# Is 1/(x^2 - 1) Lebesgue integrable on (0,1)?

• Apr 24th 2009, 09:43 AM
Amanda1990
Is 1/(x^2 - 1) Lebesgue integrable on (0,1)?
Is $\displaystyle 1/(x^2 - 1)$ Lebesgue integrable over (0,1)? What about over $\displaystyle (1, \infty)$? What kind of comparisons should I be trying?
• Apr 24th 2009, 01:52 PM
redsoxfan325
I thought that everything that was Riemann integrable was also Lebesgue integrable.
• Apr 24th 2009, 10:41 PM
Amanda1990
OK...so why is it Reimann integrable?Thanks.
• Apr 24th 2009, 11:02 PM
redsoxfan325
Oh crap...sorry, I misread the question. I thought it said $\displaystyle \frac{1}{x^2+1}$.

$\displaystyle \frac{1}{x^2-1}$ is definitely not Riemann integrable on $\displaystyle (0,1)$. I don't know enough about Lebesgue integration to solve this problem. Sorry for wasting your time.
• Apr 25th 2009, 12:10 AM
Opalg
The integral of $\displaystyle \frac1{x^2-1}$ diverges at x=1. That is to say, it diverges as an improper Riemann integral. But that implies that it does not exist as a Lebesgue integral, on either of the intervals (0,1) or (1,∞).

To prove that formally, on the interval (1,∞) you could define a sequence of functions $\displaystyle f_n(x) = \begin{cases}\frac n{n+2}&(1<x\leqslant1+\frac1n),\\ \frac1{x^2-1}&(x>1+\frac1n).\end{cases}$
Then $\displaystyle f_n(x)$ increases to $\displaystyle \frac1{x^2-1}$ as $\displaystyle n\to\infty$, but $\displaystyle \int_{(1,\infty)}f_n\to\infty$. It follows from the monotone convergence theorem that $\displaystyle \frac1{x^2-1}$ is not (Lebesgue) integrable on (1,∞). A similar argument shows that it is not integrable on (0,1).