2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ I got $\ln(4/3)$, is this the same as the answer the OP got? $\endgroup$ – Colbi Jun 17 '16 at 16:56
  • 2
    $\begingroup$ 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). $\endgroup$ – Strategy Thinker Jun 17 '16 at 17:02
  • $\begingroup$ 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 "$b = \infty$".. $\endgroup$ – user21820 Jun 18 '16 at 10:29
5
$\begingroup$

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: $$\dfrac{2}{x(x+1)(x-1)} \sim \dfrac{2}{x^3} \ \text{as}\ x \to \infty $$ $$\int_2^\infty \dfrac{2}{x^3}\; dx\ \text{converges}$$

$\endgroup$
1
$\begingroup$

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 $M\to +\infty$:

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

$\endgroup$
1
$\begingroup$

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: $$2\left(-\ln|2|+\frac{1}{2}\ln|2+1|+\frac{1}{2}\ln|2-1|\right)\Rightarrow-\ln\left(\frac{4}{3}\right)$$ $$2\left(-\ln|\infty|+\frac{1}{2}\ln|\infty+1|+\frac{1}{2}\ln|\infty-1|\right)\Rightarrow0$$ $$=0-\left(-\ln\left(\frac{4}{3}\right)\right)$$ $$=\ln\left(\frac{4}{3}\right)$$

$\endgroup$
  • $\begingroup$ The thing I don't get, is why $2\left(-\ln|\infty|+\frac{1}{2}\ln|\infty+1|+\frac{1}{2}\ln|\infty-1|\right)=0$? $\endgroup$ – hobomath Jun 17 '16 at 17:18
  • 2
    $\begingroup$ @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 $\endgroup$ – imranfat Jun 17 '16 at 17:23
0
$\begingroup$

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

$\endgroup$

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.