# Prove by test integrabilty that: $\displaystyle\int_0^{+\infty} \left|\frac{\ln x}{x^2-1}\right|\, dx < +\infty$

how we prove that the function $\dfrac{\ln x}{x^2-1}$ is in $L^1(]0,+\infty[)$ using the tests of convegence integral? I calculate is and i found that $$\displaystyle\int_0^{+\infty} \left|\frac{\ln x}{x^2-1}\right|\, dx < +\infty$$, but i want to prove that this function is in $L^1(]0,+\infty[)$ using the tests of integrability of integral.

• AT infinity use $$\lim_{x\to\infty} \frac{x^{3/2}\ln x}{x^2-1} = 0$$ That is there exists $A>0$ such that $x>A$ implies $$|\frac{\ln x}{x^2-1}|\le \frac{1}{x^{3/2}}$$ At $x=0$ use $$\lim_{x\to0} \frac{x^{1/2}\ln x}{x^2-1} = 0$$ That is there exists $a>0$ such that $0<x<a$ implies $$|\frac{\ln x}{x^2-1}|\le \frac{1}{x^{1/2}}$$ At $x=1$ there singularity is fake since, $$\lim_{x\to 1} \frac{\ln x}{x-1} = (\ln x)'|_{x=1} =1$$ Commented Nov 24, 2017 at 19:53

At infinity use

$$\lim_{x\to\infty} \frac{x^{3/2}\ln x}{x^2-1} = \lim_{x\to\infty} \frac{x^{2}}{x^2-1}\cdot\frac{\ln x}{x^{1/2}} =\lim_{x\to\infty} \frac{1}{2}\cdot\frac{\ln x^{1/2}}{x^{1/2}} = 0$$ That is there exists $A>0$ such that $x>A$ implies

$$|\frac{\ln x}{x^2-1}|\le \frac{1}{x^{3/2}}$$

At $x=0$ use

$$\lim_{x\to0} \frac{x^{1/2}\ln x}{x^2-1} = \frac12 \lim_{x\to0}x^{1/2}\ln x^{1/2} = 0$$ That is there exists $a>0$ such that $0<x<a$ implies

$$|\frac{\ln x}{x^2-1}|\le \frac{1}{x^{1/2}}$$

At $x=1$ there singularity is fake since,

$$\lim_{x\to 1} \frac{\ln x}{x-1} = (\ln x)'|_{x=1} =1$$

Notice that $|\frac{\ln x}{x^2-1}| \leq \min\{x^{1/2}, x^{3/2}\}$ and that the RHS is integrable because $p < 1$, $p > 1$ respectively. So by the comparison test, the integral in question is finite, hence $\frac{\ln x}{x^2-1} \in L^1$.

• $x^2-1$ not $x^2+1$ Commented Nov 24, 2017 at 19:40
• Please, how we prouve that $|\dfrac{\ln x}{x^2-1}| \leq \min \{x^{1/2},x^{3/2}\}$? and we have $x^2-1$ not $x^2+1$ Commented Nov 24, 2017 at 19:41
• @GuyFsone it's true either way; sign has been fixed. Commented Nov 24, 2017 at 19:42
• @user487908 see that there is a heavy singularity at $x=1$ which you forgotten. so the inequality is not enough Commented Nov 24, 2017 at 19:43
• exercise: there exists a continuous extension of $\ln(x)/(x^2 - 1)$ from $(0, 1)\cup(1,\infty)$ to $(0, \infty)$. Regardless, the upper-bound still holds, hence too the comparison test. Commented Nov 24, 2017 at 19:46