# An improper integral with parameter $\int_{1}^{+\infty}\frac{\ln(1+x^p)}{\sqrt{x^2-1}}$

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....

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 $p<0$.

• I'm so grateful to you! Your answer is splendidly clear and "tidy"!! Have a good day! – sunrise Jun 11 '15 at 21:22
• @sunrise Thank you very much. – Olivier Oloa Jun 11 '15 at 21:25
• 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! – sunrise Jun 11 '15 at 21:46
• @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. – 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 $p<0$.

NOTE:

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

• 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! – sunrise Jun 11 '15 at 22:25
• You're welcome. My pleasure. I'll edit and add to the parts for which you would like some explanation. – Mark Viola Jun 11 '15 at 22:27
• @sunrise OK. I edited. Please let me know if that helps. – Mark Viola Jun 11 '15 at 22:36
• Yes! It's clear now! I'm not used to solve exercises in this way... :) A lot of thanks for your kindness! – sunrise Jun 11 '15 at 22:43
• You're most certainly welcome. It was my pleasure. And with practice, you will do fine. – Mark Viola Jun 11 '15 at 22:49

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

• 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? – Mark Viola Jun 11 '15 at 20:25
• I see!! Sorry, you are right... it needs some more thinking. – Rogelio Molina Jun 11 '15 at 20:27
• No worry. I'm not one of those down voters. – Mark Viola Jun 11 '15 at 20:35