infty}\frac{\ln(1+x^{\alpha})}{\sqrt{x^2-1}} \ dx$$ converge?


I know this is a duplicate problem, but I don't follow the accepted answer, in particular the explanation of case 3. I understand that we should look at the cases $\alpha\in(-\infty,0), \alpha\in(0,\infty)$ and $\alpha = 0$. My questions are:

  1. In this thread, Oliver Oloa compares, in case 1 & 2, his two simplified integrals with the integral $$\int_{b}^{\infty}\frac{1}{x} \ dx, \quad (b>0).$$ But why did he cho0se the restriction $b>0$ and not $b\geq 0$? I understand that $b$ can't be negative, but surely $\int_{0}^{\infty}\frac{1}{x} \ dx$ IS divergent as well.

  2. In case 3, he goes $$ \ln(1+x^\alpha) \sim \frac1{x^{|\alpha|}} $$ $$ \frac{\ln(1+x^\alpha)}{\sqrt{x^2-1}} \sim \frac1{x^{|\alpha|+1}} $$ I really need a breakdown of the steps in the entire case 3. How did the absolute values appear?

  3. Is there another way to solve this problem?

Note: Taylor expansions/L'Hopital's are not allowed to be used.

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  • $\begingroup$ Are you allowed to use l'Hospital rule ? $\endgroup$ – Noether Nov 17 '17 at 11:25
  • $\begingroup$ No unfortunately :( $\endgroup$ – Parseval Nov 17 '17 at 14:18
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Regarding case 3, you have that for negative negative $\alpha$

$$x^\alpha = \frac{1}{x^{|\alpha|}}$$

Since you are integrating from $1$ onwards, the argument of the logarithm is close to $1$ as $1/x^{|\alpha|}$ is close to $0$. Taylor's expansion of $\log(1+t)$ near $t=0$ tells you that

$$\log(1+t) \approx t$$

On the other hand, as $x$ grows bigger $\sqrt{x^2 -1} \approx \sqrt{x^2} = x$.

Hence

$$\frac{\log\left(1+\frac{1}{x^{|\alpha|}}\right)}{\sqrt{x^2 -1}} \approx \frac{1}{x^{|\alpha|+1}}$$

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  • $\begingroup$ Hi, thanks for your post but we are not allowed to use Taylor expansions since whe haven't covered it in this course. $\endgroup$ – Parseval Nov 17 '17 at 9:14
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If $\alpha\geq 0$ the given integral is greater than $\int_{1}^{+\infty}\frac{\log 2}{\sqrt{x^2-1}}\,dx$, which is divergent. If $\alpha<0$, by setting $\beta=-\alpha$ we have $\beta>0$ and we may prove that $$ \int_{1}^{+\infty}\frac{\log\left(1+\frac{1}{x^\beta}\right)}{\sqrt{x^2-1}}\,dx\leq\int_{1}^{+\infty}\frac{dx}{x^\beta\sqrt{x^2-1}} $$ is convergent by enforcing the substitution $x=\cosh t$, for instance, turnig the last integral into $$ \int_{0}^{+\infty}\frac{dt}{(\cosh t)^{\beta}}\leq \int_{0}^{+\infty}\frac{2^\beta}{e^{\beta t}}\,dt=\frac{2^\beta}{\beta}<+\infty.$$

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