What is #int_(1)^(oo) 1/x-1/(x-1) dx #?

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Answer 1 · 2018-06-05T13:00:36

#int_1^oo (1/x-1/(x-1))dx = - oo#

Choose two numbers #a,b# with #1 < a < b#. Then:

#int_a^b (1/x-1/(x-1))dx = [lnx-ln(x-1)]_a^b = [ln(x/(x-1))]_a^b #

#int_a^b (1/x-1/(x-1))dx = ln(b/(b-1))- ln(a/(a-1))#

Now let #b->oo#: as:

#lim_(b->oo) b/(b-1) = 1#

#lim_(b->oo) ln(b/(b-1)) = 0#

then:

#int_a^oo (1/x-1/(x-1))dx = - ln(a/(a-1))#

Now:

#lim_(a->1+) a/(a-1) = +oo#

#lim_(a->1+) ln(a/(a-1)) = +oo#

so:

#int_1^oo (1/x-1/(x-1))dx = - oo#