# Evaluating Improper Integral with limits and comparison theorem

1. Mar 24, 2013

### Painguy

1. The problem statement, all variables and given/known data

evaluate the integral 1/(u^2 -36) from 0 to 6

does the integral converge?

2. Relevant equations
3. The attempt at a solution

integral 1/(u^2 -36)
integral 1/((u-6)(u+6))
Partial fraction decomposition
1/((u-6)(u+6)) = A/(u-6) + B/(u+6)
1=A(u+6) + B(u-6)
1=(A+B)u +(6A-6B)
A+B=0
A=-B
6A -6B=1
-12B=1
B=-1/12
A=1/12
1/12 int 1/(y-6) - 1/12 int 1/(y+6)
1/12 ln|y-6| - 1/12 ln|y+6|

I'm being told however by wolframalpha that it should be 1/12 ln|6-y| - 1/12 ln|y+6|
How did that happen?

I also wanted to try to use the comparison theorem to see if it converges or not.
I use the function g(x)=1/y^2 and i know that does not converge from 0 to 6, and since it is larger than f(x) then f(x) also does not converge. Did i go about that correctly?

Last edited: Mar 24, 2013
2. Mar 24, 2013

### Dick

Try and stick with one variable. I don't know why you changed from u to y, but there is no difference between |6-y| and |y-6|. And to show it diverges by a comparision test you want a function that's less than 1/((y-6)(y+6)) that diverges. 1/y^2 isn't less than 1/((y-6)(y+6)).

3. Mar 24, 2013

### Painguy

Hmm, not sure why I changed the variables myself. Oh right, I forgot I had to take the absolute value. It also seems that I have the entire comparison theorem backwards..that's not good haha. I can't think of any obvious functions less than 1/((y-6)(y+6)) that diverges. Could you point me toward the right direction of finding one? Thanks for your help I really appreciate it.

4. Mar 24, 2013

### Dick

Does 1/(y-6) diverge on [0,6]? The 1/(y+6) factor is between 1/12 and 1/6 on [0,6]. Use that info to make a comparison function.