∫ dx / (x^2 - 1) from 0 to ∞?

Question

∞
∫dx / (x^2 - 1)
0

The answer is not defined in the real domain. It may be imaginary, but it is defined.

Answer 1

This was a curious problem, but the answer is 0. The definite integral from 0 to 1 is the same as the negative of the definite integral from 1 to ∞.

The other answerers should have checked to see if adding the two definite integrals would converge as x -> 1 from both directions. Yes, imaginaries pop up, but at the end, they drop out anyway.

Advice: Sometimes you can't depend on your calculator/software for giving you the last word.

Answer 2

The Cauchy Principal Value of the integral exists and is zero.

Since there is a singularity at x=1, we need to divide the integral into two parts: the integral from 0 to 1-a and the integral from 1+a to infinity, where a is positive and small. The integral from 0 to 1-a is 1/2 ln [a / (2-a)]. The integral from 1+a to infinity is 1/2 ln [(2+a) / a]. Adding these two together, we get 1/2 ln[(2+a)/(2-a)]. Taking the limit as a -> 0 we get 0, which is the Cauchy Principal Value.

Imaginary values do not enter into the problem at all, if you do the integration properly. Hint: the integral of 1/x is ln(|x|).

Answer 3

the evaluation of the integral is -atanh(x)

evaluated from 0 to infinity--is not equal to a number-doesn't converge

Answer 4

My calculator gives the answer as undefined, so it doesn't converge/doesn't exist.