Improper Integral y = x / (x^2 - 1)

[steezebutter]

Hi all

The question is: integrate x / (x^2 - 1) dx from 0 to 1. This is an improper integral since x cannot equal 1, so I set up this problem as the limit as t approaches 1 of x / (x^2 -1) dx from 0 to t. I tried integrating this a few different ways, but I always get to a point where there is a ln(0). Is the answer here negative infinity?

I can post more work if needed. Thanks!

 

[Deleted member 4993]

Hi all

The question is: integrate x / (x^2 - 1) dx from 0 to 1. This is an improper integral since x cannot equal 1, so I set up this problem as the limit as t approaches 1 of x / (x^2 -1) dx from 0 to t. I tried integrating this a few different ways, but I always get to a point where there is a ln(0). Is the answer here negative infinity?

I can post more work if needed. Thanks!

I would state my answer to be:

"the integral does NOT converge"​

 

[steezebutter]

Thank you - is my assumption correct that non converging and negative infinity mean the same thing?

 

[topsquark]

Thank you - is my assumption correct that non converging and negative infinity mean the same thing?

"Not converging" means it doesn't converge to a real number. That means we might have [imath]\lim_{x \to 5} f(x) \to - \infty[/imath] as a possibility. For another example we might have that [imath]\lim_{x \to 5^-} f(x) \neq \lim_{x \to 5^+} f(x)[/imath]. The limit does not converge at x = 5.

-Dan