Question

How do you determine if the improper integral converges or diverges `int_0^oo 1/ (x-2)^2 dx`?

Answer 1

This integral converges to `-1/2`.

Explanation:

`int_0^∞ dx/(x-2)^2`

=`-1/(∞-2)+1/(0-2)`

=`0-1/2`

=`-1/2`

Answer 2

The integral is not convergent.

Explanation:

The function:

`f(x) = 1/(x-2)^2`

is not continuous in the interval of integration so we must split the integral as:

`int_0^oo 1/(x-2)^2dx = int_0^2 1/(x-2)^2dx+ int_2^A 1/(x-2)^2dx + int_A^oo 1/(x-2)^2dx`

with `A > 2`

Now:

`int_0^2 1/(x-2)^2dx = lim_(t->2^-) int_0^t 1/(x-2)^2dx`

`int_0^2 1/(x-2)^2dx = lim_(t->2^-) [-1/(x-2)]_0^t`

`int_0^2 1/(x-2)^2dx = lim_(t->2^-) (1/(2-t)-1/2) = +oo`

and:

`int_2^A 1/(x-2)^2dx = lim_(t->2^+) int_t^A 1/(x-2)^2dx`

`int_2^A 1/(x-2)^2dx = lim_(t->2^+) [-1/(x-2)]_t^A`

`int_2^A 1/(x-2)^2dx = lim_(t->2^+) (-1/(A-2)+1/(t-2)) = +oo`

Finally:

`int_A^oo 1/(x-2)^2dx = lim_(t->oo) int_A^t 1/(x-2)^2dx`

`int_A^oo 1/(x-2)^2dx = lim_(t->oo) [-1/(x-2)]_A^t`

`int_A^oo 1/(x-2)^2dx = lim_(t->oo) (-1/(t-2) +1/(A-2)) = 1/(A-2)`

Thus the integral is not convergent.