$\int_{2}^{\infty}\frac{dx}{x^2-1}$ converge or diverge

Question

How can I find whether the integral converge or diverge.

I did the following.

$\int_{2}^{\infty}\frac{dx}{x^2-1}$

I did the following

$\frac{a}{x-1}+\frac{b}{x+1}$

$A(x+1)+B(x-1)=1$

$t\rightarrow\infty$

$\int_{2}^{t}\frac{1/2}{x-1}-\frac{1/2}{x+1}$

$\frac{1}{2}[\ln(x-1)-\ln(x+1)$

$\ln(t-1)-ln(t+1)-\ln(2-1)-\ln(2+1)$

Then I got

$\infty-\infty-ln(1)-ln(2)$

Since I got an intermediate form I did the hospital rule.

$\frac{\ln(t+1)}{1/\ln (t-1)}$

then I took the derivative

$\frac{\frac{1}{t+1}}{\frac{1}{1/t-1}}$

Thus I got

$\frac{1}{(t-1)(t+1)}$ as t approach infity zero

thus it is convergent at at $ln(1)-ln(3)$

Answer 1

Watch the signs! Don't forget in your evaluation of the integral that we have

$$\dfrac 12[\ln(x−1)−\ln(x+1)]\Big|_2^t = \dfrac 12\Big([\ln(x-1) - \ln(x+1)] - [(\ln(2 - 1) - \ln(2 + 1)]\Big) $$

$$=\dfrac 12\left(\ln(t−1)−\ln(t+1)−\ln(1) + \ln(3)\right)$$

Now, recall that $\;\log a - \log b = \log\dfrac 1b$.

So $$\ln(t-1)-\ln(t+1)=\ln \frac{t-1}{t+1}=\ln\left(1-\frac{2}{t+1}\right)$$

And $\lim_{t\to \infty} \ln\left(1-\frac{2}{t+1}\right) = \ln 1 = 0$.

So we are left with a integral converging to $\dfrac 12(\ln(3))$.

Answer 2

Before taking limits rewrite $$\ln(t-1)-\ln(t+1)=\ln \frac{t-1}{t+1}=\ln\left(1-\frac{2}{t+1}\right)$$

This has limit $\ln 1=0$ as $t\to \infty$.

Answer 3

One can prove convergence, without evaluating, by noting that for $x\ge 2$, we have $x^2-1\gt x^2-\frac{x^2}{2}=\frac{x^2}{2}$. Thus $$0\lt \frac{1}{x^2-1}\lt \frac{2}{x^2}.$$ It is a standard fact that $\int_2^\infty \frac{1}{x^2}\,dx$ converges, so our integral does.