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

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Answer 1 · 2017-10-04T20:58:17

This integral converges to #-1/2#.

#int_0^∞ dx/(x-2)^2#

=#-1/(∞-2)+1/(0-2)#

=#0-1/2#

=#-1/2#

Answer 2 · 2017-10-04T22:25:54

The integral is not convergent.

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.