# 1/x^2 integral

1. Nov 24, 2009

### Dissonance in E

1. The problem statement, all variables and given/known data
Evaluate:
INTEGRAL(-1, 1): 1^x2

2. Relevant equations
power rule

3. The attempt at a solution
1/x^2
x^-2
x^-1
(-1/x)

plugging in values:
(-1/1) - (-1/-1)
-1 - 1
-2

Now i KNOW that -2 is an incorrect answer, the solution is infinity according to maple, nut im not 100% why. I would guess its something to do with the fact that the integral involves the value x = 0 which is undefined for f(x) = -1/x as it asymptotically approaches 0.

Could i get a nice concise, clarification. thx

2. Nov 24, 2009

### Dick

That needs to be done as an improper integral since the original function is not defined at x=0. You can't just integrate over 0, you have to approach it as a limit. Take the integral of 1/x^2 from a to 1 and let a->0 (similarly on the negative side).

3. Nov 24, 2009

### Dissonance in E

ok so could i do it like this.

INT (-1, 1) : 1/x^2
=
INT (-1, A)(Lim A--> 0, from the left): 1/x^2
+
INT(A , 1)(Lim A -->0, from the right): 1/x^2

(-1/-A) -(-1/-1)
as A approaches 0 from the left we get
INFINITY - 1 = INFINITY

(-1/1) - (-1/A)
as A approaches 0 from the right we get
-1+INFINITY = INFINITY

Now by sum of the 2 limits
inf + inf = inf

Last edited: Nov 24, 2009
4. Nov 24, 2009

### Dick

Yes. That's the idea. Don't be too casual about doing arithmetic with 'infinity' though. E.g. infinity-infinity is not necessarily 0.

5. Nov 24, 2009

### HallsofIvy

Staff Emeritus
The "Cauchy principal value" for such an integral of a function not defined at 0 is
$$\lim_{\epsilon\to 0}\left[\int_{-1}^{-\epsilon} f(x) dx+ \int_{\epsilon}^1 f(x)dx[/itex] The difference between that and the "regular" integral is that we do not take the limits independently. With f(x)= 1/x2, we have [tex]\lim_{\epsilon\to 0}\left[-x^{-1}\right]_{-1}^{-\epsilon}+ \left[-x^{-1}\right]_\epsilon^1$$
$$= \lim_{\epsilon\to 0} \left(\frac{1}{\epsilon}- 1\right)+ \left(1- \frac{1}{\epsilon}\right)= 0$$.