Asked by sarah
what is the answer for the integral of
(1/(xln(x)) from 1 to infinity?
I first found the integral using u substitution- so u=ln(x) and du=1/x*dx
this gave my du/u,leading to the ln(u)
=ln(ln(x)) from 1 to infinity
then I did the limit as t approaches infinity of 1/(tln(t)) - ln(ln(1))
my answer was that the function diverges, because the above limit is going to infinity. Is this right?
(1/(xln(x)) from 1 to infinity?
I first found the integral using u substitution- so u=ln(x) and du=1/x*dx
this gave my du/u,leading to the ln(u)
=ln(ln(x)) from 1 to infinity
then I did the limit as t approaches infinity of 1/(tln(t)) - ln(ln(1))
my answer was that the function diverges, because the above limit is going to infinity. Is this right?
Answers
Answered by
Damon
dx/(x ln x)
let u = ln x
then du = dx/x
so indeed
du/u
which is ln u
now when x = 1, u = 0
and when x = infinity , u = infinity
Yes, I get does not converge
let u = ln x
then du = dx/x
so indeed
du/u
which is ln u
now when x = 1, u = 0
and when x = infinity , u = infinity
Yes, I get does not converge
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