Question

What is the value of `int_-1^(+1) 1/x dx` ?

Justify your answer.

Answer 1

The integral diverges.

Explanation:

If you don't mind , I add an answer.

This is an improper integral as `1/x` is not defined for `x=0`

Therefore,

`int_-1^1(dx/x)=lim_(p->0^-)int_-1^p(dx)/x+lim_(q->0^+)int_q^1(dx)/x`

The first integral is

`lim_(p->0^-)int_-1^p(dx)/x=lim_(p->0^-)[ln|x|]_-1^p`

`=lim_(p->0^-)[ln-ln(1)]`

`=lim_(p->0^-)lnp-lim_(p->0^-)ln(-1)`

`=-oo-0`

`=-oo`

This integral diverges

The second integral is

`lim_(q->0^+)int_q^1(dx)/x=lim_(q->0^+)[ln|x|]_q^1`

`=lim_(q->0^+)[ln|1|-ln(q)]`

`=lim_(q->0^+)ln|1|-lim_(q->0^+)ln|q|`

`=0-(-oo)`

`=+oo`

This integral diverges