How do you show whether the improper integral #int lim (lnx) / x dx# converges or diverges from 1 to infinity?

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Answer 1 · 2016-06-06T20:04:00

#int_1^ooln(x)/xdx# diverges to #oo#.

#int_1^ooln(x)/xdx = lim_(N->oo)int_1^Nln(x)/xdx#

Let #u = ln(x) => du = 1/xdx#.
Also, #x = 1 => u = 0# and #x = N => u = ln(N)#.

Making the substitution, we have

#lim_(N->oo)int_1^Nln(x)/xdx = lim_(N->oo)int_0^ln(N)udu#

#=lim_(N->oo)(u^2/2)_0^ln(N)#

#=lim_(N->oo)(ln^2(N)/2-0^2/2)#

#=lim_(N->oo)ln^2(N)/2#

#=oo#

Thus the integral #int_1^ooln(x)/xdx# diverges to #oo#.