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?
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
Yes, your approach and answer are correct. Let's go through the steps to confirm.
You correctly started by substituting u = ln(x) and du = dx/x. So the integral becomes:
∫(1/(xln(x))) dx = ∫(1/u) du = ln(u) + C,
where C is the constant of integration. Substituting back u = ln(x), we have:
∫(1/(xln(x))) dx = ln(ln(x)) + C.
Next, evaluate the integral from 1 to infinity:
∫[1/(xln(x))] dx │[1, ∞] = [ln(ln(x)) ] │[1, ∞].
To find the limit as x approaches infinity, we can rewrite the expression as:
lim(x → ∞) ln(ln(x)) - ln(ln(1)).
Now let's calculate each term individually:
First, the limit as x approaches infinity of ln(ln(x)). The natural logarithm grows very slowly, so it is not surprising that ln(ln(x)) grows even slower. As x approaches infinity, ln(x) increases without bound, and ln(ln(x)) still grows, but very slowly. Thus, we can conclude that ln(ln(x)) approaches infinity as x approaches infinity.
Second, ln(ln(1)). The natural logarithm of 1 is 0, so ln(ln(1)) is also 0.
Putting it all together, we have:
lim(x → ∞) ln(ln(x)) - ln(ln(1)) = ∞ - 0 = ∞.
Since the limit of the integral is infinity, we can conclude that the integral:
∫[1/(xln(x))] dx │[1, ∞]
diverges. This means that the area under the curve is infinite, and the function does not have a finite value for that interval.