6
$\begingroup$

I have this integral

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

I know it diverges but I don't know how to prove it. We have not learned hyperbolic trig functions so I know this is related to inverse hyperbolic tangent but I am looking for another approach that does not rely on that knowledge.

$\endgroup$
  • 1
    $\begingroup$ Wow, what a controversial thread. Good stuff. $\endgroup$ – Randall Oct 20 '17 at 15:50
  • 1
    $\begingroup$ On $(0,1)$, for any $n\in\mathbb{N}$, we have $\frac{1}{1-x^2}\geq 1+x^2+\ldots+x^{2n}$. The harmonic series is divergent, so... $\endgroup$ – Jack D'Aurizio Oct 20 '17 at 16:40
  • $\begingroup$ $$\frac{1}{x^2-1} = -\frac{1}{2}\left(\frac{1}{x+1} +\frac{1}{1-x}\right)= -\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)'$$ Hence, $$\int_{-1}^{1} \frac{1}{x^2-1}dx=\left[-\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\right]^1_{-1}=-\infty .$$ Given that, $$\color{red}{\lim_{x\to 0^+} \ln\left(x\right)=-\infty,~~~~~\lim_{x\to \infty+} \ln\left(x\right)=\infty}$$ $\endgroup$ – Guy Fsone Nov 27 '17 at 15:42
8
$\begingroup$

Partial fractions for sure (and logs will emerge), and consider the two improper integrals $$ \int_{-1}^0 \frac{1}{x^2-1} \ dx \ \text{ with } \int_0^1 \frac{1}{x^2-1} \ dx. $$

You need BOTH to converge for your integral to converge. Carry on...

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ What is the purpose of splitting it at $0$? $\endgroup$ – user8807399 Oct 20 '17 at 15:35
  • 1
    $\begingroup$ Honestly, it depends on how you VERY CAREFULLY define improper integrals that may be improper at several points. Most people would define your integral if and only if it made sense at all possible split-points, not just $0$. But with a finite number of discontinuities you can show that this doesn't matter: examining one split-point is enough. $\endgroup$ – Randall Oct 20 '17 at 15:36
  • $\begingroup$ I don't understand. Let's say I split it and decompose into partial fractions. Both pieces will involve $\log(x+1)$ and $\log(x-1)$. The first piece will have issues with $\log(x+1)$ due to the $-1$ bound and the second piece will have issues with $\log(x-1)$ due to the $1$ bound. Wouldn't I run into this same problem anyway if I had not split at $0$? $\endgroup$ – user8807399 Oct 20 '17 at 15:38
  • 2
    $\begingroup$ I would, but it does not fit within the margins of this paper. Just kidding, I have to run and teach a class, but later I will... $\endgroup$ – Randall Oct 20 '17 at 15:52
  • 4
    $\begingroup$ @user8807399 : $\int_{-1}^1 \; \frac{1}{x} \,\mathrm{d}x$ should be treated as $\lim_{(a,b) \rightarrow (0^-,0^+)} \left( \int_{-1}^a \; \frac{1}{x} \,\mathrm{d}x + \int_{b}^1 \; \frac{1}{x} \,\mathrm{d}x \right)$, which diverges. If you take instead $a=b$, you get its Cauchy principal value, $0$. $\endgroup$ – Eric Towers Oct 20 '17 at 17:33
8
$\begingroup$

Note that for $x\in (-1,1)$, $$\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}\leq \frac{1}{(-2)(x+1)}.$$ Hence $$\int_{-1}^{1} \frac{1}{x^2-1}dx\leq -\frac{1}{2}\int_{-1}^{1} \frac{1}{x+1}dx=-\left[\frac{\ln(x+1)}{2}\right]_{-1}^1=-\infty.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What does the <= part show / prove exactly? We can evaluate convergence by using some larger integral? $\endgroup$ – user8807399 Oct 20 '17 at 15:36
  • $\begingroup$ By comparing the two integrals we are showing that $\int_{-1}^{1} \frac{1}{x^2-1}dx=-\infty$. $\endgroup$ – Robert Z Oct 20 '17 at 15:40
  • $\begingroup$ I understand the intended outcome, but I am not clear on why that comparison allows us to draw that conclusion $\endgroup$ – user8807399 Oct 20 '17 at 15:42
  • 3
    $\begingroup$ Something that it is less or equal to some other thing that goes to $-\infty$, should go to $-\infty$ too. $\endgroup$ – Robert Z Oct 20 '17 at 15:46
  • $\begingroup$ Why then the use of the $-1/2$ piece on the righthand side? How do you know to use the negative and not positive? $\endgroup$ – user8807399 Oct 20 '17 at 16:12
6
$\begingroup$

Here's a less orthodox answer. Consider just the interval $[0, 1)$, and divide it into intervals $[0, 1/2), [1/2, 2/3), \ldots, [1 - \frac{1}{n}, 1 - \frac{1}{n+1}), \ldots$, each of width $\frac{1}{n(n+1)}$. The value of $\frac{1}{1-x^2}$ at $x = 1 - \frac{1}{n}$ is $\frac{n^2}{2n-1}$. Multiplying widths by function values gives a lower Riemann sum $$\int_0^1 \frac{dx}{1-x^2} > \sum_{n=1}^\infty \frac{n}{(n+1) (2n-1)}$$ which diverges as the harmonic series.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

First of all, since the integral is improper both at $x=-1$ and at $x=1$, you should split it: $$ \int_{-1}^1 \frac{dx}{x^2-1} = \int_{-1}^0 \frac{dx}{x^2-1} + \int_0^1 \frac{dx}{x^2-1} = \lim_{a \to -1} \int_a^0 \frac{dx}{x^2-1} + \lim_{b \to 1} \int_0^b \frac{dx}{x^2-1}. $$ Consider the last integral, which I write as $$ - \int_0^b \frac{dx}{1-x^2} = -\int_0^b \frac{dx}{(1-x)(1+x)}. $$ Observe that, for $0<x<b$, $1+x<1+b<1+1=2$. Hence $$ \int_{0}^b \frac{dx}{(1+x)(1-x)} > \frac{1}{2} \int_0^b \frac{dx}{1-x} = \frac{1}{2} \left[ -\log |1-x| \right]_0^b $$ Now, let $b \to 1$ and deduce that $$ \lim_{b \to 1} \int_0^b \frac{dx}{(1+x)(1-x)} = +\infty. $$ Hence the improper integral diverges.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "First of all, since the integral is improper, you should split it" Why? $\endgroup$ – user8807399 Oct 20 '17 at 15:41
  • $\begingroup$ That's usually the definition of an improper integral containing more than one singular point. This is used to avoid the appearence of principal values of improper integrals. $\endgroup$ – Siminore Oct 20 '17 at 15:42
  • 3
    $\begingroup$ That's a bad starting point: never compute something that you are unable to define. $\endgroup$ – Siminore Oct 20 '17 at 15:52
  • 1
    $\begingroup$ Me and @Siminore are getting at the same point: you have to appeal to the true definition of convergence for an improper integral--whatever that means--and go from there. For most, this definition requires you to split it and examine the discontinuities separately. Maybe for this PARTICULAR integral it doesn't matter, but in other ones it might. $\endgroup$ – Randall Oct 20 '17 at 15:54
  • 1
    $\begingroup$ You do not want to understand. What is the definition of convergence for an improper integral of your type? $\endgroup$ – Siminore Oct 20 '17 at 16:36
1
$\begingroup$

Hin: On the interval $[0,1),$

$$\frac{1}{x^2-1}\le \frac{1}{2}\frac{1}{x-1}.$$

If the integral of the right side over $[0,1)$ is $-\infty,$ then the same is true of the left side.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

[The following is essentially @Guy Fsone's answer. It is rewritten differently.]

Observe that $$ \frac{1}{x^2-1} = -\frac{1}{2}\left(\frac{1}{x+1} +\frac{1}{1-x}\right)= -\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)' $$ Hence, $$ \int_{-1}^{1} \frac{1}{x^2-1}dx=\left[-\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\right]^1_{-1}=\frac{1}{2}[\ln(1-x)-\ln(1+x)]\big|_{-1}^1 \tag{1} $$ But note that, $$ \lim_{x\to 0^+} \ln\left(x\right)=-\infty \tag{2} $$ Combining (1) and (2) one has $$ \int_{-1}^{1} \frac{1}{x^2-1}\ dx=\frac{1}{2}\left[(-\infty-\ln 2)-\big(\ln 2-(-\infty)\big)\right]=-\infty $$ where we use the arithmetic operations of the extended real number line.

| cite | improve this answer | |
$\endgroup$

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.