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

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1 Answer
Jul 14, 2018

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