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:


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....


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 $p<0$.

  • $\begingroup$ 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

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 $p<0$.


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

  • $\begingroup$ 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

Hint: Look at the large $x$ behaviour, look at cases $p \geq 0, p<0$.

  • $\begingroup$ 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

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.