# evaluate if integral converge: $\int_2^\infty \frac{2}{x(x+1)(x-1)} dx$

Evaluate if the following integral converges: $$\int_2^\infty \frac{2}{x(x+1)(x-1)} dx$$

Here I go: $$2\int_2^\infty \frac{1}{x(x+1)(x-1)} dx$$

Partial fractions of $\dfrac{1}{x(x+1)(x-1)}$ : $$\frac{1}{x(x+1)(x-1)} =\frac{A}{x}+\frac{B}{(x+1)} + \frac{C}{(x-1)} = \frac{-1}{x}+\frac{1/2}{(x+1)} + \frac{1/2}{(x-1)}$$ Back to the integral: $$2(-\int_2^\infty\frac{1}{x}+\frac{1}{2}\int_2^\infty\frac{1}{(x+1)} + \frac{1}{2}\int_2^\infty\frac{1}{(x-1)})dx$$

$$\lim_{a\to2..{b\to\infty}}2[(-\ln(x)+\frac{\ln(x+1)}{2} + \frac{\ln(x-1)}{2}]_a^b$$

$$2[(-\ln(b)+\frac{\ln(b+1)}{2} + \frac{\ln(b-1)}{2}]-2[(-\ln(2)+\frac{\ln(3)}{2} + \frac{\ln(1)}{2}]$$

$$(-\infty+\infty+\infty) + (\ln(3/4))$$

So, the integral does not converge? I am unsure about the last part, if anyone can confirm me the answer or explain why it's wrong... thank you guys!

• I got $\ln(4/3)$, is this the same as the answer the OP got? – Colbi Jun 17 '16 at 16:56
• Note that $\infty - \infty$ is not equal to zero. Thus $\infty - \infty + \infty$ is not equal to $\infty$. In every step you must ask what rule am I using, there is a rule that says that if $\lim a = \infty$ and $\lim b = \infty$, then $\lim (a + b) = \lim a + \lim b = \infty$. Here the $+$ is important, if you change it to a $-$, then you get a completely different rule which may or may not be true (and hence it is unusable). – Strategy Thinker Jun 17 '16 at 17:02
• Your question has a serious logical flaw besides what @StrategyThinker said. How can you write "$b$" in your second last line when it is not defined? In the earlier line you have "$b \to \infty$", but in this line you didn't specify that. Furthermore, "$b \to \infty$" does not have anything to do with "$b = \infty$".. – user21820 Jun 18 '16 at 10:29

No: the "infinities" cancel.

\eqalign{-\ln(b) &+ \dfrac{\ln(b+1)}{2} + \dfrac{\ln(b-1)}{2} = \ln \left(\frac{\sqrt{b+1}\sqrt{b-1}}{b}\right) = \ln\left( \sqrt{1+1/b}\sqrt{1-1/b}\right)\cr & \to 0\ \text{as}\ b \to \infty}

But the simplest way to see this converges is by a limit comparison test: $$\dfrac{2}{x(x+1)(x-1)} \sim \dfrac{2}{x^3} \ \text{as}\ x \to \infty$$ $$\int_2^\infty \dfrac{2}{x^3}\; dx\ \text{converges}$$

The integral function is non-negative over the integration range and behaves like $\frac{1}{x^3}$ for large values of $x$, hence it is integrable for sure. Partial fraction decomposition gives: $$\frac{2}{(x-1)x(x+1)} = \frac{1}{x(x-1)}-\frac{1}{x(x+1)}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{2}{x}+\frac{1}{x+1}\right)\tag{1}$$ hence: $$\int_{2}^{M}\frac{2\,dx}{(x-1)x(x+1)}=\log\left(\frac{4}{3}\right)+\log\left(1-\frac{1}{M^2}\right)\tag{2}$$ and by letting $M\to +\infty$:

$$\int_{2}^{+\infty}\frac{2\,dx}{(x-1)x(x+1)}=\color{red}{\log\left(\frac{4}{3}\right)}.\tag{3}$$

I know you're asking if it converges, but unless I'm interpreting your answer incorrectly I think it's wrong. As others have pointed out before me you can see it converges via a limit comparison test to see $x\rightarrow\infty$. So here's how I get $\ln(4/3)$: $$\int\dfrac{2}{x(x+1)(x-1)}dx$$ $$=2\int\dfrac{1}{x(x+1)(x-1)}dx$$ $$=2\int-\frac{1}{x}+\dfrac{1}{2(x+1)}+\dfrac{1}{2(x-1)}dx$$ $$=2\left(-\ln|x|+\frac{1}{2}\ln|x+1|+\frac{1}{2}\ln|x-1|\right)+C$$ Now the bounds: $$2\left(-\ln|2|+\frac{1}{2}\ln|2+1|+\frac{1}{2}\ln|2-1|\right)\Rightarrow-\ln\left(\frac{4}{3}\right)$$ $$2\left(-\ln|\infty|+\frac{1}{2}\ln|\infty+1|+\frac{1}{2}\ln|\infty-1|\right)\Rightarrow0$$ $$=0-\left(-\ln\left(\frac{4}{3}\right)\right)$$ $$=\ln\left(\frac{4}{3}\right)$$

• The thing I don't get, is why $2\left(-\ln|\infty|+\frac{1}{2}\ln|\infty+1|+\frac{1}{2}\ln|\infty-1|\right)=0$? – hobomath Jun 17 '16 at 17:18
• @hobomath Combine the ln terms into one ln term and then take the limit inside, the answer of the inside will be $1$when $x$ goes to inf and ln1 is zero – imranfat Jun 17 '16 at 17:23

Short answer: yes, it does: the denominator doesn't cancel and the asymptotic behavior is $\sim x^{-3}$.