Question

Does `lim_(ararr0^+)(1/a^2)+lim_(brarr0-)(-1/b^2)` diverge?

I was solving `int_-1^1(1/x^3)dx` and my teacher said it diverges. I was wondering if you can do some kind of cancelling to get `lim_(ararr0^+)(1/a^2)+lim_(brarr0-)(-1/b^2)` equal to zero.

Answer 1

Please see below.

Explanation:

The standard (basic) definition of convergence of an improper integral tells us that

If `f` is a function that is not defined at `0`, then

`int_-a^a f(x) dx` converges if and only if both `int_-a^0 f(x) dx` and `int_0^a f(x) dx` converge.

There is another quantity of interest, called the Cauchy Principle Value. For the integral you are working on,
the Cauchy Principle Value is

`lim_(epsilonrarr0^+)[int_-1^(0-epsilon)1/x^3 dx + int_(0+epsilon)^1 1/x^3 dx] = 0`

The standard (or Basic) definition requires us to take the two limit separately. The Cauchy principal value takes the limit of the sum.

As for you specific question, I am not confident about any definition of convergence. that would allow us to answer the question, "Does `lim_(ararr0^+)(1/a^2)+lim_(brarr0-)(-1/b^2)` converge?"

I can say that in the extended real numbers this evaluates to `oo+(-oo)` which is not defined in the extended reals. So I would call that divergent.