Question

How do you determine if the improper integral converges or diverges `int (1/(3x)-6) dx` from negative infinity to 0?

Answer 1

`int_-oo^0(1/(3x)-6)dxrArr` diverges to `-oo`

Explanation:

To evaluate a two-sided improper integral, split it into two integrals and express each as a limit like so:

`int_-oo^0(1/(3x)-6)dx`

`=lim_(s->-oo)int_s^-1(1/(3x)-6)dx+lim_(t->0)int_-1^t(1/(3x)-6)dx`

Now compute the integrals and evaluate the limits:

`rArrlim_(s->-oo)[lnabs(3x)/3-6x]_s^-1`

`rArrlim_(s->-oo)[(ln3/3+6)-(lnabs(3s)/3-6s)]`

`rArr[(ln3/3+6)-(oo+oo)]->-oo`

Therefore the integral diverges as it approaches `-oo`.

Technically, we could stop here. If one integral diverges, the whole expression will diverge. But let's evaluate the second limit just for good measure:

`rArrlim_(t->0)[lnabs(3x)/3-6x]_-1^t`

`rArrlim_(t->0)[(lnabs(3t)/3-6t)-(ln3/3+6)]`

`rArr[(-oo-0)-(ln3/3+6)]->-oo`

And again we see the integral also diverges as it approaches 0.