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I have to analyse the convergence of this integral:

$$\int_{1}^{+\infty}\frac{\ln(1+x^p)}{\sqrt{x^2-1}}$$ where $p\in \mathbb{R}$.

I have thought to write:

$$\int_{1}^{c}\frac{\ln(1+x^p)}{\sqrt{x^2-1}}+\int_{c}^{+\infty}\frac{\ln(1+x^p)}{\sqrt{x^2-1}}$$

For the second integral, I know that $\frac{\ln(1+x^p)}{\sqrt{x^2-1}}<\frac{(1+x^p)}{\sqrt{x^2-1}}$ so I can study the second one:

if p>0:

$1+x^p\sim x^p$, $\sqrt{x^2-1}\sim x$, so I obtain: $\frac{(1+x^p)}{\sqrt{x^2-1}}\sim \frac{1}{x^{1-p}}$ that converges if p<0. But I have said that p must be >0, so I don't obtain solutions.

If p<0 the intergral function that I use to compare it with the mine, diverges. So, for the theorem on asymptotic comparision, my integral diverges.

But I don't know how to bring out something about the first integral....

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Assume $p \in \mathbb{R}$. Let's consider three cases.

  • $\color{blue}{\text{Case 1.}}$ $\quad p> 0.$

    We have, as $x \to +\infty$,

$$ \frac{\ln(1+x^p)}{\sqrt{x^2-1}} \sim p\frac{\ln x}{x} $$

and the initial integral is divergent by comparison to the divergent integral $\displaystyle \int_b^{\infty} \frac{\ln x}{x} dx \quad (b>0).$

  • $\color{blue}{\text{Case 2.}}$ $\quad p=0.$

    We have, as $x \to +\infty$,

$$ \frac{\ln(1+x^p)}{\sqrt{x^2-1}} \sim \frac{\ln 2}{x} $$

and the initial integral is divergent by comparison to the divergent integral $\displaystyle \int_b^{\infty} \frac{1}{x} dx \quad (b>0).$

  • $\color{blue}{\text{Case 3.}}$ $\quad p<0.$

    We have, as $x \to +\infty$,

$$ \ln(1+x^p) \sim \frac1{x^{|p|}} $$ $$ \frac{\ln(1+x^p)}{\sqrt{x^2-1}} \sim \frac1{x^{|p|+1}} $$

and $\displaystyle \int_b^{\infty} \frac{\ln(1+x^p)}{\sqrt{x^2-1}} dx \quad (b>0)$ is convergent by comparison to the convergent integral $\displaystyle \int_b^{\infty} \frac1{x^{|p|+1}} dx.$

We have, as $x \to 1^+$,

$$ \frac{\ln(1+x^p)}{\sqrt{x^2-1}} \sim \frac{\ln 2}{\sqrt{2} }\frac{1}{\sqrt{x-1}} $$

and $\displaystyle \int_1^{a} \frac{\ln(1+x^p)}{\sqrt{x^2-1}} dx \quad (a \to 1^+)$ is convergent by comparison to the convergent integral $\displaystyle \int_1^{a} \frac{1}{\sqrt{x-1}} dx.$

Thus your initial integral is convergent if and only if $p<0$.

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  • $\begingroup$ I'm so grateful to you! Your answer is splendidly clear and "tidy"!! Have a good day! $\endgroup$ – sunrise Jun 11 '15 at 21:22
  • $\begingroup$ @sunrise Thank you very much. $\endgroup$ – Olivier Oloa Jun 11 '15 at 21:25
  • $\begingroup$ Just a little clarification.. I haven't understood in which way you have obtained $\sqrt{x^2-1} \sim \sqrt 2 (\sqrt{x-1})$.. I have tried with "ordinary" expansion in series but I haven't obtained this result.. can you explain me? A lot of thanks again! $\endgroup$ – sunrise Jun 11 '15 at 21:46
  • $\begingroup$ @sunrise As $x \to 1^+$, we may write $ \displaystyle \sqrt{x^2-1} = \sqrt{x+1} \times \sqrt{x-1} \sim \sqrt{1+1} \times \sqrt{x-1}=\sqrt{2} \times \sqrt{x-1}$. Hoping it helps, thanks. $\endgroup$ – Olivier Oloa Jun 12 '15 at 6:14
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For any $p\ge 0$, we have

$$\frac{\log(1+x^p)}{\sqrt{x^2-1}}\ge \frac{\log 2}{x}.$$

Thus, the integral diverges for $p\ge 0$.


For $p<0$, we can write $\log(1+x^p)=\log(1+x^{-|p|})=\log \left(\frac{x^{|p|}+1}{x^{|p|}}\right)=\log(1+x^{|p|})-\log(x^{|p|})$. Therefore, we have

$$\begin{align} \left|\frac{\log (1+x^p)}{\sqrt{x^2-1}}\right|&=\frac{\log(1+x^{|p|})-\log(x^{|p|})}{\sqrt{x^2-1}}\\\\ &=\frac{1}{\sqrt{x^2-1}}\,\int_{x^{|p|}}^{1+x^{|p|}}\frac{dt}{t}\\\\ &\le \frac{2}{x^{|p|+1}} \end{align}$$

for $x>2\sqrt{3}/3$. We can see this since $\frac{1}{\sqrt{x^2-1}}=\frac{1}{x\sqrt{1-x^{-2}}}\le \frac1x \times 2$ for $x>2\sqrt{3}/3$.

Thus the integral converges for all $p<0$.


NOTE:

The singularity at $x=1$ is of order $(1-x)^{-1/2}$, and does not compromise the aforementioned analysis.

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  • $\begingroup$ Thanks for your answer. I'm sorry, I haven't understood your first step for p<0 and the origin of the "2" in the last one... Thanks again! $\endgroup$ – sunrise Jun 11 '15 at 22:25
  • $\begingroup$ You're welcome. My pleasure. I'll edit and add to the parts for which you would like some explanation. $\endgroup$ – Mark Viola Jun 11 '15 at 22:27
  • $\begingroup$ @sunrise OK. I edited. Please let me know if that helps. $\endgroup$ – Mark Viola Jun 11 '15 at 22:36
  • $\begingroup$ Yes! It's clear now! I'm not used to solve exercises in this way... :) A lot of thanks for your kindness! $\endgroup$ – sunrise Jun 11 '15 at 22:43
  • $\begingroup$ You're most certainly welcome. It was my pleasure. And with practice, you will do fine. $\endgroup$ – Mark Viola Jun 11 '15 at 22:49
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Hint: Look at the large $x$ behaviour, look at cases $p \geq 0, p<0$.

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  • $\begingroup$ Are you certain? Look at $p=-1$. The integrand is $\frac{\log(x+1)-\log x}{x}$. Do you believe the integral of that function won't converge? $\endgroup$ – Mark Viola Jun 11 '15 at 20:25
  • $\begingroup$ I see!! Sorry, you are right... it needs some more thinking. $\endgroup$ – Rogelio Molina Jun 11 '15 at 20:27
  • $\begingroup$ No worry. I'm not one of those down voters. $\endgroup$ – Mark Viola Jun 11 '15 at 20:35

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